Presentation on ‘Machine Learning in plain English – Part 2’


This presentation is a continuation of my earlier presentation Presentation on ‘Machine Learning in plain English – Part 1’. As the title suggests, the presentation is devoid of any math or programming constructs, and just focuses on the concepts and approaches to different Machine Learning algorithms. In this 2nd part, I discuss KNN regression, KNN classification, Cross Validation techniques like (LOOCV, K-Fold)   feature selection methods including best-fit,forward-fit and backward fit and finally Ridge (L2) and Lasso Regression (L1)


If you would like to see the implementations of the discussed algorithms, in this presentation, do check out my book   My book ‘Practical Machine Learning with R and Python’ on Amazon

You may also like
1.A primer on Qubits, Quantum gates and Quantum Operations
2.Deep Learning from first principles in Python, R and Octave – Part 4
3.Latency, throughput implications for the Cloud
4.Brewing a potion with Bluemix, PostgreSQL, Node.js in the cloud
5.Practical Machine Learning with R and Python – Part 6
6. Introducing cricketr! : An R package to analyze performances of cricketers

To see all post click Index of posts

Presentation on ‘Machine Learning in plain English – Part 1’


This is the first part on my series ‘Machine Learning in plain English – Part 1’ in which I discuss the intuition behind different Machine Learning algorithms, metrics and the approaches etc. These presentations will not include tiresome math or laborious programming constructs, and will instead focus on just the concepts behind the Machine Learning algorithms.  This presentation discusses what Machine Learning is, Gradient Descent, linear, multi variate & polynomial regression, bias/variance, under fit, good fit and over fit and finally logistic regression etc.

It is hoped that these presentations will trigger sufficient interest in you, to explore this fascinating field further

To see actual implementations of the most widely used Machine Learning algorithms in R and Python, check out My book ‘Practical Machine Learning with R and Python’ on Amazon

Also see
1. Practical Machine Learning with R and Python – Part 3
2.R vs Python: Different similarities and similar differences
3. Perils and pitfalls of Big Data
4. Deep Learning from first principles in Python, R and Octave – Part 2
5. Getting started with memcached-libmemcached

To see all post see “Index of posts

My book ‘Practical Machine Learning with R and Python’ on Amazon


 

My book ‘Practical Machine Learning with R and Python: Second Edition – Machine Learning in stereo’ is now available in both paperback ($10.99) and kindle ($7.99/Rs449) versions. In this book I implement some of the most common, but important Machine Learning algorithms in R and equivalent Python code. This is almost like listening to parallel channels of music in stereo!
1. Practical machine with R and Python – Machine Learning in Stereo (Paperback)
2. Practical machine with R and Python – Machine Learning in Stereo (Kindle)
This book is ideal both for beginners and the experts in R and/or Python. Those starting their journey into datascience and ML will find the first 3 chapters useful, as they touch upon the most important programming constructs in R and Python and also deal with equivalent statements in R and Python. Those who are expert in either of the languages, R or Python, will find the equivalent code ideal for brushing up on the other language. And finally,those who are proficient in both languages, can use the R and Python implementations to internalize the ML algorithms better.

Here is a look at the topics covered

Table of Contents
Essential R …………………………………….. 7
Essential Python for Datascience ………………..   54
R vs Python ……………………………………. 77
Regression of a continuous variable ………………. 96
Classification and Cross Validation ……………….113
Regression techniques and regularization …………. 134
SVMs, Decision Trees and Validation curves …………175
Splines, GAMs, Random Forests and Boosting …………202
PCA, K-Means and Hierarchical Clustering …………. 234

Pick up your copy today!!
Hope you have a great time learning as I did while implementing these algorithms!

Practical Machine Learning with R and Python – Part 5


This is the 5th and probably penultimate part of my series on ‘Practical Machine Learning with R and Python’. The earlier parts of this series included

1. Practical Machine Learning with R and Python – Part 1 In this initial post, I touch upon univariate, multivariate, polynomial regression and KNN regression in R and Python
2.Practical Machine Learning with R and Python – Part 2 In this post, I discuss Logistic Regression, KNN classification and cross validation error for both LOOCV and K-Fold in both R and Python
3.Practical Machine Learning with R and Python – Part 3 This post covered ‘feature selection’ in Machine Learning. Specifically I touch best fit, forward fit, backward fit, ridge(L2 regularization) & lasso (L1 regularization). The post includes equivalent code in R and Python.
4.Practical Machine Learning with R and Python – Part 4 In this part I discussed SVMs, Decision Trees, validation, precision recall, and roc curves

This post ‘Practical Machine Learning with R and Python – Part 5’ discusses regression with B-splines, natural splines, smoothing splines, generalized additive models (GAMS), bagging, random forest and boosting

As with my previous posts in this series, this post is largely based on the following 2 MOOC courses

1. Statistical Learning, Prof Trevor Hastie & Prof Robert Tibesherani, Online Stanford
2. Applied Machine Learning in Python Prof Kevyn-Collin Thomson, University Of Michigan, Coursera

You can download this R Markdown file and associated data files from Github at MachineLearning-RandPython-Part5

Check out my compact and minimal book  “Practical Machine Learning with R and Python:Second edition- Machine Learning in stereo”  available in Amazon in paperback($10.99) and kindle($7.99) versions. My book includes implementations of key ML algorithms and associated measures and metrics. The book is ideal for anybody who is familiar with the concepts and would like a quick reference to the different ML algorithms that can be applied to problems and how to select the best model. Pick your copy today!!

For this part I have used the data sets from UCI Machine Learning repository(Communities and Crime and Auto MPG)

1. Splines

When performing regression (continuous or logistic) between a target variable and a feature (or a set of features), a single polynomial for the entire range of the data set usually does not perform a good fit.Rather we would need to provide we could fit
regression curves for different section of the data set.

There are several techniques which do this for e.g. piecewise-constant functions, piecewise-linear functions, piecewise-quadratic/cubic/4th order polynomial functions etc. One such set of functions are the cubic splines which fit cubic polynomials to successive sections of the dataset. The points where the cubic splines join, are called ‘knots’.

Since each section has a different cubic spline, there could be discontinuities (or breaks) at these knots. To prevent these discontinuities ‘natural splines’ and ‘smoothing splines’ ensure that the seperate cubic functions have 2nd order continuity at these knots with the adjacent splines. 2nd order continuity implies that the value, 1st order derivative and 2nd order derivative at these knots are equal.

A cubic spline with knots \alpha_{k} , k=1,2,3,..K is a piece-wise cubic polynomial with continuous derivative up to order 2 at each knot. We can write y_{i} = \beta_{0} +\beta_{1}b_{1}(x_{i}) +\beta_{2}b_{2}(x_{i}) + .. + \beta_{K+3}b_{K+3}(x_{i}) + \epsilon_{i}.
For each (x{i},y{i}), b_{i} are called ‘basis’ functions, where  b_{1}(x_{i})=x_{i}b_{2}(x_{i})=x_{i}^2, b_{3}(x_{i})=x_{i}^3, b_{k+3}(x_{i})=(x_{i} -\alpha_{k})^3 where k=1,2,3… K The 1st and 2nd derivatives of cubic splines are continuous at the knots. Hence splines provide a smooth continuous fit to the data by fitting different splines to different sections of the data

1.1a Fit a 4th degree polynomial – R code

In the code below a non-linear function (a 4th order polynomial) is used to fit the data. Usually when we fit a single polynomial to the entire data set the tails of the fit tend to vary a lot particularly if there are fewer points at the ends. Splines help in reducing this variation at the extremities

library(dplyr)
library(ggplot2)
source('RFunctions-1.R')
# Read the data
df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI
df1 <- as.data.frame(sapply(df,as.numeric))
#Select specific columns
df2 <- df1 %>% dplyr::select(cylinder,displacement, horsepower,weight, acceleration, year,mpg)
auto <- df2[complete.cases(df2),]
# Fit a 4th degree polynomial
fit=lm(mpg~poly(horsepower,4),data=auto)
#Display a summary of fit
summary(fit)
## 
## Call:
## lm(formula = mpg ~ poly(horsepower, 4), data = auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -14.8820  -2.5802  -0.1682   2.2100  16.1434 
## 
## Coefficients:
##                       Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            23.4459     0.2209 106.161   <2e-16 ***
## poly(horsepower, 4)1 -120.1377     4.3727 -27.475   <2e-16 ***
## poly(horsepower, 4)2   44.0895     4.3727  10.083   <2e-16 ***
## poly(horsepower, 4)3   -3.9488     4.3727  -0.903    0.367    
## poly(horsepower, 4)4   -5.1878     4.3727  -1.186    0.236    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.373 on 387 degrees of freedom
## Multiple R-squared:  0.6893, Adjusted R-squared:  0.6861 
## F-statistic: 214.7 on 4 and 387 DF,  p-value: < 2.2e-16
#Get the range of horsepower
hp <- range(auto$horsepower)
#Create a sequence to be used for plotting
hpGrid <- seq(hp[1],hp[2],by=10)
#Predict for these values of horsepower. Set Standard error as TRUE
pred=predict(fit,newdata=list(horsepower=hpGrid),se=TRUE)
#Compute bands on either side that is 2xSE
seBands=cbind(pred$fit+2*pred$se.fit,pred$fit-2*pred$se.fit)
#Plot the fit with Standard Error bands
plot(auto$horsepower,auto$mpg,xlim=hp,cex=.5,col="black",xlab="Horsepower",
     ylab="MPG", main="Polynomial of degree 4")
lines(hpGrid,pred$fit,lwd=2,col="blue")
matlines(hpGrid,seBands,lwd=2,col="blue",lty=3)

fig1-1

1.1b Fit a 4th degree polynomial – Python code

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.preprocessing import PolynomialFeatures
from sklearn.linear_model import LinearRegression
#Read the auto data
autoDF =pd.read_csv("auto_mpg.csv",encoding="ISO-8859-1")
# Select columns
autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']]
# Convert all columns to numeric
autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce')

#Drop NAs
autoDF3=autoDF2.dropna()
autoDF3.shape
X=autoDF3[['horsepower']]
y=autoDF3['mpg']
#Create a polynomial of degree 4
poly = PolynomialFeatures(degree=4)
X_poly = poly.fit_transform(X)

# Fit a polynomial regression line
linreg = LinearRegression().fit(X_poly, y)
# Create a range of values
hpGrid = np.arange(np.min(X),np.max(X),10)
hp=hpGrid.reshape(-1,1)
# Transform to 4th degree
poly = PolynomialFeatures(degree=4)
hp_poly = poly.fit_transform(hp)

#Create a scatter plot
plt.scatter(X,y)
# Fit the prediction
ypred=linreg.predict(hp_poly)
plt.title("Poylnomial of degree 4")
fig2=plt.xlabel("Horsepower")
fig2=plt.ylabel("MPG")
# Draw the regression curve
plt.plot(hp,ypred,c="red")
plt.savefig('fig1.png', bbox_inches='tight')

fig1

1.1c Fit a B-Spline – R Code

In the code below a B- Spline is fit to data. The B-spline requires the manual selection of knots

#Splines
library(splines)
# Fit a B-spline to the data. Select knots at 60,75,100,150
fit=lm(mpg~bs(horsepower,df=6,knots=c(60,75,100,150)),data=auto)
# Use the fitted regresion to predict
pred=predict(fit,newdata=list(horsepower=hpGrid),se=T)
# Create a scatter plot
plot(auto$horsepower,auto$mpg,xlim=hp,cex=.5,col="black",xlab="Horsepower",
     ylab="MPG", main="B-Spline with 4 knots")
#Draw lines with 2 Standard Errors on either side
lines(hpGrid,pred$fit,lwd=2)
lines(hpGrid,pred$fit+2*pred$se,lty="dashed")
lines(hpGrid,pred$fit-2*pred$se,lty="dashed")
abline(v=c(60,75,100,150),lty=2,col="darkgreen")

fig2-1

1.1d Fit a Natural Spline – R Code

Here a ‘Natural Spline’ is used to fit .The Natural Spline extrapolates beyond the boundary knots and the ends of the function are much more constrained than a regular spline or a global polynomoial where the ends can wag a lot more. Natural splines do not require the explicit selection of knots

# There is no need to select the knots here. There is a smoothing parameter which
# can be specified by the degrees of freedom 'df' parameter. The natural spline

fit2=lm(mpg~ns(horsepower,df=4),data=auto)
pred=predict(fit2,newdata=list(horsepower=hpGrid),se=T)
plot(auto$horsepower,auto$mpg,xlim=hp,cex=.5,col="black",xlab="Horsepower",
     ylab="MPG", main="Natural Splines")
lines(hpGrid,pred$fit,lwd=2)
lines(hpGrid,pred$fit+2*pred$se,lty="dashed")
lines(hpGrid,pred$fit-2*pred$se,lty="dashed")

fig3-1

1.1.e Fit a Smoothing Spline – R code

Here a smoothing spline is used. Smoothing splines also do not require the explicit setting of knots. We can change the ‘degrees of freedom(df)’ paramater to get the best fit

# Smoothing spline has a smoothing parameter, the degrees of freedom
# This is too wiggly
plot(auto$horsepower,auto$mpg,xlim=hp,cex=.5,col="black",xlab="Horsepower",
     ylab="MPG", main="Smoothing Splines")

# Here df is set to 16. This has a lot of variance
fit=smooth.spline(auto$horsepower,auto$mpg,df=16)
lines(fit,col="red",lwd=2)

# We can use Cross Validation to allow the spline to pick the value of this smpopothing paramter. We do not need to set the degrees of freedom 'df'
fit=smooth.spline(auto$horsepower,auto$mpg,cv=TRUE)
lines(fit,col="blue",lwd=2)

fig4-1

1.1e Splines – Python

There isn’t as much treatment of splines in Python and SKLearn. I did find the LSQUnivariate, UnivariateSpline spline. The LSQUnivariate spline requires the explcit setting of knots

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from scipy.interpolate import LSQUnivariateSpline
autoDF =pd.read_csv("auto_mpg.csv",encoding="ISO-8859-1")
autoDF.shape
autoDF.columns
autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']]
autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce')
auto=autoDF2.dropna()
auto=auto[['horsepower','mpg']].sort_values('horsepower')

# Set the knots manually
knots=[65,75,100,150]
# Create an array for X & y
X=np.array(auto['horsepower'])
y=np.array(auto['mpg'])
# Fit a LSQunivariate spline
s = LSQUnivariateSpline(X,y,knots)

#Plot the spline
xs = np.linspace(40,230,1000)
ys = s(xs)
plt.scatter(X, y)
plt.plot(xs, ys)
plt.savefig('fig2.png', bbox_inches='tight')

fig2

1.2 Generalized Additiive models (GAMs)

Generalized Additive Models (GAMs) is a really powerful ML tool.

y_{i} = \beta_{0} + f_{1}(x_{i1}) + f_{2}(x_{i2}) + .. +f_{p}(x_{ip}) + \epsilon_{i}

In GAMs we use a different functions for each of the variables. GAMs give a much better fit since we can choose any function for the different sections

1.2a Generalized Additive Models (GAMs) – R Code

The plot below show the smooth spline that is fit for each of the features horsepower, cylinder, displacement, year and acceleration. We can use any function for example loess, 4rd order polynomial etc.

library(gam)
# Fit a smoothing spline for horsepower, cyliner, displacement and acceleration
gam=gam(mpg~s(horsepower,4)+s(cylinder,5)+s(displacement,4)+s(year,4)+s(acceleration,5),data=auto)
# Display the summary of the fit. This give the significance of each of the paramwetr
# Also an ANOVA is given for each combination of the features
summary(gam)
## 
## Call: gam(formula = mpg ~ s(horsepower, 4) + s(cylinder, 5) + s(displacement, 
##     4) + s(year, 4) + s(acceleration, 5), data = auto)
## Deviance Residuals:
##     Min      1Q  Median      3Q     Max 
## -8.3190 -1.4436 -0.0261  1.2279 12.0873 
## 
## (Dispersion Parameter for gaussian family taken to be 6.9943)
## 
##     Null Deviance: 23818.99 on 391 degrees of freedom
## Residual Deviance: 2587.881 on 370 degrees of freedom
## AIC: 1898.282 
## 
## Number of Local Scoring Iterations: 3 
## 
## Anova for Parametric Effects
##                     Df  Sum Sq Mean Sq  F value    Pr(>F)    
## s(horsepower, 4)     1 15632.8 15632.8 2235.085 < 2.2e-16 ***
## s(cylinder, 5)       1   508.2   508.2   72.666 3.958e-16 ***
## s(displacement, 4)   1   374.3   374.3   53.514 1.606e-12 ***
## s(year, 4)           1  2263.2  2263.2  323.583 < 2.2e-16 ***
## s(acceleration, 5)   1   372.4   372.4   53.246 1.809e-12 ***
## Residuals          370  2587.9     7.0                       
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Anova for Nonparametric Effects
##                    Npar Df Npar F     Pr(F)    
## (Intercept)                                    
## s(horsepower, 4)         3 13.825 1.453e-08 ***
## s(cylinder, 5)           3 17.668 9.712e-11 ***
## s(displacement, 4)       3 44.573 < 2.2e-16 ***
## s(year, 4)               3 23.364 7.183e-14 ***
## s(acceleration, 5)       4  3.848  0.004453 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
par(mfrow=c(2,3))
plot(gam,se=TRUE)

fig5-1

1.2b Generalized Additive Models (GAMs) – Python Code

I did not find the equivalent of GAMs in SKlearn in Python. There was an early prototype (2012) in Github. Looks like it is still work in progress or has probably been abandoned.

1.3 Tree based Machine Learning Models

Tree based Machine Learning are all based on the ‘bootstrapping’ technique. In bootstrapping given a sample of size N, we create datasets of size N by sampling this original dataset with replacement. Machine Learning models are built on the different bootstrapped samples and then averaged.

Decision Trees as seen above have the tendency to overfit. There are several techniques that help to avoid this namely a) Bagging b) Random Forests c) Boosting

Bagging, Random Forest and Gradient Boosting

Bagging: Bagging, or Bootstrap Aggregation decreases the variance of predictions, by creating separate Decisiion Tree based ML models on the different samples and then averaging these ML models

Random Forests: Bagging is a greedy algorithm and tries to produce splits based on all variables which try to minimize the error. However the different ML models have a high correlation. Random Forests remove this shortcoming, by using a variable and random set of features to split on. Hence the features chosen and the resulting trees are uncorrelated. When these ML models are averaged the performance is much better.

Boosting: Gradient Boosted Decision Trees also use an ensemble of trees but they don’t build Machine Learning models with random set of features at each step. Rather small and simple trees are built. Successive trees try to minimize the error from the earlier trees.

Out of Bag (OOB) Error: In Random Forest and Gradient Boosting for each bootstrap sample taken from the dataset, there will be samples left out. These are known as Out of Bag samples.Classification accuracy carried out on these OOB samples is known as OOB error

1.31a Decision Trees – R Code

The code below creates a Decision tree with the cancer training data. The summary of the fit is output. Based on the ML model, the predict function is used on test data and a confusion matrix is output.

