Category: backward fit

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

Are you wondering whether to get into the ‘R’ bus or ‘Python’ bus?
My suggestion is to you is “Why not get into the ‘R and Python’ train?”

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 ($8.99/Rs449) versions.  In the third edition all code sections have been re-formatted to use the fixed width font ‘Consolas’. This neatly organizes output which have columns like confusion matrix, dataframes etc to be columnar, making the code more readable.  There is a science to formatting too!! which improves the look and feel. It is little wonder that Steve Jobs had a keen passion for calligraphy! Additionally some typos have been fixed.

 

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 Python Third Edition – Machine Learning in Stereo(Kindle- $8.99/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

Table of Contents
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

Pick up your copy today!!
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

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

To see all post click Index of posts

Deep Learning from first principles in Python, R and Octave – Part 4

In this 4th post of my series on Deep Learning from first principles in Python, R and Octave – Part 4, I explore the details of creating a multi-class classifier using the Softmax activation unit in a neural network. The earlier posts in this series were

1. Deep Learning from first principles in Python, R and Octave – Part 1. In this post I implemented logistic regression as a simple Neural Network in vectorized Python, R and Octave
2. Deep Learning from first principles in Python, R and Octave – Part 2. This 2nd part implemented the most elementary neural network with 1 hidden layer and any number of activation units in the hidden layer with sigmoid activation at the output layer
3. Deep Learning from first principles in Python, R and Octave – Part 3. The 3rd implemented a multi-layer Deep Learning network with an arbitrary number if hidden layers and activation units per hidden layer. The output layer was for binary classification which was based on the sigmoid unit. This multi-layer deep network was implemented in vectorized Python, R and Octave.

Checkout my book ‘Deep Learning from first principles: Second Edition – In vectorized Python, R and Octave’. My book starts with the implementation of a simple 2-layer Neural Network and works its way to a generic L-Layer Deep Learning Network, with all the bells and whistles. The derivations have been discussed in detail. The code has been extensively commented and included in its entirety in the Appendix sections. My book is available on Amazon as paperback ($18.99) and in kindle version($9.99/Rs449).

This 4th part takes a swing at multi-class classification and uses the Softmax as the activation unit in the output layer. Inclusion of the Softmax activation unit in the activation layer requires us to compute the derivative of Softmax, or rather the “Jacobian” of the Softmax function, besides also computing the log loss for this Softmax activation during back propagation. Since the derivation of the Jacobian of a Softmax and the computation of the Cross Entropy/log loss is very involved, I have implemented a basic neural network with just 1 hidden layer with the Softmax activation at the output layer. I also perform multi-class classification based on the ‘spiral’ data set from CS231n Convolutional Neural Networks Stanford course, to test the performance and correctness of the implementations in Python, R and Octave. You can clone download the code for the Python, R and Octave implementations from Github at Deep Learning – Part 4

Note: A detailed discussion of the derivation below can also be seen in my video presentation Neural Networks 5

The Softmax function takes an N dimensional vector as input and generates a N dimensional vector as output.
The Softmax function is given by
S_{j}= \frac{e_{j}}{\sum_{i}^{N}e_{k}}
There is a probabilistic interpretation of the Softmax, since the sum of the Softmax values of a set of vectors will always add up to 1, given that each Softmax value is divided by the total of all values.

As mentioned earlier, the Softmax takes a vector input and returns a vector of outputs.  For e.g. the Softmax of a vector a=[1, 3, 6]  is another vector S=[0.0063,0.0471,0.9464]. Notice that vector output is proportional to the input vector.  Also, taking the derivative of a vector by another vector, is known as the Jacobian. By the way, The Matrix Calculus You Need For Deep Learning by Terence Parr and Jeremy Howard, is very good paper that distills all the main mathematical concepts for Deep Learning in one place.

