# My book ‘Practical Machine Learning in R and Python: Second edition’ on Amazon

Note: The 3rd edition of this book is now available My book ‘Practical Machine Learning in R and Python: Third edition’ on Amazon

The third edition of my book ‘Practical Machine Learning with R and Python – Machine Learning in stereo’ is now available in both paperback ($12.99) and kindle ($9.99/Rs449) versions.  This second edition includes more content,  extensive comments and formatting for better readability.

In this book I implement some of the most common, but important Machine Learning algorithms in R and equivalent Python code.
1. Practical machine with R and Python: Third Edition – Machine Learning in Stereo(Paperback-$12.99) 2. Practical machine with R and Third Edition – Machine Learning in Stereo(Kindle-$9.99/Rs449)

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

Preface …………………………………………………………………………….4
Introduction ………………………………………………………………………6
1. Essential R ………………………………………………………………… 8
2. Essential Python for Datascience ……………………………………………57
3. R vs Python …………………………………………………………………81
4. Regression of a continuous variable ……………………………………….101
5. Classification and Cross Validation ………………………………………..121
6. Regression techniques and regularization ………………………………….146
7. SVMs, Decision Trees and Validation curves ………………………………191
8. Splines, GAMs, Random Forests and Boosting ……………………………222
9. PCA, K-Means and Hierarchical Clustering ………………………………258
References ……………………………………………………………………..269

Hope you have a great time learning as I did while implementing these algorithms!

# 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

To see all post click Index of posts

# 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

1. Machine Learning in plain English-Part 1
2. Machine Learning in plain English-Part 2
3. Machine Learning in plain English-Part 3

Check out my compact and minimal book  “Practical Machine Learning with R and Python:Third edition- Machine Learning in stereo”  available in Amazon in paperback($12.99) and kindle($8.99) 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')
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)

## 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
# 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')

## 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$se,lty="dashed") abline(v=c(60,75,100,150),lty=2,col="darkgreen") ## 1.1d Fit a Natural Spline – R Code Here a ‘Natural Spline’ is used to fit .The Natural Spline extrapolates beyond the boundary knots and the ends of the function are much more constrained than a regular spline or a global polynomoial where the ends can wag a lot more. Natural splines do not require the explicit selection of knots # There is no need to select the knots here. There is a smoothing parameter which # can be specified by the degrees of freedom 'df' parameter. The natural spline fit2=lm(mpg~ns(horsepower,df=4),data=auto) pred=predict(fit2,newdata=list(horsepower=hpGrid),se=T) plot(auto$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")

## 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)

## 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.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')


## 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)

## 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 <- 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$target) ## Confusion Matrix and Statistics ## ## Reference ## Prediction 0 1 ## 0 49 7 ## 1 8 78 ## ## Accuracy : 0.8944 ## 95% CI : (0.8318, 0.9397) ## No Information Rate : 0.5986 ## P-Value [Acc > NIR] : 4.641e-15 ## ## Kappa : 0.7795 ## Mcnemar's Test P-Value : 1 ## ## Sensitivity : 0.8596 ## Specificity : 0.9176 ## Pos Pred Value : 0.8750 ## Neg Pred Value : 0.9070 ## Prevalence : 0.4014 ## Detection Rate : 0.3451 ## Detection Prevalence : 0.3944 ## Balanced Accuracy : 0.8886 ## ## 'Positive' Class : 0 ##  # Plot decision tree with labels plot(cancerStatus) text(cancerStatus,pretty=0) ## 1.31b Decision Trees – Cross Validation – R Code We can also perform a Cross Validation on the data to identify the Decision Tree which will give the minimum deviance. library(tree) cancer <- read.csv("cancer.csv",stringsAsFactors = FALSE) cancer <- cancer[,2:32] cancer$target <- as.factor(cancer$target) train_idx <- trainTestSplit(cancer,trainPercent=75,seed=5) train <- cancer[train_idx, ] test <- cancer[-train_idx, ] # Create Decision Tree cancerStatus=tree(target~.,train) # Execute 10 fold cross validation cvCancer=cv.tree(cancerStatus) plot(cvCancer) # Plot the plot(cvCancer$size,cvCancer$dev,type='b') prunedCancer=prune.tree(cancerStatus,best=4) plot(prunedCancer) text(prunedCancer,pretty=0) pred <- predict(prunedCancer,newdata=test,type="class") confusionMatrix(pred,test$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.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

(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

## 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.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')

## 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)

## 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"))

## 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

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]

## 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

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')


## 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")

# 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$medianValue)^2,2,mean) #Plot the MSE vs the number of trees plot(nTrees,boostError,pch=19,col="blue",ylab="Mean Squared Error", main="Boosting Test Error") ## 1.5b Cross Validation Boosting – R code Included below is a cross validation error vs the learning rate. The lowest error is when learning rate = 0.09 cvError <- NULL s <- c(.001,0.01,0.03,0.05,0.07,0.09,0.1) for(i in seq_along(s)){ cvBoost=gbm(medianValue~.,data=train,distribution="gaussian",n.trees=5000, shrinkage=s[i],interaction.depth=4,cv.folds=5) cvError[i] <- mean(cvBoost$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")

## 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

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.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.model_selection import cross_val_score

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:
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')

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

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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

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.

1. Machine Learning in plain English-Part 1
2. Machine Learning in plain English-Part 2
3. Machine Learning in plain English-Part 3

Check out my compact and minimal book  “Practical Machine Learning with R and Python:Third edition- Machine Learning in stereo”  available in Amazon in paperback($12.99) and kindle($8.99) 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') ## 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
}
# 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

#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')

## 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 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

#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')

## 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 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

#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')

# 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)
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()
#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)

## 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()
#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 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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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.

1. Machine Learning in plain English-Part 1
2. Machine Learning in plain English-Part 2
3. Machine Learning in plain English-Part 3

Check out my compact and minimal book  “Practical Machine Learning with R and Python:Third edition- Machine Learning in stereo”  available in Amazon in paperback($12.99) and kindle($8.99) 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)
# 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
(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
capital.loss,hours.per.week,native.country,salary)
# Set the dummy data with appropriate values

#Split as training and test

# 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
# Drop rows with NA
df1=df.dropna()
print(df1.shape)
# Select specific columns
'hours-per-week','native-country','salary']]

'hours-per-week','native-country']]
# Set approporiate values for dummy variables

random_state = 0)

# Compute and display Accuracy and Confusion matrix
print('Accuracy of Logistic regression classifier on training set: {:.2f}'
print('Accuracy of Logistic regression classifier on test set: {:.2f}'
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")
# Remove rows which have NA
df1 <- df[complete.cases(df),]
dim(df1)
## [1] 30161    16
# Select specific columns
capital.loss,hours.per.week,native.country,salary)
# Set dummy variables

#Split train and test as required by KNN classsification model
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

df1=df.dropna()
print(df1.shape)
# Select specific columns
'hours-per-week','native-country','salary']]

'hours-per-week','native-country']]

#Set values for dummy variables

random_state = 0)

# KNN classification in Python requires the data to be scaled.
# Scale the data
scaler = MinMaxScaler()
# Apply scaling to test set also
# 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.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)
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
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')


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

To see all posts see Index of posts