# Read the cancer data
library(tree)
library(caret)
library(e1071)
cancer <- read.csv("cancer.csv",stringsAsFactors = FALSE)
cancer <- cancer[,2:32]
cancer$target <- as.factor(cancer$target)
train_idx <- trainTestSplit(cancer,trainPercent=75,seed=5)
train <- cancer[train_idx, ]
test <- cancer[-train_idx, ]

# Create Decision Tree
cancerStatus=tree(target~.,train)
summary(cancerStatus)
## 
## Classification tree:
## tree(formula = target ~ ., data = train)
## Variables actually used in tree construction:
## [1] "worst.perimeter"      "worst.concave.points" "area.error"          
## [4] "worst.texture"        "mean.texture"         "mean.concave.points" 
## Number of terminal nodes:  9 
## Residual mean deviance:  0.1218 = 50.8 / 417 
## Misclassification error rate: 0.02347 = 10 / 426
pred <- predict(cancerStatus,newdata=test,type="class")
confusionMatrix(pred,test$target)
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  0  1
##          0 49  7
##          1  8 78
##                                           
##                Accuracy : 0.8944          
##                  95% CI : (0.8318, 0.9397)
##     No Information Rate : 0.5986          
##     P-Value [Acc > NIR] : 4.641e-15       
##                                           
##                   Kappa : 0.7795          
##  Mcnemar's Test P-Value : 1               
##                                           
##             Sensitivity : 0.8596          
##             Specificity : 0.9176          
##          Pos Pred Value : 0.8750          
##          Neg Pred Value : 0.9070          
##              Prevalence : 0.4014          
##          Detection Rate : 0.3451          
##    Detection Prevalence : 0.3944          
##       Balanced Accuracy : 0.8886          
##                                           
##        'Positive' Class : 0               
## 
# Plot decision tree with labels
plot(cancerStatus)
text(cancerStatus,pretty=0)

fig6-1

1.31b Decision Trees – Cross Validation – R Code

We can also perform a Cross Validation on the data to identify the Decision Tree which will give the minimum deviance.

library(tree)
cancer <- read.csv("cancer.csv",stringsAsFactors = FALSE)
cancer <- cancer[,2:32]
cancer$target <- as.factor(cancer$target)
train_idx <- trainTestSplit(cancer,trainPercent=75,seed=5)
train <- cancer[train_idx, ]
test <- cancer[-train_idx, ]

# Create Decision Tree
cancerStatus=tree(target~.,train)

# Execute 10 fold cross validation
cvCancer=cv.tree(cancerStatus)
plot(cvCancer)

fig7-1

# Plot the 
plot(cvCancer$size,cvCancer$dev,type='b')

fig1

prunedCancer=prune.tree(cancerStatus,best=4)
plot(prunedCancer)
text(prunedCancer,pretty=0)

fig2

pred <- predict(prunedCancer,newdata=test,type="class")
confusionMatrix(pred,test$target)
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  0  1
##          0 50  7
##          1  7 78
##                                          
##                Accuracy : 0.9014         
##                  95% CI : (0.8401, 0.945)
##     No Information Rate : 0.5986         
##     P-Value [Acc > NIR] : 7.988e-16      
##                                          
##                   Kappa : 0.7948         
##  Mcnemar's Test P-Value : 1              
##                                          
##             Sensitivity : 0.8772         
##             Specificity : 0.9176         
##          Pos Pred Value : 0.8772         
##          Neg Pred Value : 0.9176         
##              Prevalence : 0.4014         
##          Detection Rate : 0.3521         
##    Detection Prevalence : 0.4014         
##       Balanced Accuracy : 0.8974         
##                                          
##        'Positive' Class : 0              
## 

1.31c Decision Trees – Python Code

Below is the Python code for creating Decision Trees. The accuracy, precision, recall and F1 score is computed on the test data set.

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.metrics import confusion_matrix
from sklearn import tree
from sklearn.datasets import load_breast_cancer
from sklearn.model_selection import train_test_split
from sklearn.tree import DecisionTreeClassifier
from sklearn.datasets import make_classification, make_blobs
from sklearn.metrics import accuracy_score, precision_score, recall_score, f1_score
import graphviz 

cancer = load_breast_cancer()
(X_cancer, y_cancer) = load_breast_cancer(return_X_y = True)

X_train, X_test, y_train, y_test = train_test_split(X_cancer, y_cancer,
                                                   random_state = 0)
clf = DecisionTreeClassifier().fit(X_train, y_train)

print('Accuracy of Decision Tree classifier on training set: {:.2f}'
     .format(clf.score(X_train, y_train)))
print('Accuracy of Decision Tree classifier on test set: {:.2f}'
     .format(clf.score(X_test, y_test)))

y_predicted=clf.predict(X_test)
confusion = confusion_matrix(y_test, y_predicted)
print('Accuracy: {:.2f}'.format(accuracy_score(y_test, y_predicted)))
print('Precision: {:.2f}'.format(precision_score(y_test, y_predicted)))
print('Recall: {:.2f}'.format(recall_score(y_test, y_predicted)))
print('F1: {:.2f}'.format(f1_score(y_test, y_predicted)))

# Plot the Decision Tree
clf = DecisionTreeClassifier(max_depth=2).fit(X_train, y_train)
dot_data = tree.export_graphviz(clf, out_file=None, 
                         feature_names=cancer.feature_names,  
                         class_names=cancer.target_names,  
                         filled=True, rounded=True,  
                         special_characters=True)  
graph = graphviz.Source(dot_data)  
graph
## Accuracy of Decision Tree classifier on training set: 1.00
## Accuracy of Decision Tree classifier on test set: 0.87
## Accuracy: 0.87
## Precision: 0.97
## Recall: 0.82
## F1: 0.89

tree

1.31d Decision Trees – Cross Validation – Python Code

In the code below 5-fold cross validation is performed for different depths of the tree and the accuracy is computed. The accuracy on the test set seems to plateau when the depth is 8. But it is seen to increase again from 10 to 12. More analysis needs to be done here


import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.datasets import load_breast_cancer
from sklearn.tree import DecisionTreeClassifier
(X_cancer, y_cancer) = load_breast_cancer(return_X_y = True)
from sklearn.cross_validation import train_test_split, KFold
def computeCVAccuracy(X,y,folds):
    accuracy=[]
    foldAcc=[]
    depth=[1,2,3,4,5,6,7,8,9,10,11,12]
    nK=len(X)/float(folds)
    xval_err=0
    for i in depth: 
        kf = KFold(len(X),n_folds=folds)
        for train_index, test_index in kf:
            X_train, X_test = X.iloc[train_index], X.iloc[test_index]
            y_train, y_test = y.iloc[train_index], y.iloc[test_index]  
            clf = DecisionTreeClassifier(max_depth = i).fit(X_train, y_train)
            score=clf.score(X_test, y_test)
            accuracy.append(score)     
            
        foldAcc.append(np.mean(accuracy))  
        
    return(foldAcc)
    
    
cvAccuracy=computeCVAccuracy(pd.DataFrame(X_cancer),pd.DataFrame(y_cancer),folds=10)

df1=pd.DataFrame(cvAccuracy)
df1.columns=['cvAccuracy']
df=df1.reindex([1,2,3,4,5,6,7,8,9,10,11,12])
df.plot()
plt.title("Decision Tree - 10-fold Cross Validation Accuracy vs Depth of tree")
plt.xlabel("Depth of tree")
plt.ylabel("Accuracy")
plt.savefig('fig3.png', bbox_inches='tight')

 

 

fig3

 

1.4a Random Forest – R code

A Random Forest is fit using the Boston data. The summary shows that 4 variables were randomly chosen at each split and the resulting ML model explains 88.72% of the test data. Also the variable importance is plotted. It can be seen that ‘rooms’ and ‘status’ are the most influential features in the model

library(randomForest)
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL

# Select specific columns
Boston <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",                          "distances","highways","tax","teacherRatio","color",
                               "status","medianValue")

# Fit a Random Forest on the Boston training data
rfBoston=randomForest(medianValue~.,data=Boston)
# Display the summatu of the fit. It can be seen that the MSE is 10.88 
# and the percentage variance explained is 86.14%. About 4 variables were tried at each # #split for a maximum tree of 500.
# The MSE and percent variance is on Out of Bag trees
rfBoston
## 
## Call:
##  randomForest(formula = medianValue ~ ., data = Boston) 
##                Type of random forest: regression
##                      Number of trees: 500
## No. of variables tried at each split: 4
## 
##           Mean of squared residuals: 9.521672
##                     % Var explained: 88.72
#List and plot the variable importances
importance(rfBoston)
##              IncNodePurity
## crimeRate        2602.1550
## zone              258.8057
## indus            2599.6635
## charles           240.2879
## nox              2748.8485
## rooms           12011.6178
## age              1083.3242
## distances        2432.8962
## highways          393.5599
## tax              1348.6987
## teacherRatio     2841.5151
## color             731.4387
## status          12735.4046
varImpPlot(rfBoston)

fig8-1

1.4b Random Forest-OOB and Cross Validation Error – R code

The figure below shows the OOB error and the Cross Validation error vs the ‘mtry’. Here mtry indicates the number of random features that are chosen at each split. The lowest test error occurs when mtry = 8

library(randomForest)
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL

# Select specific columns
Boston <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",                          "distances","highways","tax","teacherRatio","color",
                               "status","medianValue")
# Split as training and tst sets
train_idx <- trainTestSplit(Boston,trainPercent=75,seed=5)
train <- Boston[train_idx, ]
test <- Boston[-train_idx, ]

#Initialize OOD and testError
oobError <- NULL
testError <- NULL
# In the code below the number of variables to consider at each split is increased
# from 1 - 13(max features) and the OOB error and the MSE is computed
for(i in 1:13){
    fitRF=randomForest(medianValue~.,data=train,mtry=i,ntree=400)
    oobError[i] <-fitRF$mse[400]
    pred <- predict(fitRF,newdata=test)
    testError[i] <- mean((pred-test$medianValue)^2)
}

# We can see the OOB and Test Error. It can be seen that the Random Forest performs
# best with the lowers MSE at mtry=6
matplot(1:13,cbind(testError,oobError),pch=19,col=c("red","blue"),
        type="b",xlab="mtry(no of varaibles at each split)", ylab="Mean Squared Error",
        main="Random Forest - OOB and Test Error")
legend("topright",legend=c("OOB","Test"),pch=19,col=c("red","blue"))

fig9-1

1.4c Random Forest – Python code

The python code for Random Forest Regression is shown below. The training and test score is computed. The variable importance shows that ‘rooms’ and ‘status’ are the most influential of the variables

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.ensemble import RandomForestRegressor
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")

X=df[['crimeRate','zone', 'indus','charles','nox','rooms', 'age','distances','highways','tax',
       'teacherRatio','color','status']]
y=df['medianValue']

X_train, X_test, y_train, y_test = train_test_split(X, y, random_state = 0)

regr = RandomForestRegressor(max_depth=4, random_state=0)
regr.fit(X_train, y_train)

print('R-squared score (training): {:.3f}'
     .format(regr.score(X_train, y_train)))
print('R-squared score (test): {:.3f}'
     .format(regr.score(X_test, y_test)))

feature_names=['crimeRate','zone', 'indus','charles','nox','rooms', 'age','distances','highways','tax',
       'teacherRatio','color','status']
print(regr.feature_importances_)
plt.figure(figsize=(10,6),dpi=80)
c_features=X_train.shape[1]
plt.barh(np.arange(c_features),regr.feature_importances_)
plt.xlabel("Feature importance")
plt.ylabel("Feature name")

plt.yticks(np.arange(c_features), feature_names)
plt.tight_layout()

plt.savefig('fig4.png', bbox_inches='tight')
## R-squared score (training): 0.917
## R-squared score (test): 0.734
## [ 0.03437382  0.          0.00580335  0.          0.00731004  0.36461548
##   0.00638577  0.03432173  0.0041244   0.01732328  0.01074148  0.0012638
##   0.51373683]

fig4

1.4d Random Forest – Cross Validation and OOB Error – Python code

As with R the ‘max_features’ determines the random number of features the random forest will use at each split. The plot shows that when max_features=8 the MSE is lowest

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.ensemble import RandomForestRegressor
from sklearn.model_selection import cross_val_score
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")

X=df[['crimeRate','zone', 'indus','charles','nox','rooms', 'age','distances','highways','tax',
       'teacherRatio','color','status']]
y=df['medianValue']

cvError=[]
oobError=[]
oobMSE=[]
for i in range(1,13):
    regr = RandomForestRegressor(max_depth=4, n_estimators=400,max_features=i,oob_score=True,random_state=0)
    mse= np.mean(cross_val_score(regr, X, y, cv=5,scoring = 'neg_mean_squared_error'))
    # Since this is neg_mean_squared_error I have inverted the sign to get MSE
    cvError.append(-mse)
    # Fit on all data to compute OOB error
    regr.fit(X, y)
    # Record the OOB error for each `max_features=i` setting
    oob = 1 - regr.oob_score_
    oobError.append(oob)
    # Get the Out of Bag prediction
    oobPred=regr.oob_prediction_ 
    # Compute the Mean Squared Error between OOB Prediction and target
    mseOOB=np.mean(np.square(oobPred-y))
    oobMSE.append(mseOOB)

# Plot the CV Error and OOB Error
# Set max_features
maxFeatures=np.arange(1,13) 
cvError=pd.DataFrame(cvError,index=maxFeatures)
oobMSE=pd.DataFrame(oobMSE,index=maxFeatures)
#Plot
fig8=df.plot()
fig8=plt.title('Random forest - CV Error and OOB Error vs max_features')
fig8.figure.savefig('fig8.png', bbox_inches='tight')

#Plot the OOB Error vs max_features
plt.plot(range(1,13),oobError)
fig2=plt.title("Random Forest - OOB Error vs max_features (variable no of features)")
fig2=plt.xlabel("max_features (variable no of features)")
fig2=plt.ylabel("OOB Error")
fig2.figure.savefig('fig7.png', bbox_inches='tight')

fig8 fig7

1.5a Boosting – R code

Here a Gradient Boosted ML Model is built with a n.trees=5000, with a learning rate of 0.01 and depth of 4. The feature importance plot also shows that rooms and status are the 2 most important features. The MSE vs the number of trees plateaus around 2000 trees

library(gbm)
# Perform gradient boosting on the Boston data set. The distribution is gaussian since we
# doing MSE. The interaction depth specifies the number of splits
boostBoston=gbm(medianValue~.,data=train,distribution="gaussian",n.trees=5000,
                shrinkage=0.01,interaction.depth=4)
#The summary gives the variable importance. The 2 most significant variables are
# number of rooms and lower status
summary(boostBoston)

##                       var    rel.inf
## rooms               rooms 42.2267200
## status             status 27.3024671
## distances       distances  7.9447972
## crimeRate       crimeRate  5.0238827
## nox                   nox  4.0616548
## teacherRatio teacherRatio  3.1991999
## age                   age  2.7909772
## color               color  2.3436295
## tax                   tax  2.1386213
## charles           charles  1.3799109
## highways         highways  0.7644026
## indus               indus  0.7236082
## zone                 zone  0.1001287
# The plots below show how each variable relates to the median value of the home. As
# the number of roomd increase the median value increases and with increase in lower status
# the median value decreases
par(mfrow=c(1,2))
#Plot the relation between the top 2 features and the target
plot(boostBoston,i="rooms")
plot(boostBoston,i="status")

fig10-2

# Create a sequence of trees between 100-5000 incremented by 50
nTrees=seq(100,5000,by=50)
# Predict the values for the test data
pred <- predict(boostBoston,newdata=test,n.trees=nTrees)
# Compute the mean for each of the MSE for each of the number of trees 
boostError <- apply((pred-test$medianValue)^2,2,mean)
#Plot the MSE vs the number of trees
plot(nTrees,boostError,pch=19,col="blue",ylab="Mean Squared Error",
     main="Boosting Test Error")

fig10-3

1.5b Cross Validation Boosting – R code

Included below is a cross validation error vs the learning rate. The lowest error is when learning rate = 0.09

cvError <- NULL
s <- c(.001,0.01,0.03,0.05,0.07,0.09,0.1)
for(i in seq_along(s)){
    cvBoost=gbm(medianValue~.,data=train,distribution="gaussian",n.trees=5000,
                shrinkage=s[i],interaction.depth=4,cv.folds=5)
    cvError[i] <- mean(cvBoost$cv.error)
}

# Create a data frame for plotting
a <- rbind(s,cvError)
b <- as.data.frame(t(a))
# It can be seen that a shrinkage parameter of 0,05 gives the lowes CV Error
ggplot(b,aes(s,cvError)) + geom_point() + geom_line(color="blue") + 
    xlab("Shrinkage") + ylab("Cross Validation Error") +
    ggtitle("Gradient boosted trees - Cross Validation error vs Shrinkage")

fig11-1

1.5c Boosting – Python code

A gradient boost ML model in Python is created below. The Rsquared score is computed on the training and test data.

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.ensemble import GradientBoostingRegressor
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")

X=df[['crimeRate','zone', 'indus','charles','nox','rooms', 'age','distances','highways','tax',
       'teacherRatio','color','status']]
y=df['medianValue']

X_train, X_test, y_train, y_test = train_test_split(X, y, random_state = 0)

regr = GradientBoostingRegressor()
regr.fit(X_train, y_train)

print('R-squared score (training): {:.3f}'
     .format(regr.score(X_train, y_train)))
print('R-squared score (test): {:.3f}'
     .format(regr.score(X_test, y_test)))
## R-squared score (training): 0.983
## R-squared score (test): 0.821

1.5c Cross Validation Boosting – Python code

the cross validation error is computed as the learning rate is varied. The minimum CV eror occurs when lr = 0.04

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.ensemble import RandomForestRegressor
from sklearn.ensemble import GradientBoostingRegressor
from sklearn.model_selection import cross_val_score
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")

X=df[['crimeRate','zone', 'indus','charles','nox','rooms', 'age','distances','highways','tax',
       'teacherRatio','color','status']]
y=df['medianValue']

cvError=[]
learning_rate =[.001,0.01,0.03,0.05,0.07,0.09,0.1]
for lr in learning_rate:
    regr = GradientBoostingRegressor(max_depth=4, n_estimators=400,learning_rate  =lr,random_state=0)
    mse= np.mean(cross_val_score(regr, X, y, cv=10,scoring = 'neg_mean_squared_error'))
    # Since this is neg_mean_squared_error I have inverted the sign to get MSE
    cvError.append(-mse)
learning_rate =[.001,0.01,0.03,0.05,0.07,0.09,0.1]
plt.plot(learning_rate,cvError)
plt.title("Gradient Boosting - 5-fold CV- Mean Squared Error vs max_features (variable no of features)")
plt.xlabel("max_features (variable no of features)")
plt.ylabel("Mean Squared Error")
plt.savefig('fig6.png', bbox_inches='tight')

fig6

Conclusion This post covered Splines and Tree based ML models like Bagging, Random Forest and Boosting. Stay tuned for further updates.

You may also like

  1. Re-introducing cricketr! : An R package to analyze performances of cricketer
  2. Designing a Social Web Portal
  3. My travels through the realms of Data Science, Machine Learning, Deep Learning and (AI)
  4. My TEDx talk on the “Internet of Things”
  5. Singularity
  6. A closer look at “Robot Horse on a Trot” in Android

To see all posts see Index of posts

Practical Machine Learning with R and Python – Part 3


In this post ‘Practical Machine Learning with R and Python – Part 3’,  I discuss ‘Feature Selection’ methods. This post is a continuation of my 2 earlier posts

  1. Practical Machine Learning with R and Python – Part 1
  2. Practical Machine Learning with R and Python – Part 2

While applying Machine Learning techniques, the data set will usually include a large number of predictors for a target variable. It is quite likely, that not all the predictors or feature variables will have an impact on the output. Hence it is becomes necessary to choose only those features which influence the output variable thus simplifying  to a reduced feature set on which to train the ML model on. The techniques that are used are the following

  • Best fit
  • Forward fit
  • Backward fit
  • Ridge Regression or L2 regularization
  • Lasso or L1 regularization

This post includes the equivalent ML code in R and Python.

All these methods remove those features which do not sufficiently influence the output. As in my previous 2 posts on “Practical Machine Learning with R and Python’, this post is largely based on the topics in the following 2 MOOC courses
1. Statistical Learning, Prof Trevor Hastie & Prof Robert Tibesherani, Online Stanford
2. Applied Machine Learning in Python Prof Kevyn-Collin Thomson, University Of Michigan, Coursera

You can download this R Markdown file and the associated data from Github – Machine Learning-RandPython-Part3. 

Check out my compact and minimal book  “Practical Machine Learning with R and Python:Second edition- Machine Learning in stereo”  available in Amazon in paperback($10.99) and kindle($7.99) versions. My book includes implementations of key ML algorithms and associated measures and metrics. The book is ideal for anybody who is familiar with the concepts and would like a quick reference to the different ML algorithms that can be applied to problems and how to select the best model. Pick your copy today!!

1.1 Best Fit

For a dataset with features f1,f2,f3…fn, the ‘Best fit’ approach, chooses all possible combinations of features and creates separate ML models for each of the different combinations. The best fit algotithm then uses some filtering criteria based on Adj Rsquared, Cp, BIC or AIC to pick out the best model among all models.

Since the Best Fit approach searches the entire solution space it is computationally infeasible. The number of models that have to be searched increase exponentially as the number of predictors increase. For ‘p’ predictors a total of 2^{p} ML models have to be searched. This can be shown as follows

There are C_{1} ways to choose single feature ML models among ‘n’ features, C_{2} ways to choose 2 feature models among ‘n’ models and so on, or
1+C_{1} + C_{2} +... + C_{n}
= Total number of models in Best Fit.  Since from Binomial theorem we have
(1+x)^{n} = 1+C_{1}x + C_{2}x^{2} +... + C_{n}x^{n}
When x=1 in the equation (1) above, this becomes
2^{n} = 1+C_{1} + C_{2} +... + C_{n}

Hence there are 2^{n} models to search amongst in Best Fit. For 10 features this is 2^{10} or ~1000 models and for 40 features this becomes 2^{40} which almost 1 trillion. Usually there are datasets with 1000 or maybe even 100000 features and Best fit becomes computationally infeasible.