Let us take a simple 2 layered neural network with just 2 activation units in the hidden layer is shown below

Z_{1}^{1} =W_{11}^{1}x_{1} + W_{21}^{1}x_{2} + b_{1}^{1}
Z_{2}^{1} =W_{12}^{1}x_{1} + W_{22}^{1}x_{2} + b_{2}^{1}
and
A_{1}^{1} = g'(Z_{1}^{1})
A_{2}^{1} = g'(Z_{2}^{1})
where g'() is the activation unit in the hidden layer which can be a relu, sigmoid or a
tanh function

Note: The superscript denotes the layer. The above denotes the equation for layer 1
of the neural network. For layer 2 with the Softmax activation, the equations are
Z_{1}^{2} =W_{11}^{2}x_{1} + W_{21}^{2}x_{2} + b_{1}^{2}
Z_{2}^{2} =W_{12}^{2}x_{1} + W_{22}^{2}x_{2} + b_{2}^{2}
and
A_{1}^{2} = S(Z_{1}^{2})
A_{2}^{2} = S(Z_{2}^{2})
where S() is the Softmax activation function
S=\begin{pmatrix} S(Z_{1}^{2})\\ S(Z_{2}^{2}) \end{pmatrix}
S=\begin{pmatrix} \frac{e^{Z1}}{e^{Z1}+e^{Z2}}\\ \frac{e^{Z2}}{e^{Z1}+e^{Z2}} \end{pmatrix}

The Jacobian of the softmax ‘S’ is given by
\begin{pmatrix} \frac {\partial S_{1}}{\partial Z_{1}} & \frac {\partial S_{1}}{\partial Z_{2}}\\ \frac {\partial S_{2}}{\partial Z_{1}} & \frac {\partial S_{2}}{\partial Z_{2}} \end{pmatrix}
\begin{pmatrix} \frac{\partial}{\partial Z_{1}} \frac {e^{Z1}}{e^{Z1}+ e^{Z2}} & \frac{\partial}{\partial Z_{2}} \frac {e^{Z1}}{e^{Z1}+ e^{Z2}}\\ \frac{\partial}{\partial Z_{1}} \frac {e^{Z2}}{e^{Z1}+ e^{Z2}} & \frac{\partial}{\partial Z_{2}} \frac {e^{Z2}}{e^{Z1}+ e^{Z2}} \end{pmatrix}     – (A)

Now the ‘division-rule’  of derivatives is as follows. If u and v are functions of x, then
\frac{d}{dx} \frac {u}{v} =\frac {vdu -udv}{v^{2}}
Using this to compute each element of the above Jacobian matrix, we see that
when i=j we have
\frac {\partial}{\partial Z1}\frac{e^{Z1}}{e^{Z1}+e^{Z2}} = \frac {\sum e^{Z1} - e^{Z1^{2}}}{\sum ^{2}}
and when i \neq j
\frac {\partial}{\partial Z1}\frac{e^{Z2}}{e^{Z1}+e^{Z2}} = \frac {0 - e^{z1}e^{Z2}}{\sum ^{2}}
This is of the general form
\frac {\partial S_{j}}{\partial z_{i}} = S_{i}( 1-S_{j})  when i=j
and
\frac {\partial S_{j}}{\partial z_{i}} = -S_{i}S_{j}  when i \neq j
Note: Since the Softmax essentially gives the probability the following
notation is also used
\frac {\partial p_{j}}{\partial z_{i}} = p_{i}( 1-p_{j}) when i=j
and
\frac {\partial p_{j}}{\partial z_{i}} = -p_{i}p_{j} when i \neq j
If you throw the “Kronecker delta” into the equation, then the above equations can be expressed even more concisely as
\frac {\partial p_{j}}{\partial z_{i}} = p_{i} (\delta_{ij} - p_{j})
where \delta_{ij} = 1 when i=j and 0 when i \neq j

