# 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")

# 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))
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")

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))
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=[]
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]]

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#### 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")
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
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


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#### 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!

To see all post click Index of posts

# Deep Learning from first principles in Python, R and Octave – Part 3

“Once upon a time, I, Chuang Tzu, dreamt I was a butterfly, fluttering hither and thither, to all intents and purposes a butterfly. I was conscious only of following my fancies as a butterfly, and was unconscious of my individuality as a man. Suddenly, I awoke, and there I lay, myself again. Now I do not know whether I was then a man dreaming I was a butterfly, or whether I am now a butterfly dreaming that I am a man.”
from The Brain: The Story of you – David Eagleman

“Thought is a great big vector of neural activity”
Prof Geoffrey Hinton

# Introduction

This is the third part in my series on Deep Learning from first principles in Python, R and Octave. In the first part Deep Learning from first principles in Python, R and Octave-Part 1, I implemented logistic regression as a 2 layer neural network. The 2nd part Deep Learning from first principles in Python, R and Octave-Part 2, dealt with the implementation of 3 layer Neural Networks with 1 hidden layer to perform classification tasks, where the 2 classes cannot be separated by a linear boundary. In this third part, I implement a multi-layer, Deep Learning (DL) network of arbitrary depth (any number of hidden layers) and arbitrary height (any number of activation units in each hidden layer). The implementations of these Deep Learning networks, in all the 3 parts, are based on vectorized versions in Python, R and Octave. The implementation in the 3rd part is for a L-layer Deep Netwwork, but without any regularization, early stopping, momentum or learning rate adaptation techniques. However even the barebones multi-layer DL, is a handful and has enough hyperparameters to fine-tune and adjust.

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).

The implementation of the vectorized L-layer Deep Learning network in Python, R and Octave were both exhausting, and exacting!! Keeping track of the indices, layer number and matrix dimensions required quite bit of focus. While the implementation was demanding, it was also very exciting to get the code to work. The trick was to be able to shift gears between the slight quirkiness between the languages. Here are some of challenges I faced.

1. Python and Octave allow multiple return values to be unpacked in a single statement. With R, unpacking multiple return values from a list, requires the list returned, to be unpacked separately. I did see that there is a package gsubfn, which does this.  I hope this feature becomes a base R feature.
2. Python and R allow dissimilar elements to be saved and returned from functions using dictionaries or lists respectively. However there is no real equivalent in Octave. The closest I got to this functionality in Octave, was the ‘cell array’. But the cell array can be accessed only by the index, and not with the key as in a Python dictionary or R list. This makes things just a bit more difficult in Octave.
3. Python and Octave include implicit broadcasting. In R, broadcasting is not implicit, but R has a nifty function, the sweep(), with which we can broadcast either by columns or by rows
4. The closest equivalent of Python’s dictionary, or R’s list, in Octave is the cell array. However I had to manage separate cell arrays for weights and biases and during gradient descent and separate gradients dW and dB
5. In Python the rank-1 numpy arrays can be annoying at times. This issue is not present in R and Octave.

Though the number of lines of code for Deep Learning functions in Python, R and Octave are about ~350 apiece, they have been some of the most difficult code I have implemented. The current vectorized implementation supports the relu, sigmoid and tanh activation functions as of now. I will be adding other activation functions like the ‘leaky relu’, ‘softmax’ and others, to the implementation in the weeks to come.

While testing with different hyper-parameters namely i) the number of hidden layers, ii) the number of activation units in each layer, iii) the activation function and iv) the number iterations, I found the L-layer Deep Learning Network to be very sensitive to these hyper-parameters. It is not easy to tune the parameters. Adding more hidden layers, or more units per layer, does not help and mostly results in gradient descent getting stuck in some local minima. It does take a fair amount of trial and error and very close observation on how the DL network performs for logical changes. We then can zero in on the most the optimal solution. Feel free to download/fork my code from Github DeepLearning-Part 3 and play around with the hyper-parameters for your own problems.