Anyways I have included the Best Fit approach as I use the Boston crime datasets which is available both the MASS package in R and Sklearn in Python and it has 13 features. Even this small feature set takes a bit of time since the Best fit needs to search among ~2^{13}= 8192  models

Initially I perform a simple Linear Regression Fit to estimate the features that are statistically insignificant. By looking at the p-values of the features it can be seen that ‘indus’ and ‘age’ features have high p-values and are not significant

1.1a Linear Regression – R code

source('RFunctions-1.R')
#Read the Boston crime data
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL
# Rename the columns
names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age",
              "distances","highways","tax","teacherRatio","color","status","cost")
# Select specific columns
df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",
                            "distances","highways","tax","teacherRatio","color","status","cost")
dim(df1)
## [1] 506  14
# Linear Regression fit
fit <- lm(cost~. ,data=df1)
summary(fit)
## 
## Call:
## lm(formula = cost ~ ., data = df1)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -15.595  -2.730  -0.518   1.777  26.199 
## 
## Coefficients:
##                Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   3.646e+01  5.103e+00   7.144 3.28e-12 ***
## crimeRate    -1.080e-01  3.286e-02  -3.287 0.001087 ** 
## zone          4.642e-02  1.373e-02   3.382 0.000778 ***
## indus         2.056e-02  6.150e-02   0.334 0.738288    
## charles       2.687e+00  8.616e-01   3.118 0.001925 ** 
## nox          -1.777e+01  3.820e+00  -4.651 4.25e-06 ***
## rooms         3.810e+00  4.179e-01   9.116  < 2e-16 ***
## age           6.922e-04  1.321e-02   0.052 0.958229    
## distances    -1.476e+00  1.995e-01  -7.398 6.01e-13 ***
## highways      3.060e-01  6.635e-02   4.613 5.07e-06 ***
## tax          -1.233e-02  3.760e-03  -3.280 0.001112 ** 
## teacherRatio -9.527e-01  1.308e-01  -7.283 1.31e-12 ***
## color         9.312e-03  2.686e-03   3.467 0.000573 ***
## status       -5.248e-01  5.072e-02 -10.347  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.745 on 492 degrees of freedom
## Multiple R-squared:  0.7406, Adjusted R-squared:  0.7338 
## F-statistic: 108.1 on 13 and 492 DF,  p-value: < 2.2e-16

Next we apply the different feature selection models to automatically remove features that are not significant below

1.1a Best Fit – R code

The Best Fit requires the ‘leaps’ R package

library(leaps)
source('RFunctions-1.R')
#Read the Boston crime data
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL
# Rename the columns
names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age",
              "distances","highways","tax","teacherRatio","color","status","cost")
# Select specific columns
df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",
                            "distances","highways","tax","teacherRatio","color","status","cost")

# Perform a best fit
bestFit=regsubsets(cost~.,df1,nvmax=13)

# Generate a summary of the fit
bfSummary=summary(bestFit)

# Plot the Residual Sum of Squares vs number of variables 
plot(bfSummary$rss,xlab="Number of Variables",ylab="RSS",type="l",main="Best fit RSS vs No of features")
# Get the index of the minimum value
a=which.min(bfSummary$rss)
# Mark this in red
points(a,bfSummary$rss[a],col="red",cex=2,pch=20)

The plot below shows that the Best fit occurs with all 13 features included. Notice that there is no significant change in RSS from 11 features onward.

# Plot the CP statistic vs Number of variables
plot(bfSummary$cp,xlab="Number of Variables",ylab="Cp",type='l',main="Best fit Cp vs No of features")
# Find the lowest CP value
b=which.min(bfSummary$cp)
# Mark this in red
points(b,bfSummary$cp[b],col="red",cex=2,pch=20)

Based on Cp metric the best fit occurs at 11 features as seen below. The values of the coefficients are also included below

# Display the set of features which provide the best fit
coef(bestFit,b)
##   (Intercept)     crimeRate          zone       charles           nox 
##  36.341145004  -0.108413345   0.045844929   2.718716303 -17.376023429 
##         rooms     distances      highways           tax  teacherRatio 
##   3.801578840  -1.492711460   0.299608454  -0.011777973  -0.946524570 
##         color        status 
##   0.009290845  -0.522553457
#  Plot the BIC value
plot(bfSummary$bic,xlab="Number of Variables",ylab="BIC",type='l',main="Best fit BIC vs No of Features")
# Find and mark the min value
c=which.min(bfSummary$bic)
points(c,bfSummary$bic[c],col="red",cex=2,pch=20)

# R has some other good plots for best fit
plot(bestFit,scale="r2",main="Rsquared vs No Features")

R has the following set of really nice visualizations. The plot below shows the Rsquared for a set of predictor variables. It can be seen when Rsquared starts at 0.74- indus, charles and age have not been included. 

plot(bestFit,scale="Cp",main="Cp vs NoFeatures")

The Cp plot below for value shows indus, charles and age as not included in the Best fit

plot(bestFit,scale="bic",main="BIC vs Features")

1.1b Best fit (Exhaustive Search ) – Python code

The Python package for performing a Best Fit is the Exhaustive Feature Selector EFS.

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from mlxtend.feature_selection import ExhaustiveFeatureSelector as EFS

# Read the Boston crime data
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")

#Rename the columns
df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status","cost"]
# Set X and y 
X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status"]]
y=df['cost']

# Perform an Exhaustive Search. The EFS and SFS packages use 'neg_mean_squared_error'. The 'mean_squared_error' seems to have been deprecated. I think this is just the MSE with the a negative sign.
lr = LinearRegression()
efs1 = EFS(lr, 
           min_features=1,
           max_features=13,
           scoring='neg_mean_squared_error',
           print_progress=True,
           cv=5)


# Create a efs fit
efs1 = efs1.fit(X.as_matrix(), y.as_matrix())

print('Best negtive mean squared error: %.2f' % efs1.best_score_)
## Print the IDX of the best features 
print('Best subset:', efs1.best_idx_)
Features: 8191/8191Best negtive mean squared error: -28.92
## ('Best subset:', (0, 1, 4, 6, 7, 8, 9, 10, 11, 12))

The indices for the best subset are shown above.

1.2 Forward fit

Forward fit is a greedy algorithm that tries to optimize the feature selected, by minimizing the selection criteria (adj Rsqaured, Cp, AIC or BIC) at every step. For a dataset with features f1,f2,f3…fn, the forward fit starts with the NULL set. It then pick the ML model with a single feature from n features which has the highest adj Rsquared, or minimum Cp, BIC or some such criteria. After picking the 1 feature from n which satisfies the criteria the most, the next feature from the remaining n-1 features is chosen. When the 2 feature model which satisfies the selection criteria the best is chosen, another feature from the remaining n-2 features are added and so on. The forward fit is a sub-optimal algorithm. There is no guarantee that the final list of features chosen will be the best among the lot. The computation required for this is of  n + n-1 + n -2 + .. 1 = n(n+1)/2 which is of the order of n^{2}. Though forward fit is a sub optimal solution it is far more computationally efficient than best fit

1.2a Forward fit – R code

Forward fit in R determines that 11 features are required for the best fit. The features are shown below

library(leaps)
# Read the data
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL
# Rename the columns
names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age",
              "distances","highways","tax","teacherRatio","color","status","cost")

# Select columns
df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",
                     "distances","highways","tax","teacherRatio","color","status","cost")

#Split as training and test 
train_idx <- trainTestSplit(df1,trainPercent=75,seed=5)
train <- df1[train_idx, ]
test <- df1[-train_idx, ]

# Find the best forward fit
fitFwd=regsubsets(cost~.,data=train,nvmax=13,method="forward")

# Compute the MSE
valErrors=rep(NA,13)
test.mat=model.matrix(cost~.,data=test)
for(i in 1:13){
    coefi=coef(fitFwd,id=i)
    pred=test.mat[,names(coefi)]%*%coefi
    valErrors[i]=mean((test$cost-pred)^2)
}

# Plot the Residual Sum of Squares
plot(valErrors,xlab="Number of Variables",ylab="Validation Error",type="l",main="Forward fit RSS vs No of features")
# Gives the index of the minimum value
a<-which.min(valErrors)
print(a)
## [1] 11
# Highlight the smallest value
points(c,valErrors[a],col="blue",cex=2,pch=20)

Forward fit R selects 11 predictors as the best ML model to predict the ‘cost’ output variable. The values for these 11 predictors are included below

#Print the 11 ccoefficients
coefi=coef(fitFwd,id=i)
coefi
##   (Intercept)     crimeRate          zone         indus       charles 
##  2.397179e+01 -1.026463e-01  3.118923e-02  1.154235e-04  3.512922e+00 
##           nox         rooms           age     distances      highways 
## -1.511123e+01  4.945078e+00 -1.513220e-02 -1.307017e+00  2.712534e-01 
##           tax  teacherRatio         color        status 
## -1.330709e-02 -8.182683e-01  1.143835e-02 -3.750928e-01

1.2b Forward fit with Cross Validation – R code

The Python package SFS includes N Fold Cross Validation errors for forward and backward fit so I decided to add this code to R. This is not available in the ‘leaps’ R package, however the implementation is quite simple. Another implementation is also available at Statistical Learning, Prof Trevor Hastie & Prof Robert Tibesherani, Online Stanford 2.

library(dplyr)
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL
names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age",
              "distances","highways","tax","teacherRatio","color","status","cost")

# Select columns
df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",
                     "distances","highways","tax","teacherRatio","color","status","cost")

set.seed(6)
# Set max number of features
nvmax<-13
cvError <- NULL
# Loop through each features
for(i in 1:nvmax){
    # Set no of folds
    noFolds=5
    # Create the rows which fall into different folds from 1..noFolds
    folds = sample(1:noFolds, nrow(df1), replace=TRUE) 
    cv<-0
    # Loop through the folds
    for(j in 1:noFolds){
        # The training is all rows for which the row is != j (k-1 folds -> training)
        train <- df1[folds!=j,]
        # The rows which have j as the index become the test set
        test <- df1[folds==j,]
        # Create a forward fitting model for this
        fitFwd=regsubsets(cost~.,data=train,nvmax=13,method="forward")
        # Select the number of features and get the feature coefficients
        coefi=coef(fitFwd,id=i)
        #Get the value of the test data
        test.mat=model.matrix(cost~.,data=test)
        # Multiply the tes data with teh fitted coefficients to get the predicted value
        # pred = b0 + b1x1+b2x2... b13x13
        pred=test.mat[,names(coefi)]%*%coefi
        # Compute mean squared error
        rss=mean((test$cost - pred)^2)
        # Add all the Cross Validation errors
        cv=cv+rss
    }
    # Compute the average of MSE for K folds for number of features 'i'
    cvError[i]=cv/noFolds
}
a <- seq(1,13)
d <- as.data.frame(t(rbind(a,cvError)))
names(d) <- c("Features","CVError")
#Plot the CV Error vs No of Features
ggplot(d,aes(x=Features,y=CVError),color="blue") + geom_point() + geom_line(color="blue") +
    xlab("No of features") + ylab("Cross Validation Error") +
    ggtitle("Forward Selection - Cross Valdation Error vs No of Features")

Forward fit with 5 fold cross validation indicates that all 13 features are required

# This gives the index of the minimum value
a=which.min(cvError)
print(a)
## [1] 13
#Print the 13 coefficients of these features
coefi=coef(fitFwd,id=a)
coefi
##   (Intercept)     crimeRate          zone         indus       charles 
##  36.650645380  -0.107980979   0.056237669   0.027016678   4.270631466 
##           nox         rooms           age     distances      highways 
## -19.000715500   3.714720418   0.019952654  -1.472533973   0.326758004 
##           tax  teacherRatio         color        status 
##  -0.011380750  -0.972862622   0.009549938  -0.582159093

1.2c Forward fit – Python code

The Backward Fit in Python uses the Sequential feature selection (SFS) package (SFS)(https://rasbt.github.io/mlxtend/user_guide/feature_selection/SequentialFeatureSelector/)

Note: The Cross validation error for SFS in Sklearn is negative, possibly because it computes the ‘neg_mean_squared_error’. The earlier ‘mean_squared_error’ in the package seems to have been deprecated. I have taken the -ve of this neg_mean_squared_error. I think this would give mean_squared_error.

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from sklearn.datasets import load_boston
from mlxtend.plotting import plot_sequential_feature_selection as plot_sfs
import matplotlib.pyplot as plt
from mlxtend.feature_selection import SequentialFeatureSelector as SFS
from sklearn.linear_model import LinearRegression


df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")
#Rename the columns
df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status","cost"]
X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status"]]
y=df['cost']
lr = LinearRegression()
# Create a forward fit model
sfs = SFS(lr, 
          k_features=(1,13), 
          forward=True, # Forward fit
          floating=False, 
          scoring='neg_mean_squared_error',
          cv=5)

# Fit this on the data
sfs = sfs.fit(X.as_matrix(), y.as_matrix())
# Get all the details of the forward fits
a=sfs.get_metric_dict()
n=[]
o=[]

# Compute the mean cross validation scores
for i in np.arange(1,13):
    n.append(-np.mean(a[i]['cv_scores']))  
m=np.arange(1,13)
# Get the index of the minimum CV score

# Plot the CV scores vs the number of features
fig1=plt.plot(m,n)
fig1=plt.title('Mean CV Scores vs No of features')
fig1.figure.savefig('fig1.png', bbox_inches='tight')

print(pd.DataFrame.from_dict(sfs.get_metric_dict(confidence_interval=0.90)).T)

idx = np.argmin(n)
print "No of features=",idx
#Get the features indices for the best forward fit and convert to list
b=list(a[idx]['feature_idx'])
print(b)

# Index the column names. 
# Features from forward fit
print("Features selected in forward fit")
print(X.columns[b])
##    avg_score ci_bound                                          cv_scores  \
## 1   -42.6185  19.0465  [-23.5582499971, -41.8215743748, -73.993608929...   
## 2   -36.0651  16.3184  [-18.002498199, -40.1507894517, -56.5286659068...   
## 3   -34.1001    20.87  [-9.43012884381, -25.9584955394, -36.184188174...   
## 4   -33.7681  20.1638  [-8.86076528781, -28.650217633, -35.7246353855...   
## 5   -33.6392  20.5271  [-8.90807628524, -28.0684679108, -35.827463022...   
## 6   -33.6276  19.0859  [-9.549485942, -30.9724602876, -32.6689523347,...   
## 7   -32.4082  19.1455  [-10.0177149635, -28.3780298492, -30.926917231...   
## 8   -32.3697   18.533  [-11.1431684243, -27.5765510172, -31.168994094...   
## 9   -32.4016  21.5561  [-10.8972555995, -25.739780653, -30.1837430353...   
## 10  -32.8504  22.6508  [-12.3909282079, -22.1533250755, -33.385407342...   
## 11  -34.1065  24.7019  [-12.6429253721, -22.1676650245, -33.956999528...   
## 12  -35.5814   25.693  [-12.7303397453, -25.0145323483, -34.211898373...   
## 13  -37.1318  23.2657  [-12.4603005692, -26.0486211062, -33.074137979...   
## 
##                                    feature_idx  std_dev  std_err  
## 1                                        (12,)  18.9042  9.45212  
## 2                                     (10, 12)  16.1965  8.09826  
## 3                                  (10, 12, 5)  20.7142  10.3571  
## 4                               (10, 3, 12, 5)  20.0132  10.0066  
## 5                            (0, 10, 3, 12, 5)  20.3738  10.1869  
## 6                         (0, 3, 5, 7, 10, 12)  18.9433  9.47167  
## 7                      (0, 2, 3, 5, 7, 10, 12)  19.0026  9.50128  
## 8                   (0, 1, 2, 3, 5, 7, 10, 12)  18.3946  9.19731  
## 9               (0, 1, 2, 3, 5, 7, 10, 11, 12)  21.3952  10.6976  
## 10           (0, 1, 2, 3, 4, 5, 7, 10, 11, 12)  22.4816  11.2408  
## 11        (0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12)  24.5175  12.2587  
## 12     (0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12)  25.5012  12.7506  
## 13  (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)  23.0919   11.546  
## No of features= 7
## [0, 2, 3, 5, 7, 10, 12]
## #################################################################################
## Features selected in forward fit
## Index([u'crimeRate', u'indus', u'chasRiver', u'rooms', u'distances',
##        u'teacherRatio', u'status'],
##       dtype='object')

The table above shows the average score, 10 fold CV errors, the features included at every step, std. deviation and std. error

The above plot indicates that 8 features provide the lowest Mean CV error

1.3 Backward Fit

Backward fit belongs to the class of greedy algorithms which tries to optimize the feature set, by dropping a feature at every stage which results in the worst performance for a given criteria of Adj RSquared, Cp, BIC or AIC. For a dataset with features f1,f2,f3…fn, the backward fit starts with the all the features f1,f2.. fn to begin with. It then pick the ML model with a n-1 features by dropping the feature,f_{j}, for e.g., the inclusion of which results in the worst performance in adj Rsquared, or minimum Cp, BIC or some such criteria. At every step 1 feature is dopped. There is no guarantee that the final list of features chosen will be the best among the lot. The computation required for this is of n + n-1 + n -2 + .. 1 = n(n+1)/2 which is of the order of n^{2}. Though backward fit is a sub optimal solution it is far more computationally efficient than best fit

1.3a Backward fit – R code

library(dplyr)
# Read the data
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL
# Rename the columns
names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age",
              "distances","highways","tax","teacherRatio","color","status","cost")

# Select columns
df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",
                     "distances","highways","tax","teacherRatio","color","status","cost")

set.seed(6)
# Set max number of features
nvmax<-13
cvError <- NULL
# Loop through each features
for(i in 1:nvmax){
    # Set no of folds
    noFolds=5
    # Create the rows which fall into different folds from 1..noFolds
    folds = sample(1:noFolds, nrow(df1), replace=TRUE) 
    cv<-0
    for(j in 1:noFolds){
        # The training is all rows for which the row is != j 
        train <- df1[folds!=j,]
        # The rows which have j as the index become the test set
        test <- df1[folds==j,]
        # Create a backward fitting model for this
        fitFwd=regsubsets(cost~.,data=train,nvmax=13,method="backward")
        # Select the number of features and get the feature coefficients
        coefi=coef(fitFwd,id=i)
        #Get the value of the test data
        test.mat=model.matrix(cost~.,data=test)
        # Multiply the tes data with teh fitted coefficients to get the predicted value
        # pred = b0 + b1x1+b2x2... b13x13
        pred=test.mat[,names(coefi)]%*%coefi
        # Compute mean squared error
        rss=mean((test$cost - pred)^2)
        # Add the Residual sum of square
        cv=cv+rss
    }
    # Compute the average of MSE for K folds for number of features 'i'
    cvError[i]=cv/noFolds
}
a <- seq(1,13)
d <- as.data.frame(t(rbind(a,cvError)))
names(d) <- c("Features","CVError")
# Plot the Cross Validation Error vs Number of features
ggplot(d,aes(x=Features,y=CVError),color="blue") + geom_point() + geom_line(color="blue") +
    xlab("No of features") + ylab("Cross Validation Error") +
    ggtitle("Backward Selection - Cross Valdation Error vs No of Features")

# This gives the index of the minimum value
a=which.min(cvError)
print(a)
## [1] 13
#Print the 13 coefficients of these features
coefi=coef(fitFwd,id=a)
coefi
##   (Intercept)     crimeRate          zone         indus       charles 
##  36.650645380  -0.107980979   0.056237669   0.027016678   4.270631466 
##           nox         rooms           age     distances      highways 
## -19.000715500   3.714720418   0.019952654  -1.472533973   0.326758004 
##           tax  teacherRatio         color        status 
##  -0.011380750  -0.972862622   0.009549938  -0.582159093

Backward selection in R also indicates the 13 features and the corresponding coefficients as providing the best fit

1.3b Backward fit – Python code

The Backward Fit in Python uses the Sequential feature selection (SFS) package (SFS)(https://rasbt.github.io/mlxtend/user_guide/feature_selection/SequentialFeatureSelector/)

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from mlxtend.plotting import plot_sequential_feature_selection as plot_sfs
import matplotlib.pyplot as plt
from mlxtend.feature_selection import SequentialFeatureSelector as SFS
from sklearn.linear_model import LinearRegression