This reduces the Jacobian of the simple 2 output softmax vectors  equation (A) as
\begin{pmatrix} p_{1}(1-p_{1}) & -p_{1}p_{2} \\ -p_{2}p_{1} & p_{2}(1-p_{2}) \end{pmatrix}
The loss of Softmax is given by
L = -\sum y_{i} log(p_{i})
For the 2 valued Softmax output this is
\frac {dL}{dp1} = -\frac {y_{1}}{p_{1}}
\frac {dL}{dp2} = -\frac {y_{2}}{p_{2}}
Using the chain rule we can write
\frac {\partial L}{\partial w_{pq}} = \sum _{i}\frac {\partial L}{\partial p_{i}} \frac {\partial p_{i}}{\partial w_{pq}} (1)
and
\frac {\partial p_{i}}{\partial w_{pq}} = \sum _{k}\frac {\partial p_{i}}{\partial z_{k}} \frac {\partial z_{k}}{\partial w_{pq}} (2)
In expanded form this is
\frac {\partial L}{\partial w_{pq}} = \sum _{i}\frac {\partial L}{\partial p_{i}} \sum _{k}\frac {\partial p_{i}}{\partial z_{k}} \frac {\partial z_{k}}{\partial w_{pq}}
Also
\frac {\partial L}{\partial Z_{i}} =\sum _{i} \frac {\partial L}{\partial p} \frac {\partial p}{\partial Z_{i}}
Therefore
\frac {\partial L}{\partial Z_{1}} =\frac {\partial L}{\partial p_{1}} \frac {\partial p_{1}}{\partial Z_{1}} +\frac {\partial L}{\partial p_{2}} \frac {\partial p_{2}}{\partial Z_{1}}
\frac {\partial L}{\partial z_{1}}=-\frac {y1}{p1} p1(1-p1) - \frac {y2}{p2}*(-p_{2}p_{1})
Since
\frac {\partial p_{j}}{\partial z_{i}} = p_{i}( 1-p_{j}) when i=j
and
\frac {\partial p_{j}}{\partial z_{i}} = -p_{i}p_{j} when i \neq j
which simplifies to
\frac {\partial L}{\partial Z_{1}} = -y_{1} + y_{1}p_{1} + y_{2}p_{1} =
p_{1}\sum (y_{1} + y_2) - y_{1}
\frac {\partial L}{\partial Z_{1}}= p_{1} - y_{1}
Since
\sum_{i} y_{i} =1
Similarly
\frac {\partial L}{\partial Z_{2}} =\frac {\partial L}{\partial p_{1}} \frac {\partial p_{1}}{\partial Z_{2}} +\frac {\partial L}{\partial p_{2}} \frac {\partial p_{2}}{\partial Z_{2}}
\frac {\partial L}{\partial z_{2}}=-\frac {y1}{p1}*(p_{1}p_{2}) - \frac {y2}{p2}*p_{2}(1-p_{2})
y_{1}p_{2} + y_{2}p_{2} - y_{2}
\frac {\partial L}{\partial Z_{2}} =p_{2}\sum (y_{1} + y_2) - y_{2}\\ = p_{2} - y_{2}
In general this is of the form
\frac {\partial L}{\partial z_{i}} = p_{i} -y_{i}
For e.g if the probabilities computed were p=[0.1, 0.7, 0.2] then this implies that the class with probability 0.7 is the likely class. This would imply that the ‘One hot encoding’ for  yi  would be yi=[0,1,0] therefore the gradient pi-yi = [0.1,-0.3,0.2]

<strong>Note: Further, we could extend this derivation for a Softmax activation output that outputs 3 classes
S=\begin{pmatrix} \frac{e^{z1}}{e^{z1}+e^{z2}+e^{z3}}\\ \frac{e^{z2}}{e^{z1}+e^{z2}+e^{z3}} \\ \frac{e^{z3}}{e^{z1}+e^{z2}+e^{z3}} \end{pmatrix}