#### Derivation of a Multi Layer Deep Learning Network

Note: A detailed discussion of the derivation below is available in my video presentation Neural Network 4
Lets take a simple 3 layer Neural network with 3 hidden layers and an output layer

In the forward propagation cycle the equations are

$Z_{1} = W_{1}A_{0} +b_{1}$  and  $A_{1} = g(Z_{1})$
$Z_{2} = W_{2}A_{1} +b_{2}$  and  $A_{2} = g(Z_{2})$
$Z_{3} = W_{3}A_{2} +b_{3}$  and $A_{3} = g(Z_{3})$

The loss function is given by
$L = -(ylogA3 + (1-y)log(1-A3))$
and $dL/dA3 = -(Y/A_{3} + (1-Y)/(1-A_{3}))$

For a binary classification the output activation function is the sigmoid function given by
$A_{3} = 1/(1+ e^{-Z3})$. It can be shown that
$dA_{3}/dZ_{3} = A_{3}(1-A_3)$ see equation 2 in Part 1

$\partial L/\partial Z_{3} = \partial L/\partial A_{3}* \partial A_{3}/\partial Z_{3} = A3-Y$ see equation (f) in  Part 1
and since
$\partial L/\partial A_{2} = \partial L/\partial Z_{3} * \partial Z_{3}/\partial A_{2} = (A_{3} -Y) * W_{3}$ because $\partial Z_{3}/\partial A_{2} = W_{3}$ -(1a)
and $\partial L/\partial Z_{2} =\partial L/\partial A_{2} * \partial A_{2}/\partial Z_{2} = (A_{3} -Y) * W_{3} *g'(Z_{2})$ -(1b)
$\partial L/\partial W_{2} = \partial L/\partial Z_{2} * A_{1}$ -(1c)
since $\partial Z_{2}/\partial W_{2} = A_{1}$
and
$\partial L/\partial b_{2} = \partial L/\partial Z_{2}$ -(1d)
because
$\partial Z_{2}/\partial b_{2} =1$

Also

$\partial L/\partial A_{1} =\partial L/\partial Z_{2} * \partial Z_{2}/\partial A_{1} = \partial L/\partial Z_{2} * W_{2}$     – (2a)
$\partial L/\partial Z_{1} =\partial L/\partial A_{1} * \partial A_{1}/\partial Z_{1} = \partial L/\partial A_{1} * W_{2} *g'(Z_{1})$          – (2b)
$\partial L/\partial W_{1} = \partial L/\partial Z_{1} * A_{0}$ – (2c)
$\partial L/\partial b_{1} = \partial L/\partial Z_{1}$ – (2d)

Inspecting the above equations (1a – 1d & 2a-2d), our ‘Uber deep, bottomless’ brain  can easily discern the pattern in these equations. The equation for any layer ‘l’ is of the form
$Z_{l} = W_{l}A_{l-1} +b_{l}$     and  $A_{l} = g(Z_{l})$
The equation for the backward propagation have the general form
$\partial L/\partial A_{l} = \partial L/\partial Z_{l+1} * W^{l+1}$
$\partial L/\partial Z_{l}=\partial L/\partial A_{l} *g'(Z_{l})$
$\partial L/\partial W_{l} =\partial L/\partial Z_{l} *A^{l-1}$
$\partial L/\partial b_{l} =\partial L/\partial Z_{l}$

Some other important results The derivatives of the activation functions in the implemented Deep Learning network
g(z) = sigmoid(z) = $1/(1+e^{-z})$ = a g’(z) = a(1-a) – See Part 1
g(z) = tanh(z) = a g’(z) = $1 - a^{2}$
g(z) = relu(z) = z  when z>0 and 0 when z 0 and 0 when z <= 0
While it appears that there is a discontinuity for the derivative at 0 the small value at the discontinuity does not present a problem

The implementation of the multi layer vectorized Deep Learning Network for Python, R and Octave is included below. For all these implementations, initially I create the size and configuration of the the Deep Learning network with the layer dimennsions So for example layersDimension Vector ‘V’ of length L indicating ‘L’ layers where

V (in Python)= $[v_{0}, v_{1}, v_{2}$, … $v_{L-1}]$
V (in R)= $c(v_{1}, v_{2}, v_{3}$ , … $v_{L})$
V (in Octave)= [ $v_{1} v_{2} v_{3}$$v_{L}]$

In all of these implementations the first element is the number of input features to the Deep Learning network and the last element is always a ‘sigmoid’ activation function since all the problems deal with binary classification.