# Read the data
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")
#Rename the columns
df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status","cost"]
X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status"]]
y=df['cost']
lr = LinearRegression()

# Create the SFS model
sfs = SFS(lr, 
          k_features=(1,13), 
          forward=False, # Backward
          floating=False, 
          scoring='neg_mean_squared_error',
          cv=5)

# Fit the model
sfs = sfs.fit(X.as_matrix(), y.as_matrix())
a=sfs.get_metric_dict()
n=[]
o=[]

# Compute the mean of the validation scores
for i in np.arange(1,13):
    n.append(-np.mean(a[i]['cv_scores'])) 
m=np.arange(1,13)

# Plot the Validation scores vs number of features
fig2=plt.plot(m,n)
fig2=plt.title('Mean CV Scores vs No of features')
fig2.figure.savefig('fig2.png', bbox_inches='tight')

print(pd.DataFrame.from_dict(sfs.get_metric_dict(confidence_interval=0.90)).T)

# Get the index of minimum cross validation error
idx = np.argmin(n)
print "No of features=",idx
#Get the features indices for the best forward fit and convert to list
b=list(a[idx]['feature_idx'])
# Index the column names. 
# Features from backward fit
print("Features selected in bacward fit")
print(X.columns[b])
##    avg_score ci_bound                                          cv_scores  \
## 1   -42.6185  19.0465  [-23.5582499971, -41.8215743748, -73.993608929...   
## 2   -36.0651  16.3184  [-18.002498199, -40.1507894517, -56.5286659068...   
## 3   -35.4992  13.9619  [-17.2329292677, -44.4178648308, -51.633177846...   
## 4    -33.463  12.4081  [-20.6415333292, -37.3247852146, -47.479302977...   
## 5   -33.1038  10.6156  [-20.2872309863, -34.6367078466, -45.931870352...   
## 6   -32.0638  10.0933  [-19.4463829372, -33.460638577, -42.726257249,...   
## 7   -30.7133  9.23881  [-19.4425181917, -31.1742902259, -40.531266671...   
## 8   -29.7432  9.84468  [-19.445277268, -30.0641187173, -40.2561247122...   
## 9   -29.0878  9.45027  [-19.3545569877, -30.094768669, -39.7506036377...   
## 10  -28.9225  9.39697  [-18.562171585, -29.968504938, -39.9586835965,...   
## 11  -29.4301  10.8831  [-18.3346152225, -30.3312847532, -45.065432793...   
## 12  -30.4589  11.1486  [-18.493389527, -35.0290639374, -45.1558231765...   
## 13  -37.1318  23.2657  [-12.4603005692, -26.0486211062, -33.074137979...   
## 
##                                    feature_idx  std_dev  std_err  
## 1                                        (12,)  18.9042  9.45212  
## 2                                     (10, 12)  16.1965  8.09826  
## 3                                  (10, 12, 7)  13.8576  6.92881  
## 4                               (12, 10, 4, 7)  12.3154  6.15772  
## 5                            (4, 7, 8, 10, 12)  10.5363  5.26816  
## 6                         (4, 7, 8, 9, 10, 12)  10.0179  5.00896  
## 7                      (1, 4, 7, 8, 9, 10, 12)  9.16981  4.58491  
## 8                  (1, 4, 7, 8, 9, 10, 11, 12)  9.77116  4.88558  
## 9               (0, 1, 4, 7, 8, 9, 10, 11, 12)  9.37969  4.68985  
## 10           (0, 1, 4, 6, 7, 8, 9, 10, 11, 12)   9.3268   4.6634  
## 11        (0, 1, 3, 4, 6, 7, 8, 9, 10, 11, 12)  10.8018  5.40092  
## 12     (0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12)  11.0653  5.53265  
## 13  (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)  23.0919   11.546  
## No of features= 9
## Features selected in bacward fit
## Index([u'crimeRate', u'zone', u'NO2', u'distances', u'idxHighways', u'taxRate',
##        u'teacherRatio', u'color', u'status'],
##       dtype='object')

The table above shows the average score, 10 fold CV errors, the features included at every step, std. deviation and std. error

Backward fit in Python indicate that 10 features provide the best fit

1.3c Sequential Floating Forward Selection (SFFS) – Python code

The Sequential Feature search also includes ‘floating’ variants which include or exclude features conditionally, once they were excluded or included. The SFFS can conditionally include features which were excluded from the previous step, if it results in a better fit. This option will tend to a better solution, than plain simple SFS. These variants are included below

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from sklearn.datasets import load_boston
from mlxtend.plotting import plot_sequential_feature_selection as plot_sfs
import matplotlib.pyplot as plt
from mlxtend.feature_selection import SequentialFeatureSelector as SFS
from sklearn.linear_model import LinearRegression


df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")
#Rename the columns
df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status","cost"]
X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status"]]
y=df['cost']
lr = LinearRegression()

# Create the floating forward search
sffs = SFS(lr, 
          k_features=(1,13), 
          forward=True,  # Forward
          floating=True,  #Floating
          scoring='neg_mean_squared_error',
          cv=5)

# Fit a model
sffs = sffs.fit(X.as_matrix(), y.as_matrix())
a=sffs.get_metric_dict()
n=[]
o=[]
# Compute mean validation scores
for i in np.arange(1,13):
    n.append(-np.mean(a[i]['cv_scores'])) 
   
    
    
m=np.arange(1,13)


# Plot the cross validation score vs number of features
fig3=plt.plot(m,n)
fig3=plt.title('SFFS:Mean CV Scores vs No of features')
fig3.figure.savefig('fig3.png', bbox_inches='tight')

print(pd.DataFrame.from_dict(sffs.get_metric_dict(confidence_interval=0.90)).T)
# Get the index of the minimum CV score
idx = np.argmin(n)
print "No of features=",idx
#Get the features indices for the best forward floating fit and convert to list
b=list(a[idx]['feature_idx'])
print(b)

print("#################################################################################")
# Index the column names. 
# Features from forward fit
print("Features selected in forward fit")
print(X.columns[b])
##    avg_score ci_bound                                          cv_scores  \
## 1   -42.6185  19.0465  [-23.5582499971, -41.8215743748, -73.993608929...   
## 2   -36.0651  16.3184  [-18.002498199, -40.1507894517, -56.5286659068...   
## 3   -34.1001    20.87  [-9.43012884381, -25.9584955394, -36.184188174...   
## 4   -33.7681  20.1638  [-8.86076528781, -28.650217633, -35.7246353855...   
## 5   -33.6392  20.5271  [-8.90807628524, -28.0684679108, -35.827463022...   
## 6   -33.6276  19.0859  [-9.549485942, -30.9724602876, -32.6689523347,...   
## 7   -32.1834  12.1001  [-17.9491036167, -39.6479234651, -45.470227740...   
## 8   -32.0908  11.8179  [-17.4389015788, -41.2453629843, -44.247557798...   
## 9   -31.0671  10.1581  [-17.2689542913, -37.4379370429, -41.366372300...   
## 10  -28.9225  9.39697  [-18.562171585, -29.968504938, -39.9586835965,...   
## 11  -29.4301  10.8831  [-18.3346152225, -30.3312847532, -45.065432793...   
## 12  -30.4589  11.1486  [-18.493389527, -35.0290639374, -45.1558231765...   
## 13  -37.1318  23.2657  [-12.4603005692, -26.0486211062, -33.074137979...   
## 
##                                    feature_idx  std_dev  std_err  
## 1                                        (12,)  18.9042  9.45212  
## 2                                     (10, 12)  16.1965  8.09826  
## 3                                  (10, 12, 5)  20.7142  10.3571  
## 4                               (10, 3, 12, 5)  20.0132  10.0066  
## 5                            (0, 10, 3, 12, 5)  20.3738  10.1869  
## 6                         (0, 3, 5, 7, 10, 12)  18.9433  9.47167  
## 7                      (0, 1, 2, 3, 7, 10, 12)  12.0097  6.00487  
## 8                   (0, 1, 2, 3, 7, 8, 10, 12)  11.7297  5.86484  
## 9                (0, 1, 2, 3, 7, 8, 9, 10, 12)  10.0822  5.04111  
## 10           (0, 1, 4, 6, 7, 8, 9, 10, 11, 12)   9.3268   4.6634  
## 11        (0, 1, 3, 4, 6, 7, 8, 9, 10, 11, 12)  10.8018  5.40092  
## 12     (0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12)  11.0653  5.53265  
## 13  (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)  23.0919   11.546  
## No of features= 9
## [0, 1, 2, 3, 7, 8, 9, 10, 12]
## #################################################################################
## Features selected in forward fit
## Index([u'crimeRate', u'zone', u'indus', u'chasRiver', u'distances',
##        u'idxHighways', u'taxRate', u'teacherRatio', u'status'],
##       dtype='object')

The table above shows the average score, 10 fold CV errors, the features included at every step, std. deviation and std. error

SFFS provides the best fit with 10 predictors

1.3d Sequential Floating Backward Selection (SFBS) – Python code

The SFBS is an extension of the SBS. Here features that are excluded at any stage can be conditionally included if the resulting feature set gives a better fit.

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from sklearn.datasets import load_boston
from mlxtend.plotting import plot_sequential_feature_selection as plot_sfs
import matplotlib.pyplot as plt
from mlxtend.feature_selection import SequentialFeatureSelector as SFS
from sklearn.linear_model import LinearRegression


df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")
#Rename the columns
df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status","cost"]
X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status"]]
y=df['cost']
lr = LinearRegression()

sffs = SFS(lr, 
          k_features=(1,13), 
          forward=False, # Backward
          floating=True, # Floating
          scoring='neg_mean_squared_error',
          cv=5)

sffs = sffs.fit(X.as_matrix(), y.as_matrix())
a=sffs.get_metric_dict()
n=[]
o=[]
# Compute the mean cross validation score
for i in np.arange(1,13):
    n.append(-np.mean(a[i]['cv_scores']))  
    
m=np.arange(1,13)

fig4=plt.plot(m,n)
fig4=plt.title('SFBS: Mean CV Scores vs No of features')
fig4.figure.savefig('fig4.png', bbox_inches='tight')

print(pd.DataFrame.from_dict(sffs.get_metric_dict(confidence_interval=0.90)).T)

# Get the index of the minimum CV score
idx = np.argmin(n)
print "No of features=",idx
#Get the features indices for the best backward floating fit and convert to list
b=list(a[idx]['feature_idx'])
print(b)

print("#################################################################################")
# Index the column names. 
# Features from forward fit
print("Features selected in backward floating fit")
print(X.columns[b])
##    avg_score ci_bound                                          cv_scores  \
## 1   -42.6185  19.0465  [-23.5582499971, -41.8215743748, -73.993608929...   
## 2   -36.0651  16.3184  [-18.002498199, -40.1507894517, -56.5286659068...   
## 3   -34.1001    20.87  [-9.43012884381, -25.9584955394, -36.184188174...   
## 4    -33.463  12.4081  [-20.6415333292, -37.3247852146, -47.479302977...   
## 5   -32.3699  11.2725  [-20.8771078371, -34.9825657934, -45.813447203...   
## 6   -31.6742  11.2458  [-20.3082500364, -33.2288990522, -45.535507868...   
## 7   -30.7133  9.23881  [-19.4425181917, -31.1742902259, -40.531266671...   
## 8   -29.7432  9.84468  [-19.445277268, -30.0641187173, -40.2561247122...   
## 9   -29.0878  9.45027  [-19.3545569877, -30.094768669, -39.7506036377...   
## 10  -28.9225  9.39697  [-18.562171585, -29.968504938, -39.9586835965,...   
## 11  -29.4301  10.8831  [-18.3346152225, -30.3312847532, -45.065432793...   
## 12  -30.4589  11.1486  [-18.493389527, -35.0290639374, -45.1558231765...   
## 13  -37.1318  23.2657  [-12.4603005692, -26.0486211062, -33.074137979...   
## 
##                                    feature_idx  std_dev  std_err  
## 1                                        (12,)  18.9042  9.45212  
## 2                                     (10, 12)  16.1965  8.09826  
## 3                                  (10, 12, 5)  20.7142  10.3571  
## 4                               (4, 10, 7, 12)  12.3154  6.15772  
## 5                            (12, 10, 4, 1, 7)  11.1883  5.59417  
## 6                        (4, 7, 8, 10, 11, 12)  11.1618  5.58088  
## 7                      (1, 4, 7, 8, 9, 10, 12)  9.16981  4.58491  
## 8                  (1, 4, 7, 8, 9, 10, 11, 12)  9.77116  4.88558  
## 9               (0, 1, 4, 7, 8, 9, 10, 11, 12)  9.37969  4.68985  
## 10           (0, 1, 4, 6, 7, 8, 9, 10, 11, 12)   9.3268   4.6634  
## 11        (0, 1, 3, 4, 6, 7, 8, 9, 10, 11, 12)  10.8018  5.40092  
## 12     (0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12)  11.0653  5.53265  
## 13  (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)  23.0919   11.546  
## No of features= 9
## [0, 1, 4, 7, 8, 9, 10, 11, 12]
## #################################################################################
## Features selected in backward floating fit
## Index([u'crimeRate', u'zone', u'NO2', u'distances', u'idxHighways', u'taxRate',
##        u'teacherRatio', u'color', u'status'],
##       dtype='object')

The table above shows the average score, 10 fold CV errors, the features included at every step, std. deviation and std. error

SFBS indicates that 10 features are needed for the best fit

1.4 Ridge regression

In Linear Regression the Residual Sum of Squares (RSS) is given as

RSS = \sum_{i=1}^{n} (y_{i} - \beta_{0} - \sum_{j=1}^{p}\beta_jx_{ij})^{2}
Ridge regularization =\sum_{i=1}^{n} (y_{i} - \beta_{0} - \sum_{j=1}^{p}\beta_jx_{ij})^{2} + \lambda \sum_{j=1}^{p}\beta^{2}

where is the regularization or tuning parameter. Increasing increases the penalty on the coefficients thus shrinking them. However in Ridge Regression features that do not influence the target variable will shrink closer to zero but never become zero except for very large values of

Ridge regression in R requires the ‘glmnet’ package

1.4a Ridge Regression – R code

library(glmnet)
library(dplyr)
# Read the data
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL
#Rename the columns
names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age",
              "distances","highways","tax","teacherRatio","color","status","cost")
# Select specific columns
df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",
                            "distances","highways","tax","teacherRatio","color","status","cost")

# Set X and y as matrices
X=as.matrix(df1[,1:13])
y=df1$cost

# Fit a Ridge model
fitRidge <-glmnet(X,y,alpha=0)

#Plot the model where the coefficient shrinkage is plotted vs log lambda
plot(fitRidge,xvar="lambda",label=TRUE,main= "Ridge regression coefficient shrikage vs log lambda")

The plot below shows how the 13 coefficients for the 13 predictors vary when lambda is increased. The x-axis includes log (lambda). We can see that increasing lambda from 10^{2} to 10^{6} significantly shrinks the coefficients. We can draw a vertical line from the x-axis and read the values of the 13 coefficients. Some of them will be close to zero

# Compute the cross validation error
cvRidge=cv.glmnet(X,y,alpha=0)

#Plot the cross validation error
plot(cvRidge, main="Ridge regression Cross Validation Error (10 fold)")

This gives the 10 fold Cross Validation  Error with respect to log (lambda) As lambda increase the MSE increases

1.4a Ridge Regression – Python code

The coefficient shrinkage for Python can be plotted like R using Least Angle Regression model a.k.a. LARS package. This is included below

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split


df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")
#Rename the columns
df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status","cost"]
X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status"]]
y=df['cost']
from sklearn.preprocessing import MinMaxScaler
scaler = MinMaxScaler()

from sklearn.linear_model import Ridge
X_train, X_test, y_train, y_test = train_test_split(X, y,
                                                   random_state = 0)

# Scale the X_train and X_test
X_train_scaled = scaler.fit_transform(X_train)
X_test_scaled = scaler.transform(X_test)

# Fit a ridge regression with alpha=20
linridge = Ridge(alpha=20.0).fit(X_train_scaled, y_train)

# Print the training R squared
print('R-squared score (training): {:.3f}'
     .format(linridge.score(X_train_scaled, y_train)))
# Print the test Rsquared
print('R-squared score (test): {:.3f}'
     .format(linridge.score(X_test_scaled, y_test)))
print('Number of non-zero features: {}'
     .format(np.sum(linridge.coef_ != 0)))

trainingRsquared=[]
testRsquared=[]
# Plot the effect of alpha on the test Rsquared
print('Ridge regression: effect of alpha regularization parameter\n')
# Choose a list of alpha values
for this_alpha in [0.001,.01,.1,0, 1, 10, 20, 50, 100, 1000]:
    linridge = Ridge(alpha = this_alpha).fit(X_train_scaled, y_train)
    # Compute training rsquared
    r2_train = linridge.score(X_train_scaled, y_train)
    # Compute test rsqaured
    r2_test = linridge.score(X_test_scaled, y_test)
    num_coeff_bigger = np.sum(abs(linridge.coef_) > 1.0)
    trainingRsquared.append(r2_train)
    testRsquared.append(r2_test)
    
# Create a dataframe
alpha=[0.001,.01,.1,0, 1, 10, 20, 50, 100, 1000]    
trainingRsquared=pd.DataFrame(trainingRsquared,index=alpha)
testRsquared=pd.DataFrame(testRsquared,index=alpha)

# Plot training and test R squared as a function of alpha
df3=pd.concat([trainingRsquared,testRsquared],axis=1)
df3.columns=['trainingRsquared','testRsquared']

fig5=df3.plot()
fig5=plt.title('Ridge training and test squared error vs Alpha')
fig5.figure.savefig('fig5.png', bbox_inches='tight')

# Plot the coefficient shrinage using the LARS package

from sklearn import linear_model
# #############################################################################
# Compute paths

n_alphas = 200
alphas = np.logspace(0, 8, n_alphas)

coefs = []
for a in alphas:
    ridge = linear_model.Ridge(alpha=a, fit_intercept=False)
    ridge.fit(X_train_scaled, y_train)
    coefs.append(ridge.coef_)

# #############################################################################
# Display results

ax = plt.gca()

fig6=ax.plot(alphas, coefs)
fig6=ax.set_xscale('log')
fig6=ax.set_xlim(ax.get_xlim()[::-1])  # reverse axis
fig6=plt.xlabel('alpha')
fig6=plt.ylabel('weights')
fig6=plt.title('Ridge coefficients as a function of the regularization')
fig6=plt.axis('tight')
plt.savefig('fig6.png', bbox_inches='tight')
## R-squared score (training): 0.620
## R-squared score (test): 0.438
## Number of non-zero features: 13
## Ridge regression: effect of alpha regularization parameter

The plot below shows the training and test error when increasing the tuning or regularization parameter ‘alpha’

For Python the coefficient shrinkage with LARS must be viewed from right to left, where you have increasing alpha. As alpha increases the coefficients shrink to 0.