We could derive
\frac {\partial L}{\partial z1}= \frac {\partial L}{\partial p_{1}} \frac {\partial p_{1}}{\partial z_{1}} +\frac {\partial L}{\partial p_{2}} \frac {\partial p_{2}}{\partial z_{1}} +\frac {\partial L}{\partial p_{3}} \frac {\partial p_{3}}{\partial z_{1}} which similarly reduces to
\frac {\partial L}{\partial z_{1}}=-\frac {y1}{p1} p1(1-p1) - \frac {y2}{p2}*(-p_{2}p_{1}) - \frac {y3}{p3}*(-p_{3}p_{1})
-y_{1}+ y_{1}p_{1} + y_{2}p_{1} + y_{3}p1 = p_{1}\sum (y_{1} + y_2 + y_3) - y_{1} = p_{1} - y_{1}
Interestingly, despite the lengthy derivations the final result is simple and intuitive!

As seen in my post ‘Deep Learning from first principles with Python, R and Octave – Part 3 the key equations for forward and backward propagation are

Forward propagation equations layer 1
Z_{1} = W_{1}X +b_{1}     and  A_{1} = g(Z_{1})
Forward propagation equations layer 1
Z_{2} = W_{2}A_{1} +b_{2}  and  A_{2} = S(Z_{2})

Using the result (A) in the back propagation equations below we have
Backward propagation equations layer 2
\partial L/\partial W_{2} =\partial L/\partial Z_{2}*A_{1}=(p_{2}-y_{2})*A_{1}
\partial L/\partial b_{2} =\partial L/\partial Z_{2}=p_{2}-y_{2}
\partial L/\partial A_{1} = \partial L/\partial Z_{2} * W_{2}=(p_{2}-y_{2})*W_{2}
Backward propagation equations layer 1
\partial L/\partial W_{1} =\partial L/\partial Z_{1} *A_{0}=(p_{1}-y_{1})*A_{0}
\partial L/\partial b_{1} =\partial L/\partial Z_{1}=(p_{1}-y_{1})

2.0 Spiral data set

As I mentioned earlier, I will be using the ‘spiral’ data from CS231n Convolutional Neural Networks to ensure that my vectorized implementations in Python, R and Octave are correct. Here is the ‘spiral’ data set.

import numpy as np
import matplotlib.pyplot as plt
import os
os.chdir("C:/junk/dl-4/dl-4")
exec(open("././DLfunctions41.py").read())

# Create an input data set - Taken from CS231n Convolutional Neural networks
# http://cs231n.github.io/neural-networks-case-study/
N = 100 # number of points per class
D = 2 # dimensionality
K = 3 # number of classes
X = np.zeros((N*K,D)) # data matrix (each row = single example)
y = np.zeros(N*K, dtype='uint8') # class labels
for j in range(K):
  ix = range(N*j,N*(j+1))
  r = np.linspace(0.0,1,N) # radius
  t = np.linspace(j*4,(j+1)*4,N) + np.random.randn(N)*0.2 # theta
  X[ix] = np.c_[r*np.sin(t), r*np.cos(t)]
  y[ix] = j
# Plot the data
plt.scatter(X[:, 0], X[:, 1], c=y, s=40, cmap=plt.cm.Spectral)
plt.savefig("fig1.png", bbox_inches='tight')


The implementations of the vectorized Python, R and Octave code are shown diagrammatically below

2.1 Multi-class classification with Softmax – Python code

A simple 2 layer Neural network with a single hidden layer , with 100 Relu activation units in the hidden layer and the Softmax activation unit in the output layer is used for multi-class classification. This Deep Learning Network, plots the non-linear boundary of the 3 classes as shown below

import numpy as np
import matplotlib.pyplot as plt
import os
os.chdir("C:/junk/dl-4/dl-4")
exec(open("././DLfunctions41.py").read())