The number of elements between the first and the last element are the number of hidden layers and the magnitude of each $v_{i}$ is the number of activation units in each hidden layer, which is specified while actually executing the Deep Learning network using the function L_Layer_DeepModel(), in all the implementations Python, R and Octave

## 1a. Classification with Multi layer Deep Learning Network – Relu activation(Python)

In the code below a 4 layer Neural Network is trained to generate a non-linear boundary between the classes. In the code below the ‘Relu’ Activation function is used. The number of activation units in each layer is 9. The cost vs iterations is plotted in addition to the decision boundary. Further the accuracy, precision, recall and F1 score are also computed

import os
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.colors
import sklearn.linear_model

from sklearn.model_selection import train_test_split
from sklearn.datasets import make_classification, make_blobs
from matplotlib.colors import ListedColormap
import sklearn
import sklearn.datasets

#from DLfunctions import plot_decision_boundary
execfile("./DLfunctions34.py") #
os.chdir("C:\\software\\DeepLearning-Posts\\part3")

# Create clusters of 2 classes
X1, Y1 = make_blobs(n_samples = 400, n_features = 2, centers = 9,
cluster_std = 1.3, random_state = 4)
#Create 2 classes
Y1=Y1.reshape(400,1)
Y1 = Y1 % 2
X2=X1.T
Y2=Y1.T
# Set the dimensions of DL Network
#  Below we have
#  2 - 2 input features
#  9,9 - 2 hidden layers with 9 activation units per layer and
#  1 - 1 sigmoid activation unit in the output layer as this is a binary classification
# The activation in the hidden layer is the 'relu' specified in L_Layer_DeepModel

layersDimensions = [2, 9, 9,1] #  4-layer model
parameters = L_Layer_DeepModel(X2, Y2, layersDimensions,hiddenActivationFunc='relu', learning_rate = 0.3,num_iterations = 2500, fig="fig1.png")
#Plot the decision boundary
plot_decision_boundary(lambda x: predict(parameters, x.T), X2,Y2,str(0.3),"fig2.png")

# Compute the confusion matrix
yhat = predict(parameters,X2)
from sklearn.metrics import confusion_matrix
a=confusion_matrix(Y2.T,yhat.T)
from sklearn.metrics import accuracy_score, precision_score, recall_score, f1_score
print('Accuracy: {:.2f}'.format(accuracy_score(Y2.T, yhat.T)))
print('Precision: {:.2f}'.format(precision_score(Y2.T, yhat.T)))
print('Recall: {:.2f}'.format(recall_score(Y2.T, yhat.T)))
print('F1: {:.2f}'.format(f1_score(Y2.T, yhat.T)))
## Accuracy: 0.90
## Precision: 0.91
## Recall: 0.87
## F1: 0.89

For more details on metrics like Accuracy, Recall, Precision etc. used in classification take a look at my post Practical Machine Learning with R and Python – Part 2. More details about these and other metrics besides implementation of the most common machine learning algorithms are available in my book My book ‘Practical Machine Learning with R and Python’ on Amazon

## 1b. Classification with Multi layer Deep Learning Network – Relu activation(R)

In the code below, binary classification is performed on the same data set as above using the Relu activation function. The DL network is same as above

library(ggplot2)
x <- z[,1:2]
y <- z[,3]
X1 <- t(x)
Y1 <- t(y)

# Set the dimensions of the Deep Learning network
# No of input features =2, 2 hidden layers with 9 activation units and 1 output layer
layersDimensions = c(2, 9, 9,1)
# Execute the Deep Learning Neural Network
retvals = L_Layer_DeepModel(X1, Y1, layersDimensions,
hiddenActivationFunc='relu',
learningRate = 0.3,
numIterations = 5000,
print_cost = True)
library(ggplot2)
source("DLfunctions33.R")
# Get the computed costs
costs <- retvals[['costs']]
# Create a sequence of iterations
numIterations=5000
iterations <- seq(0,numIterations,by=1000)
df <-data.frame(iterations,costs)
# Plot the Costs vs number of iterations
ggplot(df,aes(x=iterations,y=costs)) + geom_point() +geom_line(color="blue") +
xlab('No of iterations') + ylab('Cost') + ggtitle("Cost vs No of iterations")