1.5 Lasso regularization

The Lasso is another form of regularization, also known as L1 regularization. Unlike the Ridge Regression where the coefficients of features which do not influence the target tend to zero, in the lasso regualrization the coefficients become 0. The general form of Lasso is as follows

\sum_{i=1}^{n} (y_{i} - \beta_{0} - \sum_{j=1}^{p}\beta_jx_{ij})^{2} + \lambda \sum_{j=1}^{p}|\beta|

1.5a Lasso regularization – R code

library(glmnet)
library(dplyr)
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - SL
names(df) <-c("no","crimeRate","zone","indus","charles","nox","rooms","age",
              "distances","highways","tax","teacherRatio","color","status","cost")
df1 <- df %>% dplyr::select("crimeRate","zone","indus","charles","nox","rooms","age",
                            "distances","highways","tax","teacherRatio","color","status","cost")

# Set X and y as matrices
X=as.matrix(df1[,1:13])
y=df1$cost

# Fit the lasso model
fitLasso <- glmnet(X,y)
# Plot the coefficient shrinkage as a function of log(lambda)
plot(fitLasso,xvar="lambda",label=TRUE,main="Lasso regularization - Coefficient shrinkage vs log lambda")

The plot below shows that in L1 regularization the coefficients actually become zero with increasing lambda

# Compute the cross validation error (10 fold)
cvLasso=cv.glmnet(X,y,alpha=0)
# Plot the cross validation error
plot(cvLasso)

This gives the MSE for the lasso model

1.5 b Lasso regularization – Python code

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import Lasso
from sklearn.preprocessing import MinMaxScaler
from sklearn import linear_model

scaler = MinMaxScaler()
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")
#Rename the columns
df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status","cost"]
X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status"]]
y=df['cost']
X_train, X_test, y_train, y_test = train_test_split(X, y,
                                                   random_state = 0)

X_train_scaled = scaler.fit_transform(X_train)
X_test_scaled = scaler.transform(X_test)

linlasso = Lasso(alpha=0.1, max_iter = 10).fit(X_train_scaled, y_train)

print('Non-zero features: {}'
     .format(np.sum(linlasso.coef_ != 0)))
print('R-squared score (training): {:.3f}'
     .format(linlasso.score(X_train_scaled, y_train)))
print('R-squared score (test): {:.3f}\n'
     .format(linlasso.score(X_test_scaled, y_test)))
print('Features with non-zero weight (sorted by absolute magnitude):')

for e in sorted (list(zip(list(X), linlasso.coef_)),
                key = lambda e: -abs(e[1])):
    if e[1] != 0:
        print('\t{}, {:.3f}'.format(e[0], e[1]))
        

print('Lasso regression: effect of alpha regularization\n\
parameter on number of features kept in final model\n')

trainingRsquared=[]
testRsquared=[]
#for alpha in [0.01,0.05,0.1, 1, 2, 3, 5, 10, 20, 50]:
for alpha in [0.01,0.07,0.05, 0.1, 1,2, 3, 5, 10]:
    linlasso = Lasso(alpha, max_iter = 10000).fit(X_train_scaled, y_train)
    r2_train = linlasso.score(X_train_scaled, y_train)
    r2_test = linlasso.score(X_test_scaled, y_test)
    trainingRsquared.append(r2_train)
    testRsquared.append(r2_test)
    
alpha=[0.01,0.07,0.05, 0.1, 1,2, 3, 5, 10]    
#alpha=[0.01,0.05,0.1, 1, 2, 3, 5, 10, 20, 50]
trainingRsquared=pd.DataFrame(trainingRsquared,index=alpha)
testRsquared=pd.DataFrame(testRsquared,index=alpha)

df3=pd.concat([trainingRsquared,testRsquared],axis=1)
df3.columns=['trainingRsquared','testRsquared']

fig7=df3.plot()
fig7=plt.title('LASSO training and test squared error vs Alpha')
fig7.figure.savefig('fig7.png', bbox_inches='tight')



## Non-zero features: 7
## R-squared score (training): 0.726
## R-squared score (test): 0.561
## 
## Features with non-zero weight (sorted by absolute magnitude):
##  status, -18.361
##  rooms, 18.232
##  teacherRatio, -8.628
##  taxRate, -2.045
##  color, 1.888
##  chasRiver, 1.670
##  distances, -0.529
## Lasso regression: effect of alpha regularization
## parameter on number of features kept in final model
## 
## Computing regularization path using the LARS ...
## .C:\Users\Ganesh\ANACON~1\lib\site-packages\sklearn\linear_model\coordinate_descent.py:484: ConvergenceWarning: Objective did not converge. You might want to increase the number of iterations. Fitting data with very small alpha may cause precision problems.
##   ConvergenceWarning)

The plot below gives the training and test R squared error

1.5c Lasso coefficient shrinkage – Python code

To plot the coefficient shrinkage for Lasso the Least Angle Regression model a.k.a. LARS package. This is shown below

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import Lasso
from sklearn.preprocessing import MinMaxScaler
from sklearn import linear_model
scaler = MinMaxScaler()
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")
#Rename the columns
df.columns=["no","crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status","cost"]
X=df[["crimeRate","zone","indus","chasRiver","NO2","rooms","age",
              "distances","idxHighways","taxRate","teacherRatio","color","status"]]
y=df['cost']
X_train, X_test, y_train, y_test = train_test_split(X, y,
                                                   random_state = 0)

X_train_scaled = scaler.fit_transform(X_train)
X_test_scaled = scaler.transform(X_test)


print("Computing regularization path using the LARS ...")
alphas, _, coefs = linear_model.lars_path(X_train_scaled, y_train, method='lasso', verbose=True)

xx = np.sum(np.abs(coefs.T), axis=1)
xx /= xx[-1]

fig8=plt.plot(xx, coefs.T)

ymin, ymax = plt.ylim()
fig8=plt.vlines(xx, ymin, ymax, linestyle='dashed')
fig8=plt.xlabel('|coef| / max|coef|')
fig8=plt.ylabel('Coefficients')
fig8=plt.title('LASSO Path - Coefficient Shrinkage vs L1')
fig8=plt.axis('tight')
plt.savefig('fig8.png', bbox_inches='tight')
This plot show the coefficient shrinkage for lasso.
This 3rd part of the series covers the main ‘feature selection’ methods. I hope these posts serve as a quick and useful reference to ML code both for R and Python!
Stay tuned for further updates to this series!
Watch this space!

 

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6. R vs Python: Different similarities and similar differences

To see all posts see Index of posts

Practical Machine Learning with R and Python – Part 2


In this 2nd part of the series “Practical Machine Learning with R and Python – Part 2”, I continue where I left off in my first post Practical Machine Learning with R and Python – Part 2. In this post I cover the some classification algorithmns and cross validation. Specifically I touch
-Logistic Regression
-K Nearest Neighbors (KNN) classification
-Leave out one Cross Validation (LOOCV)
-K Fold Cross Validation
in both R and Python.

As in my initial post the algorithms are based on the following courses.

You can download this R Markdown file along with the data from Github. I hope these posts can be used as a quick reference in R and Python and Machine Learning.I have tried to include the coolest part of either course in this post.

Check out my compact and minimal book  “Practical Machine Learning with R and Python:Second edition- Machine Learning in stereo”  available in Amazon in paperback($10.99) and kindle($7.99) versions. My book includes implementations of key ML algorithms and associated measures and metrics. The book is ideal for anybody who is familiar with the concepts and would like a quick reference to the different ML algorithms that can be applied to problems and how to select the best model. Pick your copy today!!

The following classification problem is based on Logistic Regression. The data is an included data set in Scikit-Learn, which I have saved as csv and use it also for R. The fit of a classification Machine Learning Model depends on how correctly classifies the data. There are several measures of testing a model’s classification performance. They are

Accuracy = TP + TN / (TP + TN + FP + FN) – Fraction of all classes correctly classified
Precision = TP / (TP + FP) – Fraction of correctly classified positives among those classified as positive
Recall = TP / (TP + FN) Also known as sensitivity, or True Positive Rate (True positive) – Fraction of correctly classified as positive among all positives in the data
F1 = 2 * Precision * Recall / (Precision + Recall)

1a. Logistic Regression – R code

The caret and e1071 package is required for using the confusionMatrix call

source("RFunctions.R")
library(dplyr)
library(caret)
library(e1071)
# Read the data (from sklearn)
cancer <- read.csv("cancer.csv")
# Rename the target variable
names(cancer) <- c(seq(1,30),"output")
# Split as training and test sets
train_idx <- trainTestSplit(cancer,trainPercent=75,seed=5)
train <- cancer[train_idx, ]
test <- cancer[-train_idx, ]

# Fit a generalized linear logistic model, 
fit=glm(output~.,family=binomial,data=train,control = list(maxit = 50))
# Predict the output from the model
a=predict(fit,newdata=train,type="response")
# Set response >0.5 as 1 and <=0.5 as 0
b=ifelse(a>0.5,1,0)
# Compute the confusion matrix for training data
confusionMatrix(b,train$output)
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction   0   1
##          0 154   0
##          1   0 272
##                                      
##                Accuracy : 1          
##                  95% CI : (0.9914, 1)
##     No Information Rate : 0.6385     
##     P-Value [Acc > NIR] : < 2.2e-16  
##                                      
##                   Kappa : 1          
##  Mcnemar's Test P-Value : NA         
##                                      
##             Sensitivity : 1.0000     
##             Specificity : 1.0000     
##          Pos Pred Value : 1.0000     
##          Neg Pred Value : 1.0000     
##              Prevalence : 0.3615     
##          Detection Rate : 0.3615     
##    Detection Prevalence : 0.3615     
##       Balanced Accuracy : 1.0000     
##                                      
##        'Positive' Class : 0          
## 
m=predict(fit,newdata=test,type="response")
n=ifelse(m>0.5,1,0)
# Compute the confusion matrix for test output
confusionMatrix(n,test$output)
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  0  1
##          0 52  4
##          1  5 81
##                                           
##                Accuracy : 0.9366          
##                  95% CI : (0.8831, 0.9706)
##     No Information Rate : 0.5986          
##     P-Value [Acc > NIR] : <2e-16          
##                                           
##                   Kappa : 0.8677          
##  Mcnemar's Test P-Value : 1               
##                                           
##             Sensitivity : 0.9123          
##             Specificity : 0.9529          
##          Pos Pred Value : 0.9286          
##          Neg Pred Value : 0.9419          
##              Prevalence : 0.4014          
##          Detection Rate : 0.3662          
##    Detection Prevalence : 0.3944          
##       Balanced Accuracy : 0.9326          
##                                           
##        'Positive' Class : 0               
## 

1b. Logistic Regression – Python code

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LogisticRegression
os.chdir("C:\\Users\\Ganesh\\RandPython")
from sklearn.datasets import make_classification, make_blobs

from sklearn.metrics import confusion_matrix
from matplotlib.colors import ListedColormap
from sklearn.datasets import load_breast_cancer
# Load the cancer data
(X_cancer, y_cancer) = load_breast_cancer(return_X_y = True)
X_train, X_test, y_train, y_test = train_test_split(X_cancer, y_cancer,
                                                   random_state = 0)
# Call the Logisitic Regression function
clf = LogisticRegression().fit(X_train, y_train)
fig, subaxes = plt.subplots(1, 1, figsize=(7, 5))
# Fit a model
clf = LogisticRegression().fit(X_train, y_train)

# Compute and print the Accuray scores
print('Accuracy of Logistic regression classifier on training set: {:.2f}'
     .format(clf.score(X_train, y_train)))
print('Accuracy of Logistic regression classifier on test set: {:.2f}'
     .format(clf.score(X_test, y_test)))
y_predicted=clf.predict(X_test)
# Compute and print confusion matrix
confusion = confusion_matrix(y_test, y_predicted)
from sklearn.metrics import accuracy_score, precision_score, recall_score, f1_score
print('Accuracy: {:.2f}'.format(accuracy_score(y_test, y_predicted)))
print('Precision: {:.2f}'.format(precision_score(y_test, y_predicted)))
print('Recall: {:.2f}'.format(recall_score(y_test, y_predicted)))
print('F1: {:.2f}'.format(f1_score(y_test, y_predicted)))
## Accuracy of Logistic regression classifier on training set: 0.96
## Accuracy of Logistic regression classifier on test set: 0.96
## Accuracy: 0.96
## Precision: 0.99
## Recall: 0.94
## F1: 0.97

2. Dummy variables

The following R and Python code show how dummy variables are handled in R and Python. Dummy variables are categorival variables which have to be converted into appropriate values before using them in Machine Learning Model For e.g. if we had currency as ‘dollar’, ‘rupee’ and ‘yen’ then the dummy variable will convert this as
dollar 0 0 0
rupee 0 0 1
yen 0 1 0

2a. Logistic Regression with dummy variables- R code

# Load the dummies library
library(dummies) 
df <- read.csv("adult1.csv",stringsAsFactors = FALSE,na.strings = c(""," "," ?"))

# Remove rows which have NA
df1 <- df[complete.cases(df),]
dim(df1)
## [1] 30161    16
# Select specific columns
adult <- df1 %>% dplyr::select(age,occupation,education,educationNum,capitalGain,
                               capital.loss,hours.per.week,native.country,salary)
# Set the dummy data with appropriate values
adult1 <- dummy.data.frame(adult, sep = ".")

#Split as training and test
train_idx <- trainTestSplit(adult1,trainPercent=75,seed=1111)
train <- adult1[train_idx, ]
test <- adult1[-train_idx, ]

# Fit a binomial logistic regression
fit=glm(salary~.,family=binomial,data=train)
# Predict response
a=predict(fit,newdata=train,type="response")
# If response >0.5 then it is a 1 and 0 otherwise
b=ifelse(a>0.5,1,0)
confusionMatrix(b,train$salary)
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction     0     1
##          0 16065  3145
##          1   968  2442
##                                           
##                Accuracy : 0.8182          
##                  95% CI : (0.8131, 0.8232)
##     No Information Rate : 0.753           
##     P-Value [Acc > NIR] : < 2.2e-16       
##                                           
##                   Kappa : 0.4375          
##  Mcnemar's Test P-Value : < 2.2e-16       
##                                           
##             Sensitivity : 0.9432          
##             Specificity : 0.4371          
##          Pos Pred Value : 0.8363          
##          Neg Pred Value : 0.7161          
##              Prevalence : 0.7530          
##          Detection Rate : 0.7102          
##    Detection Prevalence : 0.8492          
##       Balanced Accuracy : 0.6901          
##                                           
##        'Positive' Class : 0               
## 
# Compute and display confusion matrix
m=predict(fit,newdata=test,type="response")
## Warning in predict.lm(object, newdata, se.fit, scale = 1, type =
## ifelse(type == : prediction from a rank-deficient fit may be misleading
n=ifelse(m>0.5,1,0)
confusionMatrix(n,test$salary)
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction    0    1
##          0 5263 1099
##          1  357  822
##                                           
##                Accuracy : 0.8069          
##                  95% CI : (0.7978, 0.8158)
##     No Information Rate : 0.7453          
##     P-Value [Acc > NIR] : < 2.2e-16       
##                                           
##                   Kappa : 0.4174          
##  Mcnemar's Test P-Value : < 2.2e-16       
##                                           
##             Sensitivity : 0.9365          
##             Specificity : 0.4279          
##          Pos Pred Value : 0.8273          
##          Neg Pred Value : 0.6972          
##              Prevalence : 0.7453          
##          Detection Rate : 0.6979          
##    Detection Prevalence : 0.8437          
##       Balanced Accuracy : 0.6822          
##                                           
##        'Positive' Class : 0               
## 

2b. Logistic Regression with dummy variables- Python code

Pandas has a get_dummies function for handling dummies

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LogisticRegression
from sklearn.metrics import confusion_matrix
from sklearn.metrics import accuracy_score, precision_score, recall_score, f1_score
# Read data
df =pd.read_csv("adult1.csv",encoding="ISO-8859-1",na_values=[""," "," ?"])
# Drop rows with NA
df1=df.dropna()
print(df1.shape)
# Select specific columns
adult = df1[['age','occupation','education','educationNum','capitalGain','capital-loss', 
             'hours-per-week','native-country','salary']]

X=adult[['age','occupation','education','educationNum','capitalGain','capital-loss', 
             'hours-per-week','native-country']]
# Set approporiate values for dummy variables
X_adult=pd.get_dummies(X,columns=['occupation','education','native-country'])
y=adult['salary']

X_adult_train, X_adult_test, y_train, y_test = train_test_split(X_adult, y,
                                                   random_state = 0)
clf = LogisticRegression().fit(X_adult_train, y_train)

# Compute and display Accuracy and Confusion matrix
print('Accuracy of Logistic regression classifier on training set: {:.2f}'
     .format(clf.score(X_adult_train, y_train)))
print('Accuracy of Logistic regression classifier on test set: {:.2f}'
     .format(clf.score(X_adult_test, y_test)))
y_predicted=clf.predict(X_adult_test)
confusion = confusion_matrix(y_test, y_predicted)
print('Accuracy: {:.2f}'.format(accuracy_score(y_test, y_predicted)))
print('Precision: {:.2f}'.format(precision_score(y_test, y_predicted)))
print('Recall: {:.2f}'.format(recall_score(y_test, y_predicted)))
print('F1: {:.2f}'.format(f1_score(y_test, y_predicted)))
## (30161, 16)
## Accuracy of Logistic regression classifier on training set: 0.82
## Accuracy of Logistic regression classifier on test set: 0.81
## Accuracy: 0.81
## Precision: 0.68
## Recall: 0.41
## F1: 0.51

3a – K Nearest Neighbors Classification – R code

The Adult data set is taken from UCI Machine Learning Repository

source("RFunctions.R")
df <- read.csv("adult1.csv",stringsAsFactors = FALSE,na.strings = c(""," "," ?"))
# Remove rows which have NA
df1 <- df[complete.cases(df),]
dim(df1)
## [1] 30161    16
# Select specific columns
adult <- df1 %>% dplyr::select(age,occupation,education,educationNum,capitalGain,
                               capital.loss,hours.per.week,native.country,salary)
# Set dummy variables
adult1 <- dummy.data.frame(adult, sep = ".")

#Split train and test as required by KNN classsification model
train_idx <- trainTestSplit(adult1,trainPercent=75,seed=1111)
train <- adult1[train_idx, ]
test <- adult1[-train_idx, ]
train.X <- train[,1:76]
train.y <- train[,77]
test.X <- test[,1:76]
test.y <- test[,77]

# Fit a model for 1,3,5,10 and 15 neighbors
cMat <- NULL
neighbors <-c(1,3,5,10,15)
for(i in seq_along(neighbors)){
    fit =knn(train.X,test.X,train.y,k=i)
    table(fit,test.y)
    a<-confusionMatrix(fit,test.y)
    cMat[i] <- a$overall[1]
    print(a$overall[1])
}
##  Accuracy 
## 0.7835831 
##  Accuracy 
## 0.8162047 
##  Accuracy 
## 0.8089113 
##  Accuracy 
## 0.8209787 
##  Accuracy 
## 0.8184591
#Plot the Accuracy for each of the KNN models
df <- data.frame(neighbors,Accuracy=cMat)
ggplot(df,aes(x=neighbors,y=Accuracy)) + geom_point() +geom_line(color="blue") +
    xlab("Number of neighbors") + ylab("Accuracy") +
    ggtitle("KNN regression - Accuracy vs Number of Neighors (Unnormalized)")

3b – K Nearest Neighbors Classification – Python code

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.metrics import confusion_matrix
from sklearn.metrics import accuracy_score, precision_score, recall_score, f1_score
from sklearn.neighbors import KNeighborsClassifier
from sklearn.preprocessing import MinMaxScaler

# Read data
df =pd.read_csv("adult1.csv",encoding="ISO-8859-1",na_values=[""," "," ?"])
df1=df.dropna()
print(df1.shape)
# Select specific columns
adult = df1[['age','occupation','education','educationNum','capitalGain','capital-loss', 
             'hours-per-week','native-country','salary']]

X=adult[['age','occupation','education','educationNum','capitalGain','capital-loss', 
             'hours-per-week','native-country']]
             
#Set values for dummy variables
X_adult=pd.get_dummies(X,columns=['occupation','education','native-country'])
y=adult['salary']

X_adult_train, X_adult_test, y_train, y_test = train_test_split(X_adult, y,
                                                   random_state = 0)
                                                   
# KNN classification in Python requires the data to be scaled. 
# Scale the data
scaler = MinMaxScaler()
X_train_scaled = scaler.fit_transform(X_adult_train)
# Apply scaling to test set also
X_test_scaled = scaler.transform(X_adult_test)
# Compute the KNN model for 1,3,5,10 & 15 neighbors
accuracy=[]
neighbors=[1,3,5,10,15]
for i in neighbors:
    knn = KNeighborsClassifier(n_neighbors = i)
    knn.fit(X_train_scaled, y_train)
    accuracy.append(knn.score(X_test_scaled, y_test))
    print('Accuracy test score: {:.3f}'
        .format(knn.score(X_test_scaled, y_test)))

# Plot the models with the Accuracy attained for each of these models    
fig1=plt.plot(neighbors,accuracy)
fig1=plt.title("KNN regression - Accuracy vs Number of neighbors")
fig1=plt.xlabel("Neighbors")
fig1=plt.ylabel("Accuracy")
fig1.figure.savefig('foo1.png', bbox_inches='tight')
## (30161, 16)
## Accuracy test score: 0.749
## Accuracy test score: 0.779
## Accuracy test score: 0.793
## Accuracy test score: 0.804
## Accuracy test score: 0.803

Output image:

4 MPG vs Horsepower

The following scatter plot shows the non-linear relation between mpg and horsepower. This will be used as the data input for computing K Fold Cross Validation Error

4a MPG vs Horsepower scatter plot – R Code

df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI
df1 <- as.data.frame(sapply(df,as.numeric))
df2 <- df1 %>% dplyr::select(cylinder,displacement, horsepower,weight, acceleration, year,mpg)
df3 <- df2[complete.cases(df2),]
ggplot(df3,aes(x=horsepower,y=mpg)) + geom_point() + xlab("Horsepower") + 
    ylab("Miles Per gallon") + ggtitle("Miles per Gallon vs Hosrsepower")

4b MPG vs Horsepower scatter plot – Python Code

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
autoDF =pd.read_csv("auto_mpg.csv",encoding="ISO-8859-1")
autoDF.shape
autoDF.columns
autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']]
autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce')
autoDF3=autoDF2.dropna()
autoDF3.shape
#X=autoDF3[['cylinder','displacement','horsepower','weight']]
X=autoDF3[['horsepower']]
y=autoDF3['mpg']

fig11=plt.scatter(X,y)
fig11=plt.title("KNN regression - Accuracy vs Number of neighbors")
fig11=plt.xlabel("Neighbors")
fig11=plt.ylabel("Accuracy")
fig11.figure.savefig('foo11.png', bbox_inches='tight')

5 K Fold Cross Validation

K Fold Cross Validation is a technique in which the data set is divided into K Folds or K partitions. The Machine Learning model is trained on K-1 folds and tested on the Kth fold i.e.
we will have K-1 folds for training data and 1 for testing the ML model. Since we can partition this as C_{1}^{K} or K choose 1, there will be K such partitions. The K Fold Cross
Validation estimates the average validation error that we can expect on a new unseen test data.