# Read the input data
N = 100 # number of points per class
D = 2 # dimensionality
K = 3 # number of classes
X = np.zeros((N*K,D)) # data matrix (each row = single example)
y = np.zeros(N*K, dtype='uint8') # class labels
for j in range(K):
  ix = range(N*j,N*(j+1))
  r = np.linspace(0.0,1,N) # radius
  t = np.linspace(j*4,(j+1)*4,N) + np.random.randn(N)*0.2 # theta
  X[ix] = np.c_[r*np.sin(t), r*np.cos(t)]
  y[ix] = j
  
# Set the number of features, hidden units in hidden layer and number of classess
numHidden=100 # No of hidden units in hidden layer
numFeats= 2 # dimensionality
numOutput = 3 # number of classes

# Initialize the model
parameters=initializeModel(numFeats,numHidden,numOutput)
W1= parameters['W1']
b1= parameters['b1']
W2= parameters['W2']
b2= parameters['b2']

# Set the learning rate
learningRate=0.6 

# Initialize losses
losses=[]
# Perform Gradient descent
for i in range(10000):
    # Forward propagation through hidden layer with Relu units
    A1,cache1= layerActivationForward(X.T,W1,b1,'relu')
    
    # Forward propagation through output layer with Softmax
    A2,cache2 = layerActivationForward(A1,W2,b2,'softmax')
    
    # No of training examples
    numTraining = X.shape[0]
    # Compute log probs. Take the log prob of correct class based on output y
    correct_logprobs = -np.log(A2[range(numTraining),y])
    # Conpute loss
    loss = np.sum(correct_logprobs)/numTraining
    
    # Print the loss
    if i % 1000 == 0:
        print("iteration %d: loss %f" % (i, loss))
        losses.append(loss)

    dA=0

    # Backward  propagation through output layer with Softmax
    dA1,dW2,db2 = layerActivationBackward(dA, cache2, y, activationFunc='softmax')
    # Backward  propagation through hidden layer with Relu unit
    dA0,dW1,db1 = layerActivationBackward(dA1.T, cache1, y, activationFunc='relu')
    
    #Update paramaters with the learning rate
    W1 += -learningRate * dW1
    b1 += -learningRate * db1
    W2 += -learningRate * dW2.T
    b2 += -learningRate * db2.T

#Plot losses vs iterations  
i=np.arange(0,10000,1000)
plt.plot(i,losses)

plt.xlabel('Iterations')
plt.ylabel('Loss')
plt.title('Losses vs Iterations')
plt.savefig("fig2.png", bbox="tight")

#Compute the multi-class Confusion Matrix
from sklearn.metrics import confusion_matrix
from sklearn.metrics import accuracy_score, precision_score, recall_score, f1_score

# We need to determine the predicted values from the learnt data
# Forward propagation through hidden layer with Relu units
A1,cache1= layerActivationForward(X.T,W1,b1,'relu')
    
# Forward propagation through output layer with Softmax
A2,cache2 = layerActivationForward(A1,W2,b2,'softmax')
#Compute predicted values from weights and biases
yhat=np.argmax(A2, axis=1)

a=confusion_matrix(y.T,yhat.T)
print("Multi-class Confusion Matrix")
print(a)
## iteration 0: loss 1.098507
## iteration 1000: loss 0.214611
## iteration 2000: loss 0.043622
## iteration 3000: loss 0.032525
## iteration 4000: loss 0.025108
## iteration 5000: loss 0.021365
## iteration 6000: loss 0.019046
## iteration 7000: loss 0.017475
## iteration 8000: loss 0.016359
## iteration 9000: loss 0.015703
## Multi-class Confusion Matrix
## [[ 99   1   0]
##  [  0 100   0]
##  [  0   1  99]]