# Plot the decision boundary
plotDecisionBoundary(z,retvals,hiddenActivationFunc="relu",0.3)

library(caret)
# Predict the output for the data values
yhat <-predict(retvals$parameters,X1,hiddenActivationFunc="relu") yhat[yhat==FALSE]=0 yhat[yhat==TRUE]=1 # Compute the confusion matrix confusionMatrix(yhat,Y1) ## Confusion Matrix and Statistics ## ## Reference ## Prediction 0 1 ## 0 201 10 ## 1 21 168 ## ## Accuracy : 0.9225 ## 95% CI : (0.8918, 0.9467) ## No Information Rate : 0.555 ## P-Value [Acc > NIR] : < 2e-16 ## ## Kappa : 0.8441 ## Mcnemar's Test P-Value : 0.07249 ## ## Sensitivity : 0.9054 ## Specificity : 0.9438 ## Pos Pred Value : 0.9526 ## Neg Pred Value : 0.8889 ## Prevalence : 0.5550 ## Detection Rate : 0.5025 ## Detection Prevalence : 0.5275 ## Balanced Accuracy : 0.9246 ## ## 'Positive' Class : 0 ##  ## 1c. Classification with Multi layer Deep Learning Network – Relu activation(Octave) Included below is the code for performing classification. Incidentally Octave does not seem to have implemented the confusion matrix, but confusionmat is available in Matlab. # Read the data data=csvread("data.csv"); X=data(:,1:2); Y=data(:,3); # Set layer dimensions layersDimensions = [2 9 7 1] #tanh=-0.5(ok), #relu=0.1 best! # Execute Deep Network [weights biases costs]=L_Layer_DeepModel(X', Y', layersDimensions, hiddenActivationFunc='relu', learningRate = 0.1, numIterations = 10000); plotCostVsIterations(10000,costs); plotDecisionBoundary(data,weights, biases,hiddenActivationFunc="tanh")  ## 2a. Classification with Multi layer Deep Learning Network – Tanh activation(Python) Below the Tanh activation function is used to perform the same classification. I found the Tanh activation required a simpler Neural Network of 3 layers. # Tanh activation import os import numpy as np import matplotlib.pyplot as plt import matplotlib.colors import sklearn.linear_model from sklearn.model_selection import train_test_split from sklearn.datasets import make_classification, make_blobs from matplotlib.colors import ListedColormap import sklearn import sklearn.datasets #from DLfunctions import plot_decision_boundary os.chdir("C:\\software\\DeepLearning-Posts\\part3") execfile("./DLfunctions34.py") # Create the dataset X1, Y1 = make_blobs(n_samples = 400, n_features = 2, centers = 9, cluster_std = 1.3, random_state = 4) #Create 2 classes Y1=Y1.reshape(400,1) Y1 = Y1 % 2 X2=X1.T Y2=Y1.T # Set the dimensions of the Neural Network layersDimensions = [2, 4, 1] # 3-layer model # Compute the DL network parameters = L_Layer_DeepModel(X2, Y2, layersDimensions, hiddenActivationFunc='tanh', learning_rate = .5,num_iterations = 2500,fig="fig3.png") #Plot the decision boundary plot_decision_boundary(lambda x: predict(parameters, x.T), X2,Y2,str(0.5),"fig4.png")  ## 2b. Classification with Multi layer Deep Learning Network – Tanh activation(R) R performs better with a Tanh activation than the Relu as can be seen below  #Set the dimensions of the Neural Network layersDimensions = c(2, 9, 9,1) library(ggplot2) # Read the data z <- as.matrix(read.csv("data.csv",header=FALSE)) x <- z[,1:2] y <- z[,3] X1 <- t(x) Y1 <- t(y) # Execute the Deep Model retvals = L_Layer_DeepModel(X1, Y1, layersDimensions, hiddenActivationFunc='tanh', learningRate = 0.3, numIterations = 5000, print_cost = True) # Get the costs costs <- retvals[['costs']] iterations <- seq(0,numIterations,by=1000) df <-data.frame(iterations,costs) # Plot Cost vs number of iterations ggplot(df,aes(x=iterations,y=costs)) + geom_point() +geom_line(color="blue") + xlab('No of iterations') + ylab('Cost') + ggtitle("Cost vs No of iterations") #Plot the decision boundary plotDecisionBoundary(z,retvals,hiddenActivationFunc="tanh",0.3) ## 2c. Classification with Multi layer Deep Learning Network – Tanh activation(Octave) The code below uses the Tanh activation in the hidden layers for Octave # Read the data