The formula for K Fold Cross validation is as follows

MSE_{K} = \frac{\sum (y-yhat)^{2}}{n_{K}}
and
n_{K} = \frac{N}{K}
and
CV_{K} = \sum_{K=1}^{K} (\frac{n_{K}}{N}) MSE_{K}

where n_{K} is the number of elements in partition ‘K’ and N is the total number of elements
CV_{K} =\sum_{K=1}^{K} MSE_{K}

CV_{K} =\frac{\sum_{K=1}^{K} MSE_{K}}{K}
Leave Out one Cross Validation (LOOCV) is a special case of K Fold Cross Validation where N-1 data points are used to train the model and 1 data point is used to test the model. There are N such paritions of N-1 & 1 that are possible. The mean error is measured The Cross Valifation Error for LOOCV is

CV_{N} = \frac{1}{n} *\frac{\sum_{1}^{n}(y-yhat)^{2}}{1-h_{i}}
where h_{i} is the diagonal hat matrix

see [Statistical Learning]

The above formula is also included in this blog post

It took me a day and a half to implement the K Fold Cross Validation formula. I think it is correct. In any case do let me know if you think it is off

5a. Leave out one cross validation (LOOCV) – R Code

R uses the package ‘boot’ for performing Cross Validation error computation

library(boot)
library(reshape2)
# Read data
df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI
df1 <- as.data.frame(sapply(df,as.numeric))
# Select complete cases
df2 <- df1 %>% dplyr::select(cylinder,displacement, horsepower,weight, acceleration, year,mpg)
df3 <- df2[complete.cases(df2),]
set.seed(17)
cv.error=rep(0,10)
# For polynomials 1,2,3... 10 fit a LOOCV model
for (i in 1:10){
    glm.fit=glm(mpg~poly(horsepower,i),data=df3)
    cv.error[i]=cv.glm(df3,glm.fit)$delta[1]
    
}
cv.error
##  [1] 24.23151 19.24821 19.33498 19.42443 19.03321 18.97864 18.83305
##  [8] 18.96115 19.06863 19.49093
# Create and display a plot
folds <- seq(1,10)
df <- data.frame(folds,cvError=cv.error)
ggplot(df,aes(x=folds,y=cvError)) + geom_point() +geom_line(color="blue") +
    xlab("Degree of Polynomial") + ylab("Cross Validation Error") +
    ggtitle("Leave one out Cross Validation - Cross Validation Error vs Degree of Polynomial")

5b. Leave out one cross validation (LOOCV) – Python Code

In Python there is no available function to compute Cross Validation error and we have to compute the above formula. I have done this after several hours. I think it is now in reasonable shape. Do let me know if you think otherwise. For LOOCV I use the K Fold Cross Validation with K=N

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
from sklearn.cross_validation import train_test_split, KFold
from sklearn.preprocessing import PolynomialFeatures
from sklearn.metrics import mean_squared_error
# Read data
autoDF =pd.read_csv("auto_mpg.csv",encoding="ISO-8859-1")
autoDF.shape
autoDF.columns
autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']]
autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce')
# Remove rows with NAs
autoDF3=autoDF2.dropna()
autoDF3.shape
X=autoDF3[['horsepower']]
y=autoDF3['mpg']

# For polynomial degree 1,2,3... 10
def computeCVError(X,y,folds):
    deg=[]
    mse=[]
    degree1=[1,2,3,4,5,6,7,8,9,10]
    
    nK=len(X)/float(folds)
    xval_err=0
    # For degree 'j'
    for j in degree1: 
        # Split as 'folds'
        kf = KFold(len(X),n_folds=folds)
        for train_index, test_index in kf:
            # Create the appropriate train and test partitions from the fold index
            X_train, X_test = X.iloc[train_index], X.iloc[test_index]
            y_train, y_test = y.iloc[train_index], y.iloc[test_index]  

            # For the polynomial degree 'j'
            poly = PolynomialFeatures(degree=j)        
            # Transform the X_train and X_test
            X_train_poly = poly.fit_transform(X_train)
            X_test_poly = poly.fit_transform(X_test)
            # Fit a model on the transformed data
            linreg = LinearRegression().fit(X_train_poly, y_train)
            # Compute yhat or ypred
            y_pred = linreg.predict(X_test_poly)   
            # Compute MSE * n_K/N
            test_mse = mean_squared_error(y_test, y_pred)*float(len(X_train))/float(len(X))     
            # Add the test_mse for this partition of the data
            mse.append(test_mse)
        # Compute the mean of all folds for degree 'j'   
        deg.append(np.mean(mse))
        
    return(deg)


df=pd.DataFrame()
print(len(X))
# Call the function once. For LOOCV K=N. hence len(X) is passed as number of folds
cvError=computeCVError(X,y,len(X))

# Create and plot LOOCV
df=pd.DataFrame(cvError)
fig3=df.plot()
fig3=plt.title("Leave one out Cross Validation - Cross Validation Error vs Degree of Polynomial")
fig3=plt.xlabel("Degree of Polynomial")
fig3=plt.ylabel("Cross validation Error")
fig3.figure.savefig('foo3.png', bbox_inches='tight')

 

6a K Fold Cross Validation – R code

Here K Fold Cross Validation is done for 4, 5 and 10 folds using the R package boot and the glm package

library(boot)
library(reshape2)
set.seed(17)
#Read data
df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI
df1 <- as.data.frame(sapply(df,as.numeric))
df2 <- df1 %>% dplyr::select(cylinder,displacement, horsepower,weight, acceleration, year,mpg)
df3 <- df2[complete.cases(df2),]
a=matrix(rep(0,30),nrow=3,ncol=10)
set.seed(17)
# Set the folds as 4,5 and 10
folds<-c(4,5,10)
for(i in seq_along(folds)){
    cv.error.10=rep(0,10)
    for (j in 1:10){
        # Fit a generalized linear model
        glm.fit=glm(mpg~poly(horsepower,j),data=df3)
        # Compute K Fold Validation error
        a[i,j]=cv.glm(df3,glm.fit,K=folds[i])$delta[1]
        
    }
    
}

# Create and display the K Fold Cross Validation Error
b <- t(a)
df <- data.frame(b)
df1 <- cbind(seq(1,10),df)
names(df1) <- c("PolynomialDegree","4-fold","5-fold","10-fold")

df2 <- melt(df1,id="PolynomialDegree")
ggplot(df2) + geom_line(aes(x=PolynomialDegree, y=value, colour=variable),size=2) +
    xlab("Degree of Polynomial") + ylab("Cross Validation Error") +
    ggtitle("K Fold Cross Validation - Cross Validation Error vs Degree of Polynomial")

6b. K Fold Cross Validation – Python code

The implementation of K-Fold Cross Validation Error has to be implemented and I have done this below. There is a small discrepancy in the shapes of the curves with the R plot above. Not sure why!

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
from sklearn.cross_validation import train_test_split, KFold
from sklearn.preprocessing import PolynomialFeatures
from sklearn.metrics import mean_squared_error
# Read data
autoDF =pd.read_csv("auto_mpg.csv",encoding="ISO-8859-1")
autoDF.shape
autoDF.columns
autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']]
autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce')
# Drop NA rows
autoDF3=autoDF2.dropna()
autoDF3.shape
#X=autoDF3[['cylinder','displacement','horsepower','weight']]
X=autoDF3[['horsepower']]
y=autoDF3['mpg']

# Create Cross Validation function
def computeCVError(X,y,folds):
    deg=[]
    mse=[]
    # For degree 1,2,3,..10
    degree1=[1,2,3,4,5,6,7,8,9,10]
    
    nK=len(X)/float(folds)
    xval_err=0
    for j in degree1: 
        # Split the data into 'folds'
        kf = KFold(len(X),n_folds=folds)
        for train_index, test_index in kf:
            # Partition the data acccording the fold indices generated
            X_train, X_test = X.iloc[train_index], X.iloc[test_index]
            y_train, y_test = y.iloc[train_index], y.iloc[test_index]  

            # Scale the X_train and X_test as per the polynomial degree 'j'
            poly = PolynomialFeatures(degree=j)             
            X_train_poly = poly.fit_transform(X_train)
            X_test_poly = poly.fit_transform(X_test)
            # Fit a polynomial regression
            linreg = LinearRegression().fit(X_train_poly, y_train)
            # Compute yhat or ypred
            y_pred = linreg.predict(X_test_poly)  
            # Compute MSE *(nK/N)
            test_mse = mean_squared_error(y_test, y_pred)*float(len(X_train))/float(len(X))  
            # Append to list for different folds
            mse.append(test_mse)
        # Compute the mean for poylnomial 'j' 
        deg.append(np.mean(mse))
        
    return(deg)

# Create and display a plot of K -Folds
df=pd.DataFrame()
for folds in [4,5,10]:
    cvError=computeCVError(X,y,folds)
    #print(cvError)
    df1=pd.DataFrame(cvError)
    df=pd.concat([df,df1],axis=1)
    #print(cvError)
    
df.columns=['4-fold','5-fold','10-fold']
df=df.reindex([1,2,3,4,5,6,7,8,9,10])
df
fig2=df.plot()
fig2=plt.title("K Fold Cross Validation - Cross Validation Error vs Degree of Polynomial")
fig2=plt.xlabel("Degree of Polynomial")
fig2=plt.ylabel("Cross validation Error")
fig2.figure.savefig('foo2.png', bbox_inches='tight')

output

This concludes this 2nd part of this series. I will look into model tuning and model selection in R and Python in the coming parts. Comments, suggestions and corrections are welcome!
To be continued….
Watch this space!

Also see

  1. Design Principles of Scalable, Distributed Systems
  2. Re-introducing cricketr! : An R package to analyze performances of cricketers
  3. Spicing up a IBM Bluemix cloud app with MongoDB and NodeExpress
  4. Using Linear Programming (LP) for optimizing bowling change or batting lineup in T20 cricket
  5. Simulating an Edge Shape in Android

To see all posts see Index of posts

Practical Machine Learning with R and Python – Part 1


Introduction

This is the 1st part of a series of posts I intend to write on some common Machine Learning Algorithms in R and Python. In this first part I cover the following Machine Learning Algorithms

  • Univariate Regression
  • Multivariate Regression
  • Polynomial Regression
  • K Nearest Neighbors Regression

The code includes the implementation in both R and Python. This series of posts are based on the following 2 MOOC courses I did at Stanford Online and at Coursera

  1. Statistical Learning, Prof Trevor Hastie & Prof Robert Tibesherani, Online Stanford
  2. Applied Machine Learning in Python Prof Kevyn-Collin Thomson, University Of Michigan, Coursera

I have used the data sets from UCI Machine Learning repository(Communities and Crime and Auto MPG). I also use the Boston data set from MASS package

Check out my compact and minimal book  “Practical Machine Learning with R and Python:Second edition- Machine Learning in stereo”  available in Amazon in paperback($10.99) and kindle($7.99) versions. My book includes implementations of key ML algorithms and associated measures and metrics. The book is ideal for anybody who is familiar with the concepts and would like a quick reference to the different ML algorithms that can be applied to problems and how to select the best model. Pick your copy today!!

 

While coding in R and Python I found that there were some aspects that were more convenient in one language and some in the other. For example, plotting the fit in R is straightforward in R, while computing the R squared, splitting as Train & Test sets etc. are already available in Python. In any case, these minor inconveniences can be easily be implemented in either language.

R squared computation in R is computed as follows
RSS=\sum (y-yhat)^{2}
TSS= \sum(y-mean(y))^{2}
Rsquared- 1-\frac{RSS}{TSS}

Note: You can download this R Markdown file and the associated data sets from Github at MachineLearning-RandPython
Note 1: This post was created as an R Markdown file in RStudio which has a cool feature of including R and Python snippets. The plot of matplotlib needs a workaround but otherwise this is a real cool feature of RStudio!

1.1a Univariate Regression – R code

Here a simple linear regression line is fitted between a single input feature and the target variable

# Source in the R function library
source("RFunctions.R")
# Read the Boston data file
df=read.csv("Boston.csv",stringsAsFactors = FALSE) # Data from MASS - Statistical Learning

# Split the data into training and test sets (75:25)
train_idx <- trainTestSplit(df,trainPercent=75,seed=5)
train <- df[train_idx, ]
test <- df[-train_idx, ]

# Fit a linear regression line between 'Median value of owner occupied homes' vs 'lower status of 
# population'
fit=lm(medv~lstat,data=df)
# Display details of fir
summary(fit)
## 
## Call:
## lm(formula = medv ~ lstat, data = df)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -15.168  -3.990  -1.318   2.034  24.500 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 34.55384    0.56263   61.41   <2e-16 ***
## lstat       -0.95005    0.03873  -24.53   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.216 on 504 degrees of freedom
## Multiple R-squared:  0.5441, Adjusted R-squared:  0.5432 
## F-statistic: 601.6 on 1 and 504 DF,  p-value: < 2.2e-16
# Display the confidence intervals
confint(fit)
##                 2.5 %     97.5 %
## (Intercept) 33.448457 35.6592247
## lstat       -1.026148 -0.8739505
plot(df$lstat,df$medv, xlab="Lower status (%)",ylab="Median value of owned homes ($1000)", main="Median value of homes ($1000) vs Lowe status (%)")
abline(fit)
abline(fit,lwd=3)
abline(fit,lwd=3,col="red")

rsquared=Rsquared(fit,test,test$medv)
sprintf("R-squared for uni-variate regression (Boston.csv)  is : %f", rsquared)
## [1] "R-squared for uni-variate regression (Boston.csv)  is : 0.556964"

1.1b Univariate Regression – Python code

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
#os.chdir("C:\\software\\machine-learning\\RandPython")

# Read the CSV file
df = pd.read_csv("Boston.csv",encoding = "ISO-8859-1")
# Select the feature variable
X=df['lstat']

# Select the target 
y=df['medv']

# Split into train and test sets (75:25)
X_train, X_test, y_train, y_test = train_test_split(X, y,random_state = 0)
X_train=X_train.values.reshape(-1,1)
X_test=X_test.values.reshape(-1,1)

# Fit a linear model
linreg = LinearRegression().fit(X_train, y_train)

# Print the training and test R squared score
print('R-squared score (training): {:.3f}'.format(linreg.score(X_train, y_train)))
print('R-squared score (test): {:.3f}'.format(linreg.score(X_test, y_test)))
     
# Plot the linear regression line
fig=plt.scatter(X_train,y_train)

# Create a range of points. Compute yhat=coeff1*x + intercept and plot
x=np.linspace(0,40,20)
fig1=plt.plot(x, linreg.coef_ * x + linreg.intercept_, color='red')
fig1=plt.title("Median value of homes ($1000) vs Lowe status (%)")
fig1=plt.xlabel("Lower status (%)")
fig1=plt.ylabel("Median value of owned homes ($1000)")
fig.figure.savefig('foo.png', bbox_inches='tight')
fig1.figure.savefig('foo1.png', bbox_inches='tight')
print "Finished"
## R-squared score (training): 0.571
## R-squared score (test): 0.458
## Finished

1.2a Multivariate Regression – R code

# Read crimes data
crimesDF <- read.csv("crimes.csv",stringsAsFactors = FALSE)

# Remove the 1st 7 columns which do not impact output
crimesDF1 <- crimesDF[,7:length(crimesDF)]

# Convert all to numeric
crimesDF2 <- sapply(crimesDF1,as.numeric)

# Check for NAs
a <- is.na(crimesDF2)
# Set to 0 as an imputation
crimesDF2[a] <-0
#Create as a dataframe
crimesDF2 <- as.data.frame(crimesDF2)
#Create a train/test split
train_idx <- trainTestSplit(crimesDF2,trainPercent=75,seed=5)
train <- crimesDF2[train_idx, ]
test <- crimesDF2[-train_idx, ]

# Fit a multivariate regression model between crimesPerPop and all other features
fit <- lm(ViolentCrimesPerPop~.,data=train)

# Compute and print R Squared
rsquared=Rsquared(fit,test,test$ViolentCrimesPerPop)
sprintf("R-squared for multi-variate regression (crimes.csv)  is : %f", rsquared)
## [1] "R-squared for multi-variate regression (crimes.csv)  is : 0.653940"

1.2b Multivariate Regression – Python code

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
# Read the data
crimesDF =pd.read_csv("crimes.csv",encoding="ISO-8859-1")
#Remove the 1st 7 columns
crimesDF1=crimesDF.iloc[:,7:crimesDF.shape[1]]
# Convert to numeric
crimesDF2 = crimesDF1.apply(pd.to_numeric, errors='coerce')
# Impute NA to 0s
crimesDF2.fillna(0, inplace=True)

# Select the X (feature vatiables - all)
X=crimesDF2.iloc[:,0:120]

# Set the target
y=crimesDF2.iloc[:,121]

X_train, X_test, y_train, y_test = train_test_split(X, y,random_state = 0)
# Fit a multivariate regression model
linreg = LinearRegression().fit(X_train, y_train)

# compute and print the R Square
print('R-squared score (training): {:.3f}'.format(linreg.score(X_train, y_train)))
print('R-squared score (test): {:.3f}'.format(linreg.score(X_test, y_test)))
## R-squared score (training): 0.699
## R-squared score (test): 0.677

1.3a Polynomial Regression – R

For Polynomial regression , polynomials of degree 1,2 & 3 are used and R squared is computed. It can be seen that the quadaratic model provides the best R squared score and hence the best fit

 # Polynomial degree 1
df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI
df1 <- as.data.frame(sapply(df,as.numeric))

# Select key columns
df2 <- df1 %>% select(cylinder,displacement, horsepower,weight, acceleration, year,mpg)
df3 <- df2[complete.cases(df2),]

# Split as train and test sets
train_idx <- trainTestSplit(df3,trainPercent=75,seed=5)
train <- df3[train_idx, ]
test <- df3[-train_idx, ]

# Fit a model of degree 1
fit <- lm(mpg~. ,data=train)
rsquared1 <-Rsquared(fit,test,test$mpg)
sprintf("R-squared for Polynomial regression of degree 1 (auto_mpg.csv)  is : %f", rsquared1)
## [1] "R-squared for Polynomial regression of degree 1 (auto_mpg.csv)  is : 0.763607"
# Polynomial degree 2 - Quadratic
x = as.matrix(df3[1:6])
# Make a  polynomial  of degree 2 for feature variables before split
df4=as.data.frame(poly(x,2,raw=TRUE))
df5 <- cbind(df4,df3[7])

# Split into train and test set
train_idx <- trainTestSplit(df5,trainPercent=75,seed=5)
train <- df5[train_idx, ]
test <- df5[-train_idx, ]