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

2.2 Multi-class classification with Softmax – R code

The spiral data set created with Python was saved, and is used as the input with R code. The R Neural Network seems to perform much,much slower than both Python and Octave. Not sure why! Incidentally the computation of loss and the softmax derivative are identical for both R and Octave. yet R is much slower. To compute the softmax derivative I create matrices for the One Hot Encoded yi and then stack them before subtracting pi-yi. I am sure there is a more elegant and more efficient way to do this, much like Python. Any suggestions?

library(ggplot2)
library(dplyr)
library(RColorBrewer)
source("DLfunctions41.R")
# Read the spiral dataset
Z <- as.matrix(read.csv("spiral.csv",header=FALSE)) 
Z1=data.frame(Z)
#Plot the dataset
ggplot(Z1,aes(x=V1,y=V2,col=V3)) +geom_point() + 
  scale_colour_gradientn(colours = brewer.pal(10, "Spectral"))

# Setup the data
X <- Z[,1:2]
y <- Z[,3]
X1 <- t(X)
Y1 <- t(y)

# Initialize number of features, number of hidden units in hidden layer and
# number of classes
numFeats<-2 # No features
numHidden<-100 # No of hidden units
numOutput<-3 # No of classes

# Initialize model
parameters <-initializeModel(numFeats, numHidden,numOutput)

W1 <-parameters[['W1']]
b1 <-parameters[['b1']]
W2 <-parameters[['W2']]
b2 <-parameters[['b2']]

# Set the learning rate
learningRate <- 0.5
# Initialize losses
losses <- NULL
# Perform gradient descent
for(i in 0:9000){

# Forward propagation through hidden layer with Relu units
retvals <- layerActivationForward(X1,W1,b1,'relu')
A1 <- retvals[['A']]
cache1 <- retvals[['cache']]
forward_cache1 <- cache1[['forward_cache1']]
activation_cache <- cache1[['activation_cache']]

# Forward propagation through output layer with Softmax units
retvals = layerActivationForward(A1,W2,b2,'softmax')
A2 <- retvals[['A']]
cache2 <- retvals[['cache']]
forward_cache2 <- cache2[['forward_cache1']]
activation_cache2 <- cache2[['activation_cache']]

# No oftraining examples
numTraining <- dim(X)[1]
dA <-0

# Select the elements where the y values are 0, 1 or 2 and make a vector
a=c(A2[y==0,1],A2[y==1,2],A2[y==2,3])
# Take log
correct_probs = -log(a)
# Compute loss
loss= sum(correct_probs)/numTraining

if(i %% 1000 == 0){
sprintf("iteration %d: loss %f",i, loss)
print(loss)
}
# Backward propagation through output layer with Softmax units
retvals = layerActivationBackward(dA, cache2, y, activationFunc='softmax')
dA1 = retvals[['dA_prev']]
dW2= retvals[['dW']]
db2= retvals[['db']]
# Backward propagation through hidden layer with Relu units
retvals = layerActivationBackward(t(dA1), cache1, y, activationFunc='relu')
dA0 = retvals[['dA_prev']]
dW1= retvals[['dW']]
db1= retvals[['db']]

# Update parameters
W1 <- W1 - learningRate * dW1
b1 <- b1 - learningRate * db1
W2 <- W2 - learningRate * t(dW2)
b2 <- b2 - learningRate * t(db2)
}
## [1] 1.212487
## [1] 0.5740867
## [1] 0.4048824
## [1] 0.3561941
## [1] 0.2509576
## [1] 0.7351063
## [1] 0.2066114
## [1] 0.2065875
## [1] 0.2151943
## [1] 0.1318807

 

#Create iterations
iterations <- seq(0,10)
#df=data.frame(iterations,losses)
ggplot(df,aes(x=iterations,y=losses)) + geom_point() + geom_line(color="blue") +
    ggtitle("Losses vs iterations") + xlab("Iterations") + ylab("Loss")

plotDecisionBoundary(Z,W1,b1,W2,b2)



Multi-class Confusion Matrix

library(caret)
library(e1071)

# Forward propagation through hidden layer with Relu units
retvals <- layerActivationForward(X1,W1,b1,'relu')
A1 <- retvals[['A']]

# Forward propagation through output layer with Softmax units
retvals = layerActivationForward(A1,W2,b2,'softmax')
A2 <- retvals[['A']]
yhat <- apply(A2, 1,which.max) -1
Confusion Matrix and Statistics
          Reference
Prediction  0  1  2
         0 97  0  1
         1  2 96  4
         2  1  4 95

Overall Statistics                                        
               Accuracy : 0.96            
                 95% CI : (0.9312, 0.9792)
    No Information Rate : 0.3333          
    P-Value [Acc > NIR] : <2e-16          
                                          
                  Kappa : 0.94            
 Mcnemar's Test P-Value : 0.5724          
Statistics by Class:

                     Class: 0 Class: 1 Class: 2
Sensitivity            0.9700   0.9600   0.9500
Specificity            0.9950   0.9700   0.9750
Pos Pred Value         0.9898   0.9412   0.9500
Neg Pred Value         0.9851   0.9798   0.9750
Prevalence             0.3333   0.3333   0.3333
Detection Rate         0.3233   0.3200   0.3167
Detection Prevalence   0.3267   0.3400   0.3333
Balanced Accuracy      0.9825   0.9650   0.9625

My book “Practical Machine Learning with R and Python” includes the implementation for many Machine Learning algorithms and associated metrics. Pick up your copy today!

2.3 Multi-class classification with Softmax – Octave code

A 2 layer Neural network with the Softmax activation unit in the output layer is constructed in Octave. The same spiral data set is used for Octave also

source("DL41functions.m")
# Read the spiral data
data=csvread("spiral.csv");
# Setup the data
X=data(:,1:2);
Y=data(:,3);
# Set the number of features, number of hidden units in hidden layer and number of classes
numFeats=2; #No features
numHidden=100; # No of hidden units
numOutput=3; # No of classes
# Initialize model
[W1 b1 W2 b2] = initializeModel(numFeats,numHidden,numOutput);
# Initialize losses
losses=[]
#Initialize learningRate
learningRate=0.5;
for k =1:10000
# Forward propagation through hidden layer with Relu units
[A1,cache1 activation_cache1]= layerActivationForward(X',W1,b1,activationFunc ='relu');
# Forward propagation through output layer with Softmax units
[A2,cache2 activation_cache2] =
layerActivationForward(A1,W2,b2,activationFunc='softmax');
# No of training examples
numTraining = size(X)(1);
# Select rows where Y=0,1,and 2 and concatenate to a long vector
a=[A2(Y==0,1) ;A2(Y==1,2) ;A2(Y==2,3)];
#Select the correct column for log prob
correct_probs = -log(a);
#Compute log loss
loss= sum(correct_probs)/numTraining;
if(mod(k,1000) == 0)
disp(loss);
losses=[losses loss];
endif
dA=0;
# Backward propagation through output layer with Softmax units
[dA1 dW2 db2] = layerActivationBackward(dA, cache2, activation_cache2,Y,activationFunc='softmax');
# Backward propagation through hidden layer with Relu units
[dA0,dW1,db1] = layerActivationBackward(dA1', cache1, activation_cache1, Y, activationFunc='relu');
#Update parameters
W1 += -learningRate * dW1;
b1 += -learningRate * db1;
W2 += -learningRate * dW2';
b2 += -learningRate * db2';
endfor
# Plot Losses vs Iterations
iterations=0:1000:9000
plotCostVsIterations(iterations,losses)
# Plot the decision boundary
plotDecisionBoundary( X,Y,W1,b1,W2,b2)

The code for the Python, R and Octave implementations can be downloaded from Github at Deep Learning – Part 4

Conclusion

In this post I have implemented a 2 layer Neural Network with the Softmax classifier. In Part 3, I implemented a multi-layer Deep Learning Network. I intend to include the Softmax activation unit into the generalized multi-layer Deep Network along with the other activation units of sigmoid,tanh and relu.