# Fit the quadratic model
fit <- lm(mpg~. ,data=train)
# Compute R squared
rsquared2=Rsquared(fit,test,test$mpg)
sprintf("R-squared for Polynomial regression of degree 2 (auto_mpg.csv)  is : %f", rsquared2)
## [1] "R-squared for Polynomial regression of degree 2 (auto_mpg.csv)  is : 0.831372"
#Polynomial degree 3
x = as.matrix(df3[1:6])
# Make polynomial of degree 4  of feature variables before split
df4=as.data.frame(poly(x,3,raw=TRUE))
df5 <- cbind(df4,df3[7])
train_idx <- trainTestSplit(df5,trainPercent=75,seed=5)

train <- df5[train_idx, ]
test <- df5[-train_idx, ]
# Fit a model of degree 3
fit <- lm(mpg~. ,data=train)
# Compute R squared
rsquared3=Rsquared(fit,test,test$mpg)
sprintf("R-squared for Polynomial regression of degree 2 (auto_mpg.csv)  is : %f", rsquared3)
## [1] "R-squared for Polynomial regression of degree 2 (auto_mpg.csv)  is : 0.773225"
df=data.frame(degree=c(1,2,3),Rsquared=c(rsquared1,rsquared2,rsquared3))
# Make a plot of Rsquared and degree
ggplot(df,aes(x=degree,y=Rsquared)) +geom_point() + geom_line(color="blue") +
    ggtitle("Polynomial regression - R squared vs Degree of polynomial") +
    xlab("Degree") + ylab("R squared")

1.3a Polynomial Regression – Python

For Polynomial regression , polynomials of degree 1,2 & 3 are used and R squared is computed. It can be seen that the quadaratic model provides the best R squared score and hence the best fit

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from sklearn.preprocessing import PolynomialFeatures
autoDF =pd.read_csv("auto_mpg.csv",encoding="ISO-8859-1")
autoDF.shape
autoDF.columns
# Select key columns
autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']]
# Convert columns to numeric
autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce')
# Drop NAs
autoDF3=autoDF2.dropna()
autoDF3.shape
X=autoDF3[['cylinder','displacement','horsepower','weight','acceleration','year']]
y=autoDF3['mpg']

# Polynomial degree 1
X_train, X_test, y_train, y_test = train_test_split(X, y,random_state = 0)
linreg = LinearRegression().fit(X_train, y_train)
print('R-squared score - Polynomial degree 1 (training): {:.3f}'.format(linreg.score(X_train, y_train)))
# Compute R squared     
rsquared1 =linreg.score(X_test, y_test)
print('R-squared score - Polynomial degree 1 (test): {:.3f}'.format(linreg.score(X_test, y_test)))

# Polynomial degree 2
poly = PolynomialFeatures(degree=2)
X_poly = poly.fit_transform(X)
X_train, X_test, y_train, y_test = train_test_split(X_poly, y,random_state = 0)
linreg = LinearRegression().fit(X_train, y_train)

# Compute R squared
print('R-squared score - Polynomial degree 2 (training): {:.3f}'.format(linreg.score(X_train, y_train)))
rsquared2 =linreg.score(X_test, y_test)
print('R-squared score - Polynomial degree 2 (test): {:.3f}\n'.format(linreg.score(X_test, y_test)))

#Polynomial degree 3

poly = PolynomialFeatures(degree=3)
X_poly = poly.fit_transform(X)
X_train, X_test, y_train, y_test = train_test_split(X_poly, y,random_state = 0)
linreg = LinearRegression().fit(X_train, y_train)
print('(R-squared score -Polynomial degree 3  (training): {:.3f}'
     .format(linreg.score(X_train, y_train)))
# Compute R squared     
rsquared3 =linreg.score(X_test, y_test)
print('R-squared score Polynomial degree 3 (test): {:.3f}\n'.format(linreg.score(X_test, y_test)))
degree=[1,2,3]
rsquared =[rsquared1,rsquared2,rsquared3]
fig2=plt.plot(degree,rsquared)
fig2=plt.title("Polynomial regression - R squared vs Degree of polynomial")
fig2=plt.xlabel("Degree")
fig2=plt.ylabel("R squared")
fig2.figure.savefig('foo2.png', bbox_inches='tight')
print "Finished plotting and saving"
## R-squared score - Polynomial degree 1 (training): 0.811
## R-squared score - Polynomial degree 1 (test): 0.799
## R-squared score - Polynomial degree 2 (training): 0.861
## R-squared score - Polynomial degree 2 (test): 0.847
## 
## (R-squared score -Polynomial degree 3  (training): 0.933
## R-squared score Polynomial degree 3 (test): 0.710
## 
## Finished plotting and saving

1.4 K Nearest Neighbors

The code below implements KNN Regression both for R and Python. This is done for different neighbors. The R squared is computed in each case. This is repeated after performing feature scaling. It can be seen the model fit is much better after feature scaling. Normalization refers to

X_{normalized} = \frac{X-min(X)}{max(X-min(X))}

Another technique that is used is Standardization which is

X_{standardized} = \frac{X-mean(X)}{sd(X)}

1.4a K Nearest Neighbors Regression – R( Unnormalized)

The R code below does not use feature scaling

# KNN regression requires the FNN package
df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI
df1 <- as.data.frame(sapply(df,as.numeric))
df2 <- df1 %>% select(cylinder,displacement, horsepower,weight, acceleration, year,mpg)
df3 <- df2[complete.cases(df2),]

# Split train and test
train_idx <- trainTestSplit(df3,trainPercent=75,seed=5)
train <- df3[train_idx, ]
test <- df3[-train_idx, ]
#  Select the feature variables
train.X=train[,1:6]
# Set the target for training
train.Y=train[,7]
# Do the same for test set
test.X=test[,1:6]
test.Y=test[,7]

rsquared <- NULL
# Create a list of neighbors
neighbors <-c(1,2,4,8,10,14)
for(i in seq_along(neighbors)){
    # Perform a KNN regression fit
    knn=knn.reg(train.X,test.X,train.Y,k=neighbors[i])
    # Compute R sqaured
    rsquared[i]=knnRSquared(knn$pred,test.Y)
}

# Make a dataframe for plotting
df <- data.frame(neighbors,Rsquared=rsquared)
# Plot the number of neighors vs the R squared
ggplot(df,aes(x=neighbors,y=Rsquared)) + geom_point() +geom_line(color="blue") +
    xlab("Number of neighbors") + ylab("R squared") +
    ggtitle("KNN regression - R squared vs Number of Neighors (Unnormalized)")

1.4b K Nearest Neighbors Regression – Python( Unnormalized)

The Python code below does not use feature scaling

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from sklearn.preprocessing import PolynomialFeatures
from sklearn.neighbors import KNeighborsRegressor
autoDF =pd.read_csv("auto_mpg.csv",encoding="ISO-8859-1")
autoDF.shape
autoDF.columns
autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']]
autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce')
autoDF3=autoDF2.dropna()
autoDF3.shape
X=autoDF3[['cylinder','displacement','horsepower','weight','acceleration','year']]
y=autoDF3['mpg']

# Perform a train/test split
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state = 0)
# Create a list of neighbors
rsquared=[]
neighbors=[1,2,4,8,10,14]
for i in neighbors:
        # Fit a KNN model
        knnreg = KNeighborsRegressor(n_neighbors = i).fit(X_train, y_train)
        # Compute R squared
        rsquared.append(knnreg.score(X_test, y_test))
        print('R-squared test score: {:.3f}'
        .format(knnreg.score(X_test, y_test)))
# Plot the number of neighors vs the R squared        
fig3=plt.plot(neighbors,rsquared)
fig3=plt.title("KNN regression - R squared vs Number of neighbors(Unnormalized)")
fig3=plt.xlabel("Neighbors")
fig3=plt.ylabel("R squared")
fig3.figure.savefig('foo3.png', bbox_inches='tight')
print "Finished plotting and saving"
## R-squared test score: 0.527
## R-squared test score: 0.678
## R-squared test score: 0.707
## R-squared test score: 0.684
## R-squared test score: 0.683
## R-squared test score: 0.670
## Finished plotting and saving

1.4c K Nearest Neighbors Regression – R( Normalized)

It can be seen that R squared improves when the features are normalized.

df=read.csv("auto_mpg.csv",stringsAsFactors = FALSE) # Data from UCI
df1 <- as.data.frame(sapply(df,as.numeric))
df2 <- df1 %>% select(cylinder,displacement, horsepower,weight, acceleration, year,mpg)
df3 <- df2[complete.cases(df2),]

# Perform MinMaxScaling of feature variables 
train.X.scaled=MinMaxScaler(train.X)
test.X.scaled=MinMaxScaler(test.X)

# Create a list of neighbors
rsquared <- NULL
neighbors <-c(1,2,4,6,8,10,12,15,20,25,30)
for(i in seq_along(neighbors)){
    # Fit a KNN model
    knn=knn.reg(train.X.scaled,test.X.scaled,train.Y,k=i)
    # Compute R ssquared
    rsquared[i]=knnRSquared(knn$pred,test.Y)
    
}

df <- data.frame(neighbors,Rsquared=rsquared)
# Plot the number of neighors vs the R squared 
ggplot(df,aes(x=neighbors,y=Rsquared)) + geom_point() +geom_line(color="blue") +
    xlab("Number of neighbors") + ylab("R squared") +
    ggtitle("KNN regression - R squared vs Number of Neighors(Normalized)")

1.4d K Nearest Neighbors Regression – Python( Normalized)

R squared improves when the features are normalized with MinMaxScaling

import numpy as np
import pandas as pd
import os
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from sklearn.preprocessing import PolynomialFeatures
from sklearn.neighbors import KNeighborsRegressor
from sklearn.preprocessing import MinMaxScaler
autoDF =pd.read_csv("auto_mpg.csv",encoding="ISO-8859-1")
autoDF.shape
autoDF.columns
autoDF1=autoDF[['mpg','cylinder','displacement','horsepower','weight','acceleration','year']]
autoDF2 = autoDF1.apply(pd.to_numeric, errors='coerce')
autoDF3=autoDF2.dropna()
autoDF3.shape
X=autoDF3[['cylinder','displacement','horsepower','weight','acceleration','year']]
y=autoDF3['mpg']

# Perform a train/ test  split
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state = 0)
# Use MinMaxScaling
scaler = MinMaxScaler()
X_train_scaled = scaler.fit_transform(X_train)
# Apply scaling on test set
X_test_scaled = scaler.transform(X_test)

# Create a list of neighbors
rsquared=[]
neighbors=[1,2,4,6,8,10,12,15,20,25,30]
for i in neighbors:
    # Fit a KNN model
    knnreg = KNeighborsRegressor(n_neighbors = i).fit(X_train_scaled, y_train)
    # Compute R squared
    rsquared.append(knnreg.score(X_test_scaled, y_test))
    print('R-squared test score: {:.3f}'
        .format(knnreg.score(X_test_scaled, y_test)))

# Plot the number of neighors vs the R squared 
fig4=plt.plot(neighbors,rsquared)
fig4=plt.title("KNN regression - R squared vs Number of neighbors(Normalized)")
fig4=plt.xlabel("Neighbors")
fig4=plt.ylabel("R squared")
fig4.figure.savefig('foo4.png', bbox_inches='tight')
print "Finished plotting and saving"
## R-squared test score: 0.703
## R-squared test score: 0.810
## R-squared test score: 0.830
## R-squared test score: 0.838
## R-squared test score: 0.834
## R-squared test score: 0.828
## R-squared test score: 0.827
## R-squared test score: 0.826
## R-squared test score: 0.816
## R-squared test score: 0.815
## R-squared test score: 0.809
## Finished plotting and saving

Conclusion

In this initial post I cover the regression models when the output is continous. I intend to touch upon other Machine Learning algorithms.
Comments, suggestions and corrections are welcome.

Watch this this space!

To be continued….

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2. Neural Networks: The mechanics of backpropagation
3. More book, more cricket! 2nd edition of my books now on Amazon
4. Spicing up a IBM Bluemix cloud app with MongoDB and NodeExpress
5. Introducing cricket package yorkr:Part 4-In the block hole!

To see all posts see Index of posts

Revisiting crimes against women in India


Here I go again, raking the muck about crimes against women in India. My earlier post “A crime map of India in R: Crimes against women in India” garnered a lot of responses from readers. In fact one of the readers even volunteered to create the only choropleth map in that post. The data for this post is taken from http://data.gov.in. You can download the data from the link “Crimes against women in India

I was so impressed by the choropleth map that I decided to do that for all crimes against women.(Wikipedia definition: A choropleth map is a thematic map in which areas are shaded or patterned in proportion to the measurement of the statistical variable being displayed on the map). Personally, I think pictures tell the story better. I am sure you will agree!

So here, I have it a Shiny app which will plot choropleth maps for a chosen crime in a given year.

You can try out my interactive Shiny app at  Crimes against women in India

Checkout out my book  on Amazon available in both  Paperback ($9.99) and a Kindle version($6.99/Rs449/). (see ‘Practical Machine Learning with R and Python – Machine Learning in stereo‘)

The following technique can be used to determine the ‘goodness’ of a hypothesis or how well the hypothesis can fit the data and can also generalize to new examples not in the training set.

In the picture below  are the details of  ‘Rape” in the year 2015.
1

Interestingly the ‘Total Crime against women’ in 2001 shows the Top 5 as
1) Uttar Pradresh 2) Andhra Pradesh 3) Madhya Pradesh 4) Maharashtra 5) Rajasthan

2

But in 2015 West Bengal tops the list, as the real heavy weight in crimes against women. The new pecking order in 2015 for ‘Total Crimes against Women’ is

1) West Bengal 2) Andhra Pradesh 3) Uttar Pradesh  4) Rajasthan 5) Maharashtra

3

Similarly for rapes, West Bengal is nowhere in the top 5 list in 2001. In 2015, it is in second only to the national rape leader Madhya Pradesh.  Also in 2001 West Bengal is not in the top 5 for any of 6 crime heads. But in 2015, West Bengal is in the top 5 of 6 crime heads. The emergence of West Bengal as the leader in Crimes against Women is due to the steep increase in crime rate  over the years.Clearly the law and order situation in West Bengal is heading south.

In Dowry Deaths, UP, Bihar, MP, West Bengal lead the pack, and in that order in 2015.

The usual suspects for most crime categories are West Bengal, UP, MP, AP & Maharashtra.

The state-wise crime charts plot the incidence of the crime (rape, dowry death, assault on women etc) over the years. Data for each state and for each crime was available from 2001-2013. The data for period 2014-2018 are projected using linear regression. The shaded portion in the plots indicate the 95% confidence level in the prediction (i.e in other words we can be 95% certain that the true mean of the crime rate in the projected years will lie within the shaded region)

4

Several  interesting requests came from readers to my earlier post. Some of them were to to plot the crimes as function of population and per capita income of the State/Union Territory to see if the plots  throw up new crime leaders. I have not got the relevant state-wise population distribution data yet. I intend to update this when I get my hands on this data.

I have included the crimes.csv which has been used to generate the visualization. However for the Shiny app I save this as .RData for better performance of the app.

You can clone/download  the code for the Shiny app from GitHub at  crimesAgainWomenIndia

Please checkout my Shiny app : Crimes against women

I also intend to add further interactivity to my visualizations in a future version. Watch this space. I’ll be back!

You may like
1. My book ‘Practical Machine Learning with R and Python’ on Amazon
2. Natural Language Processing: What would Shakespeare say?
3. Introducing cricketr! : An R package to analyze performances of cricketers
4. A peek into literacy in India: Statistical Learning with R
5. Informed choices through Machine Learning : Analyzing Kohli, Tendulkar and Dravid
6. Re-working the Lucy-Richardson Algorithm in OpenCV
7.  What’s up Watson? Using IBM Watson’s QAAPI with Bluemix, NodeExpress – Part 1
8.  Bend it like Bluemix, MongoDB with autoscaling – Part 2
9. TWS-4: Gossip protocol: Epidemics and rumors to the rescue
10. Thinking Web Scale (TWS-3): Map-Reduce – Bring compute to data
11.  Simulating an Edge Shape in Android

cricketr plays the ODIs!


Published in R bloggers: cricketr plays the ODIs

Introduction

In this post my package ‘cricketr’ takes a swing at One Day Internationals(ODIs). Like test batsman who adapt to ODIs with some innovative strokes, the cricketr package has some additional functions and some modified functions to handle the high strike and economy rates in ODIs. As before I have chosen my top 4 ODI batsmen and top 4 ODI bowlers.

Unititled2

If you are passionate about cricket, and love analyzing cricket performances, then check out my 2 racy books on cricket! In my books, I perform detailed yet compact analysis of performances of both batsmen, bowlers besides evaluating team & match performances in Tests , ODIs, T20s & IPL. You can buy my books on cricket from Amazon at $12.99 for the paperback and $4.99/$6.99 respectively for the kindle versions. The books can be accessed at Cricket analytics with cricketr  and Beaten by sheer pace-Cricket analytics with yorkr  A must read for any cricket lover! Check it out!!

1

d $4.99/Rs 320 and $6.99/Rs448 respectively

Do check out my interactive Shiny app implementation using the cricketr package – Sixer – R package cricketr’s new Shiny avatar

You can also read this post at Rpubs as odi-cricketr. Dowload this report as a PDF file from odi-cricketr.pdf

Note: If you would like to do a similar analysis for a different set of batsman and bowlers, you can clone/download my skeleton cricketr template from Github (which is the R Markdown file I have used for the analysis below). You will only need to make appropriate changes for the players you are interested in. Just a familiarity with R and R Markdown only is needed.
Batsmen

  1. Virendar Sehwag (Ind)
  2. AB Devilliers (SA)
  3. Chris Gayle (WI)
  4. Glenn Maxwell (Aus)

Bowlers

  1. Mitchell Johnson (Aus)
  2. Lasith Malinga (SL)
  3. Dale Steyn (SA)
  4. Tim Southee (NZ)

I have sprinkled the plots with a few of my comments. Feel free to draw your conclusions! The analysis is included below

The profile for Virender Sehwag is 35263. This can be used to get the ODI data for Sehwag. For a batsman the type should be “batting” and for a bowler the type should be “bowling” and the function is getPlayerDataOD()

The package can be installed directly from CRAN

if (!require("cricketr")){ 
    install.packages("cricketr",lib = "c:/test") 
} 
library(cricketr)

or from Github

library(devtools)
install_github("tvganesh/cricketr")
library(cricketr)

The One day data for a particular player can be obtained with the getPlayerDataOD() function. To do you will need to go to ESPN CricInfo Player and type in the name of the player for e.g Virendar Sehwag, etc. This will bring up a page which have the profile number for the player e.g. for Virendar Sehwag this would be http://www.espncricinfo.com/india/content/player/35263.html. Hence, Sehwag’s profile is 35263. This can be used to get the data for Virat Sehwag as shown below

sehwag <- getPlayerDataOD(35263,dir="..",file="sehwag.csv",type="batting")

Analyses of Batsmen

The following plots gives the analysis of the 4 ODI batsmen

  1. Virendar Sehwag (Ind) – Innings – 245, Runs = 8586, Average=35.05, Strike Rate= 104.33
  2. AB Devilliers (SA) – Innings – 179, Runs= 7941, Average=53.65, Strike Rate= 99.12
  3. Chris Gayle (WI) – Innings – 264, Runs= 9221, Average=37.65, Strike Rate= 85.11
  4. Glenn Maxwell (Aus) – Innings – 45, Runs= 1367, Average=35.02, Strike Rate= 126.69

Plot of 4s, 6s and the scoring rate in ODIs

The 3 charts below give the number of

  1. 4s vs Runs scored
  2. 6s vs Runs scored
  3. Balls faced vs Runs scored

A regression line is fitted in each of these plots for each of the ODI batsmen A. Virender Sehwag

par(mfrow=c(1,3))
par(mar=c(4,4,2,2))
batsman4s("./sehwag.csv","Sehwag")
batsman6s("./sehwag.csv","Sehwag")
batsmanScoringRateODTT("./sehwag.csv","Sehwag")

sehwag-4s6sSR-1

dev.off()
## null device 
##           1

B. AB Devilliers

par(mfrow=c(1,3))
par(mar=c(4,4,2,2))
batsman4s("./devilliers.csv","Devillier")
batsman6s("./devilliers.csv","Devillier")
batsmanScoringRateODTT("./devilliers.csv","Devillier")

devillier-4s6SR-1

dev.off()
## null device 
##           1

C. Chris Gayle

par(mfrow=c(1,3))
par(mar=c(4,4,2,2))
batsman4s("./gayle.csv","Gayle")
batsman6s("./gayle.csv","Gayle")
batsmanScoringRateODTT("./gayle.csv","Gayle")

gayle-4s6sSR-1

dev.off()
## null device 
##           1

D. Glenn Maxwell

par(mfrow=c(1,3))
par(mar=c(4,4,2,2))
batsman4s("./maxwell.csv","Maxwell")
batsman6s("./maxwell.csv","Maxwell")
batsmanScoringRateODTT("./maxwell.csv","Maxwell")

maxwell-4s6sout-1

dev.off()
## null device 
##           1

Relative Mean Strike Rate

In this first plot I plot the Mean Strike Rate of the batsmen. It can be seen that Maxwell has a awesome strike rate in ODIs. However we need to keep in mind that Maxwell has relatively much fewer (only 45 innings) innings. He is followed by Sehwag who(most innings- 245) also has an excellent strike rate till 100 runs and then we have Devilliers who roars ahead. This is also seen in the overall strike rate in above

par(mar=c(4,4,2,2))
frames <- list("./sehwag.csv","./devilliers.csv","gayle.csv","maxwell.csv")
names <- list("Sehwag","Devilliers","Gayle","Maxwell")
relativeBatsmanSRODTT(frames,names)

plot-1-1

Relative Runs Frequency Percentage

Sehwag leads in the percentage of runs in 10 run ranges upto 50 runs. Maxwell and Devilliers lead in 55-66 & 66-85 respectively.