Stick around, I’ll be back!!
Watch this space!

References
1. Deep Learning Specialization
2. Neural Networks for Machine Learning
3. CS231 Convolutional Neural Networks for Visual Recognition
4. Eli Bendersky’s Website – The Softmax function and its derivative
5. Cross Validated – Backpropagation with Softmax / Cross Entropy
6. Stackoverflow – CS231n: How to calculate gradient for Softmax loss function?
7. Math Stack Exchange – Derivative of Softmax
8. The Matrix Calculus for Deep Learning

You may like
1.My book ‘Practical Machine Learning with R and Python’ on Amazon
2. My travels through the realms of Data Science, Machine Learning, Deep Learning and (AI)
3. Deblurring with OpenCV: Weiner filter reloaded
4. A method to crowd source pothole marking on (Indian) roads
5. Rock N’ Roll with Bluemix, Cloudant & NodeExpress
6. Sea shells on the seashore
7. Design Principles of Scalable, Distributed Systems

To see all post click Index of posts

My book ‘Practical Machine Learning with R and Python’ 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

My book ‘Practical Machine Learning with R and Python: Second Edition – Machine Learning in stereo’ is now available in both paperback ($10.99) and kindle ($7.99/Rs449) versions. In this book I implement some of the most common, but important Machine Learning algorithms in R and equivalent Python code. This is almost like listening to parallel channels of music in stereo!
1. Practical machine with R and Python: Third Edition – Machine Learning in Stereo(Paperback-$12.99)
2. Practical machine with R and Python Third Edition – Machine Learning in Stereo(Kindle- $8.99/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

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

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

Practical Machine Learning with R and Python – Part 3

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

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

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

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

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

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

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

Note: Please listen to my video presentations Machine Learning in youtube
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')

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

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

1.3 Backward Fit

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

1.3a Backward fit – R code

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

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

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

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

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

1.3b Backward fit – Python code

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

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

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

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

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

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

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

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

# Get the index of minimum cross validation error
idx = np.argmin(n)
print "No of features=",idx
#Get the features indices for the best forward fit and convert to list
b=list(a[idx]['feature_idx'])
# Index the column names. 
# Features from backward fit
print("Features selected in bacward fit")
print(X.columns[b])

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

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

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

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

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

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


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

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

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


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

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

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

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

SFFS provides the best fit with 10 predictors

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

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

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


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

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

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

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

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

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

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

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

SFBS indicates that 10 features are needed for the best fit

1.4 Ridge regression

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

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

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

Ridge regression in R requires the ‘glmnet’ package

1.4a Ridge Regression – R code

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

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

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

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

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

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

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

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

1.4a Ridge Regression – Python code

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

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


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

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

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

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

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

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

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

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

# Plot the coefficient shrinage using the LARS package

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

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

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

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

ax = plt.gca()

fig6=ax.plot(alphas, coefs)
fig6=ax.set_xscale('log')
fig6=ax.set_xlim(ax.get_xlim()[::-1])  # reverse axis
fig6=plt.xlabel('alpha')
fig6=plt.ylabel('weights')
fig6=plt.title('Ridge coefficients as a function of the regularization')
fig6=plt.axis('tight')
plt.savefig('fig6.png', bbox_inches='tight')

## R-squared score (training): 0.620
## R-squared score (test): 0.438
## Number of non-zero features: 13
## Ridge regression: effect of alpha regularization parameter

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

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

1.5 Lasso regularization

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

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

1.5a Lasso regularization – R code

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

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

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

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

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

This gives the MSE for the lasso model

1.5 b Lasso regularization – Python code

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

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

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

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

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

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

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

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

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

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




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

The plot below gives the training and test R squared error

1.5c Lasso coefficient shrinkage – Python code

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

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

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


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

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

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

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

 

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