frames <- list("./sehwag.csv","./devilliers.csv","gayle.csv","maxwell.csv")
names <- list("Sehwag","Devilliers","Gayle","Maxwell")
relativeRunsFreqPerfODTT(frames,names)

plot-2-1

Percentage of 4s,6s in the runs scored

The plot below shows the percentage of runs made by the batsmen by ways of 1s,2s,3s, 4s and 6s. It can be seen that Sehwag has the higheest percent of 4s (33.36%) in his overall runs in ODIs. Maxwell has the highest percentage of 6s (13.36%) in his ODI career. If we take the overall 4s+6s then Sehwag leads with (33.36 +5.95 = 39.31%),followed by Gayle (27.80+10.15=37.95%)

Percent 4’s,6’s in total runs scored

The plot below shows the contrib

frames <- list("./sehwag.csv","./devilliers.csv","gayle.csv","maxwell.csv")
names <- list("Sehwag","Devilliers","Gayle","Maxwell")
runs4s6s <-batsman4s6s(frames,names)

plot-46s-1

print(runs4s6s)
##                Sehwag Devilliers Gayle Maxwell
## Runs(1s,2s,3s)  60.69      67.39 62.05   62.11
## 4s              33.36      24.28 27.80   24.53
## 6s               5.95       8.32 10.15   13.36
 

Runs forecast

The forecast for the batsman is shown below.

par(mfrow=c(2,2))
par(mar=c(4,4,2,2))
batsmanPerfForecast("./sehwag.csv","Sehwag")
batsmanPerfForecast("./devilliers.csv","Devilliers")
batsmanPerfForecast("./gayle.csv","Gayle")
batsmanPerfForecast("./maxwell.csv","Maxwell")

swcr-perf-1

dev.off()
## null device 
##           1

3D plot of Runs vs Balls Faced and Minutes at Crease

The plot is a scatter plot of Runs vs Balls faced and Minutes at Crease. A prediction plane is fitted

par(mfrow=c(1,2))
par(mar=c(4,4,2,2))
battingPerf3d("./sehwag.csv","V Sehwag")
battingPerf3d("./devilliers.csv","AB Devilliers")

plot-3-1

dev.off()
## null device 
##           1
par(mfrow=c(1,2))
par(mar=c(4,4,2,2))
battingPerf3d("./gayle.csv","C Gayle")
battingPerf3d("./maxwell.csv","G Maxwell")

plot-4-1

dev.off()
## null device 
##           1

Predicting Runs given Balls Faced and Minutes at Crease

A multi-variate regression plane is fitted between Runs and Balls faced +Minutes at crease.

BF <- seq( 10, 200,length=10)
Mins <- seq(30,220,length=10)
newDF <- data.frame(BF,Mins)

sehwag <- batsmanRunsPredict("./sehwag.csv","Sehwag",newdataframe=newDF)
devilliers <- batsmanRunsPredict("./devilliers.csv","Devilliers",newdataframe=newDF)
gayle <- batsmanRunsPredict("./gayle.csv","Gayle",newdataframe=newDF)
maxwell <- batsmanRunsPredict("./maxwell.csv","Maxwell",newdataframe=newDF)

The fitted model is then used to predict the runs that the batsmen will score for a hypotheticial Balls faced and Minutes at crease. It can be seen that Maxwell sets a searing pace in the predicted runs for a given Balls Faced and Minutes at crease followed by Sehwag. But we have to keep in mind that Maxwell has only around 1/5th of the innings of Sehwag (45 to Sehwag’s 245 innings). They are followed by Devilliers and then finally Gayle

batsmen <-cbind(round(sehwag$Runs),round(devilliers$Runs),round(gayle$Runs),round(maxwell$Runs))
colnames(batsmen) <- c("Sehwag","Devilliers","Gayle","Maxwell")
newDF <- data.frame(round(newDF$BF),round(newDF$Mins))
colnames(newDF) <- c("BallsFaced","MinsAtCrease")
predictedRuns <- cbind(newDF,batsmen)
predictedRuns
##    BallsFaced MinsAtCrease Sehwag Devilliers Gayle Maxwell
## 1          10           30     11         12    11      18
## 2          31           51     33         32    28      43
## 3          52           72     55         52    46      67
## 4          73           93     77         71    63      92
## 5          94          114    100         91    81     117
## 6         116          136    122        111    98     141
## 7         137          157    144        130   116     166
## 8         158          178    167        150   133     191
## 9         179          199    189        170   151     215
## 10        200          220    211        190   168     240

Highest runs likelihood

The plots below the runs likelihood of batsman. This uses K-Means It can be seen that Devilliers has almost 27.75% likelihood to make around 90+ runs. Gayle and Sehwag have 34% to make 40+ runs. A. Virender Sehwag

A. Virender Sehwag

batsmanRunsLikelihood("./sehwag.csv","Sehwag")

smith-1

## Summary of  Sehwag 's runs scoring likelihood
## **************************************************
## 
## There is a 35.22 % likelihood that Sehwag  will make  46 Runs in  44 balls over 67  Minutes 
## There is a 9.43 % likelihood that Sehwag  will make  119 Runs in  106 balls over  158  Minutes 
## There is a 55.35 % likelihood that Sehwag  will make  12 Runs in  13 balls over 18  Minutes

B. AB Devilliers

batsmanRunsLikelihood("./devilliers.csv","Devilliers")

warner-1

## Summary of  Devilliers 's runs scoring likelihood
## **************************************************
## 
## There is a 30.65 % likelihood that Devilliers  will make  44 Runs in  43 balls over 60  Minutes 
## There is a 29.84 % likelihood that Devilliers  will make  91 Runs in  88 balls over  124  Minutes 
## There is a 39.52 % likelihood that Devilliers  will make  11 Runs in  15 balls over 21  Minutes

C. Chris Gayle

batsmanRunsLikelihood("./gayle.csv","Gayle")

cook,cache-TRUE-1

## Summary of  Gayle 's runs scoring likelihood
## **************************************************
## 
## There is a 32.69 % likelihood that Gayle  will make  47 Runs in  51 balls over 72  Minutes 
## There is a 54.49 % likelihood that Gayle  will make  10 Runs in  15 balls over  20  Minutes 
## There is a 12.82 % likelihood that Gayle  will make  109 Runs in  119 balls over 172  Minutes

D. Glenn Maxwell

batsmanRunsLikelihood("./maxwell.csv","Maxwell")

oot-1

## Summary of  Maxwell 's runs scoring likelihood
## **************************************************
## 
## There is a 34.38 % likelihood that Maxwell  will make  39 Runs in  29 balls over 35  Minutes 
## There is a 15.62 % likelihood that Maxwell  will make  89 Runs in  55 balls over  69  Minutes 
## There is a 50 % likelihood that Maxwell  will make  6 Runs in  7 balls over 9  Minutes

Average runs at ground and against opposition

A. Virender Sehwag

par(mfrow=c(1,2))
par(mar=c(4,4,2,2))
batsmanAvgRunsGround("./sehwag.csv","Sehwag")
batsmanAvgRunsOpposition("./sehwag.csv","Sehwag")

avgrg-1-1

dev.off()
## null device 
##           1

B. AB Devilliers

par(mfrow=c(1,2))
par(mar=c(4,4,2,2))
batsmanAvgRunsGround("./devilliers.csv","Devilliers")
batsmanAvgRunsOpposition("./devilliers.csv","Devilliers")

avgrg-2-1

dev.off()
## null device 
##           1

C. Chris Gayle

par(mfrow=c(1,2))
par(mar=c(4,4,2,2))
batsmanAvgRunsGround("./gayle.csv","Gayle")
batsmanAvgRunsOpposition("./gayle.csv","Gayle")

avgrg-3-1

dev.off()
## null device 
##           1

D. Glenn Maxwell

par(mfrow=c(1,2))
par(mar=c(4,4,2,2))
batsmanAvgRunsGround("./maxwell.csv","Maxwell")
batsmanAvgRunsOpposition("./maxwell.csv","Maxwell")

avgrg-4-1

dev.off()
## null device 
##           1

Moving Average of runs over career

The moving average for the 4 batsmen indicate the following

1. The moving average of Devilliers and Maxwell is on the way up.
2. Sehwag shows a slight downward trend from his 2nd peak in 2011
3. Gayle maintains a consistent 45 runs for the last few years

par(mfrow=c(2,2))
par(mar=c(4,4,2,2))
batsmanMovingAverage("./sehwag.csv","Sehwag")
batsmanMovingAverage("./devilliers.csv","Devilliers")
batsmanMovingAverage("./gayle.csv","Gayle")
batsmanMovingAverage("./maxwell.csv","Maxwell")

sdgm-ma-1

dev.off()
## null device 
##           1

Check batsmen in-form, out-of-form

  1. Maxwell, Devilliers, Sehwag are in-form. This is also evident from the moving average plot
  2. Gayle is out-of-form
checkBatsmanInForm("./sehwag.csv","Sehwag")
## *******************************************************************************************
## 
## Population size: 143  Mean of population: 33.76 
## Sample size: 16  Mean of sample: 37.44 SD of sample: 55.15 
## 
## Null hypothesis H0 : Sehwag 's sample average is within 95% confidence interval 
##         of population average
## Alternative hypothesis Ha : Sehwag 's sample average is below the 95% confidence
##         interval of population average
## 
## [1] "Sehwag 's Form Status: In-Form because the p value: 0.603525  is greater than alpha=  0.05"
## *******************************************************************************************
checkBatsmanInForm("./devilliers.csv","Devilliers")
## *******************************************************************************************
## 
## Population size: 111  Mean of population: 43.5 
## Sample size: 13  Mean of sample: 57.62 SD of sample: 40.69 
## 
## Null hypothesis H0 : Devilliers 's sample average is within 95% confidence interval 
##         of population average
## Alternative hypothesis Ha : Devilliers 's sample average is below the 95% confidence
##         interval of population average
## 
## [1] "Devilliers 's Form Status: In-Form because the p value: 0.883541  is greater than alpha=  0.05"
## *******************************************************************************************
checkBatsmanInForm("./gayle.csv","Gayle")
## *******************************************************************************************
## 
## Population size: 140  Mean of population: 37.1 
## Sample size: 16  Mean of sample: 17.25 SD of sample: 20.25 
## 
## Null hypothesis H0 : Gayle 's sample average is within 95% confidence interval 
##         of population average
## Alternative hypothesis Ha : Gayle 's sample average is below the 95% confidence
##         interval of population average
## 
## [1] "Gayle 's Form Status: Out-of-Form because the p value: 0.000609  is less than alpha=  0.05"
## *******************************************************************************************
checkBatsmanInForm("./maxwell.csv","Maxwell")
## *******************************************************************************************
## 
## Population size: 28  Mean of population: 25.25 
## Sample size: 4  Mean of sample: 64.25 SD of sample: 36.97 
## 
## Null hypothesis H0 : Maxwell 's sample average is within 95% confidence interval 
##         of population average
## Alternative hypothesis Ha : Maxwell 's sample average is below the 95% confidence
##         interval of population average
## 
## [1] "Maxwell 's Form Status: In-Form because the p value: 0.948744  is greater than alpha=  0.05"
## *******************************************************************************************

Analysis of bowlers

  1. Mitchell Johnson (Aus) – Innings-150, Wickets – 239, Econ Rate : 4.83
  2. Lasith Malinga (SL)- Innings-182, Wickets – 287, Econ Rate : 5.26
  3. Dale Steyn (SA)- Innings-103, Wickets – 162, Econ Rate : 4.81
  4. Tim Southee (NZ)- Innings-96, Wickets – 135, Econ Rate : 5.33

Malinga has the highest number of innings and wickets followed closely by Mitchell. Steyn and Southee have relatively fewer innings.

To get the bowler’s data use

malinga <- getPlayerDataOD(49758,dir=".",file="malinga.csv",type="bowling")

Wicket Frequency percentage

This plot gives the percentage of wickets for each wickets (1,2,3…etc)

par(mfrow=c(1,4))
par(mar=c(4,4,2,2))
bowlerWktsFreqPercent("./mitchell.csv","J Mitchell")
bowlerWktsFreqPercent("./malinga.csv","Malinga")
bowlerWktsFreqPercent("./steyn.csv","Steyn")
bowlerWktsFreqPercent("./southee.csv","southee")

relBowlFP-1

dev.off()
## null device 
##           1

Wickets Runs plot

The plot below gives a boxplot of the runs ranges for each of the wickets taken by the bowlers. M Johnson and Steyn are more economical than Malinga and Southee corroborating the figures above

par(mfrow=c(1,4))
par(mar=c(4,4,2,2))

bowlerWktsRunsPlot("./mitchell.csv","J Mitchell")
bowlerWktsRunsPlot("./malinga.csv","Malinga")
bowlerWktsRunsPlot("./steyn.csv","Steyn")
bowlerWktsRunsPlot("./southee.csv","southee")

wktsrun-1

dev.off()
## null device 
##           1

Average wickets in different grounds and opposition

A. Mitchell Johnson

par(mfrow=c(1,2))
par(mar=c(4,4,2,2))
bowlerAvgWktsGround("./mitchell.csv","J Mitchell")
bowlerAvgWktsOpposition("./mitchell.csv","J Mitchell")

gr-1-1

dev.off()
## null device 
##           1

B. Lasith Malinga

par(mfrow=c(1,2))
par(mar=c(4,4,2,2))
bowlerAvgWktsGround("./malinga.csv","Malinga")
bowlerAvgWktsOpposition("./malinga.csv","Malinga")

gr-2-1

dev.off()
## null device 
##           1

C. Dale Steyn

par(mfrow=c(1,2))
par(mar=c(4,4,2,2))
bowlerAvgWktsGround("./steyn.csv","Steyn")
bowlerAvgWktsOpposition("./steyn.csv","Steyn")

gr-3-1

dev.off()
## null device 
##           1

D. Tim Southee

par(mfrow=c(1,2))
par(mar=c(4,4,2,2))
bowlerAvgWktsGround("./southee.csv","southee")
bowlerAvgWktsOpposition("./southee.csv","southee")

avgrg-4-1

dev.off()
## null device 
##           1

Relative bowling performance

The plot below shows that Mitchell Johnson and Southee have more wickets in 3-4 wickets range while Steyn and Malinga in 1-2 wicket range

frames <- list("./mitchell.csv","./malinga.csv","steyn.csv","southee.csv")
names <- list("M Johnson","Malinga","Steyn","Southee")
relativeBowlingPerf(frames,names)

relBowlPerf-1

Relative Economy Rate against wickets taken

Steyn had the best economy rate followed by M Johnson. Malinga and Southee have a poorer economy rate

frames <- list("./mitchell.csv","./malinga.csv","steyn.csv","southee.csv")
names <- list("M Johnson","Malinga","Steyn","Southee")
relativeBowlingERODTT(frames,names)

relBowlER-1

Moving average of wickets over career

Johnson and Steyn career vs wicket graph is on the up-swing. Southee is maintaining a reasonable record while Malinga shows a decline in ODI performance

par(mfrow=c(2,2))
par(mar=c(4,4,2,2))
bowlerMovingAverage("./mitchell.csv","M Johnson")
bowlerMovingAverage("./malinga.csv","Malinga")
bowlerMovingAverage("./steyn.csv","Steyn")
bowlerMovingAverage("./southee.csv","Southee")

jmss-bowlma-1

dev.off()
## null device 
##           1

Wickets forecast

par(mfrow=c(2,2))
par(mar=c(4,4,2,2))
bowlerPerfForecast("./mitchell.csv","M Johnson")
bowlerPerfForecast("./malinga.csv","Malinga")
bowlerPerfForecast("./steyn.csv","Steyn")
bowlerPerfForecast("./southee.csv","southee")

jsba-pfcst-1

dev.off()
## null device 
##           1

Check bowler in-form, out-of-form

All the bowlers are shown to be still in-form

checkBowlerInForm("./mitchell.csv","J Mitchell")
## *******************************************************************************************
## 
## Population size: 135  Mean of population: 1.55 
## Sample size: 15  Mean of sample: 2 SD of sample: 1.07 
## 
## Null hypothesis H0 : J Mitchell 's sample average is within 95% confidence interval 
##         of population average
## Alternative hypothesis Ha : J Mitchell 's sample average is below the 95% confidence
##         interval of population average
## 
## [1] "J Mitchell 's Form Status: In-Form because the p value: 0.937917  is greater than alpha=  0.05"
## *******************************************************************************************
checkBowlerInForm("./malinga.csv","Malinga")
## *******************************************************************************************
## 
## Population size: 163  Mean of population: 1.58 
## Sample size: 19  Mean of sample: 1.58 SD of sample: 1.22 
## 
## Null hypothesis H0 : Malinga 's sample average is within 95% confidence interval 
##         of population average
## Alternative hypothesis Ha : Malinga 's sample average is below the 95% confidence
##         interval of population average
## 
## [1] "Malinga 's Form Status: In-Form because the p value: 0.5  is greater than alpha=  0.05"
## *******************************************************************************************
checkBowlerInForm("./steyn.csv","Steyn")
## *******************************************************************************************
## 
## Population size: 93  Mean of population: 1.59 
## Sample size: 11  Mean of sample: 1.45 SD of sample: 0.69 
## 
## Null hypothesis H0 : Steyn 's sample average is within 95% confidence interval 
##         of population average
## Alternative hypothesis Ha : Steyn 's sample average is below the 95% confidence
##         interval of population average
## 
## [1] "Steyn 's Form Status: In-Form because the p value: 0.257438  is greater than alpha=  0.05"
## *******************************************************************************************
checkBowlerInForm("./southee.csv","southee")
## *******************************************************************************************
## 
## Population size: 86  Mean of population: 1.48 
## Sample size: 10  Mean of sample: 0.8 SD of sample: 1.14 
## 
## Null hypothesis H0 : southee 's sample average is within 95% confidence interval 
##         of population average
## Alternative hypothesis Ha : southee 's sample average is below the 95% confidence
##         interval of population average
## 
## [1] "southee 's Form Status: Out-of-Form because the p value: 0.044302  is less than alpha=  0.05"
## *******************************************************************************************

***************

Key findings

Here are some key conclusions ODI batsmen

  1. AB Devilliers has high frequency of runs in the 60-120 range and the highest average
  2. Sehwag has the most number of innings and good strike rate
  3. Maxwell has the best strike rate but it should be kept in mind that he has 1/5 of the innings of Sehwag. We need to see how he progress further
  4. Sehwag has the highest percentage of 4s in the runs scored, while Maxwell has the most 6s
  5. For a hypothetical Balls Faced and Minutes at creases Maxwell will score the most runs followed by Sehwag
  6. The moving average of indicates that the best is yet to come for Devilliers and Maxwell. Sehwag has a few more years in him while Gayle shows a decline in ODI performance and an out of form is indicated.

ODI bowlers

  1. Malinga has the highest played the highest innings and also has the highest wickets though he has poor economy rate
  2. M Johnson is the most effective in the 3-4 wicket range followed by Southee
  3. M Johnson and Steyn has the best overall economy rate followed by Malinga and Steyn 4 M Johnson and Steyn’s career is on the up-swing,Southee maintains a steady consistent performance, while Malinga shows a downward trend

Hasta la vista! I’ll be back!
Watch this space!

Also see my other posts in R

  1. Introducing cricketr! : An R package to analyze performances of cricketers
  2. cricketr digs the Ashes!
  3. A peek into literacy in India: Statistical Learning with R
  4. A crime map of India in R – Crimes against women
  5. Analyzing cricket’s batting legends – Through the mirage with R
  6. Mirror, mirror . the best batsman of them all?

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