data=csvread("data.csv"); X=data(:,1:2); Y=data(:,3); # Set layer dimensions layersDimensions = [2 9 7 1] #tanh=-0.5(ok), #relu=0.1 best! # Execute Deep Network [weights biases costs]=L_Layer_DeepModel(X', Y', layersDimensions, hiddenActivationFunc='tanh', learningRate = 0.1, numIterations = 10000); plotCostVsIterations(10000,costs); plotDecisionBoundary(data,weights, biases,hiddenActivationFunc="tanh")  ## 3. Bernoulli’s Lemniscate To make things more interesting, I create a 2D figure of the Bernoulli’s lemniscate to perform non-linear classification. The Lemniscate is given by the equation $(x^{2} + y^{2})^{2}$ = $2a^{2}*(x^{2}-y^{2})$ ## 3a. Classifying a lemniscate with Deep Learning Network – Relu activation(Python) import os import numpy as np import matplotlib.pyplot as plt os.chdir("C:\\software\\DeepLearning-Posts\\part3") execfile("./DLfunctions33.py") x1=np.random.uniform(0,10,2000).reshape(2000,1) x2=np.random.uniform(0,10,2000).reshape(2000,1) X=np.append(x1,x2,axis=1) X.shape # Create a subset of values where squared is <0,4. Perform ravel() to flatten this vector # Create the equation # (x^{2} + y^{2})^2 - 2a^2*(x^{2}-y^{2}) <= 0 a=np.power(np.power(X[:,0]-5,2) + np.power(X[:,1]-5,2),2) b=np.power(X[:,0]-5,2) - np.power(X[:,1]-5,2) c= a - (b*np.power(4,2)) <=0 Y=c.reshape(2000,1) # Create a scatter plot of the lemniscate plt.scatter(X[:,0], X[:,1], c=Y, marker= 'o', s=15,cmap="viridis") Z=np.append(X,Y,axis=1) plt.savefig("fig50.png",bbox_inches='tight') plt.clf() # Set the data for classification X2=X.T Y2=Y.T # These settings work the best # Set the Deep Learning layer dimensions for a Relu activation layersDimensions = [2,7,4,1] #Execute the DL network parameters = L_Layer_DeepModel(X2, Y2, layersDimensions, hiddenActivationFunc='relu', learning_rate = 0.5,num_iterations = 10000, fig="fig5.png") #Plot the decision boundary plot_decision_boundary(lambda x: predict(parameters, x.T), X2, Y2,str(2.2),"fig6.png") # Compute the Confusion matrix yhat = predict(parameters,X2) from sklearn.metrics import confusion_matrix a=confusion_matrix(Y2.T,yhat.T) from sklearn.metrics import accuracy_score, precision_score, recall_score, f1_score print('Accuracy: {:.2f}'.format(accuracy_score(Y2.T, yhat.T))) print('Precision: {:.2f}'.format(precision_score(Y2.T, yhat.T))) print('Recall: {:.2f}'.format(recall_score(Y2.T, yhat.T))) print('F1: {:.2f}'.format(f1_score(Y2.T, yhat.T))) ## Accuracy: 0.93 ## Precision: 0.77 ## Recall: 0.76 ## F1: 0.76 We could get better performance by tuning further. Do play around if you fork the code. Note:: The lemniscate data is saved as a CSV and then read in R and also in Octave. I do this instead of recreating the lemniscate shape ## 3b. Classifying a lemniscate with Deep Learning Network – Relu activation(R code) The R decision boundary for the Bernoulli’s lemniscate is shown below Z <- as.matrix(read.csv("lemniscate.csv",header=FALSE)) Z1=data.frame(Z) # Create a scatter plot of the lemniscate ggplot(Z1,aes(x=V1,y=V2,col=V3)) +geom_point() #Set the data for the DL network X=Z[,1:2] Y=Z[,3] X1=t(X) Y1=t(Y) # Set the layer dimensions for the tanh activation function layersDimensions = c(2,5,4,1) # Execute the Deep Learning network with Tanh activation retvals = L_Layer_DeepModel(X1, Y1, layersDimensions, hiddenActivationFunc='tanh', learningRate = 0.3, numIterations = 20000, print_cost = True) # Plot cost vs iteration costs <- retvals[['costs']] numIterations = 20000 iterations <- seq(0,numIterations,by=1000) df <-data.frame(iterations,costs) ggplot(df,aes(x=iterations,y=costs)) + geom_point() +geom_line(color="blue") + xlab('No of iterations') + ylab('Cost') + ggtitle("Cost vs No of iterations") #Plot the decision boundary plotDecisionBoundary(Z,retvals,hiddenActivationFunc="tanh",0.3) ## 3c. Classifying a lemniscate with Deep Learning Network – Relu activation(Octave code) Octave is used to generate the non-linear lemniscate boundary.  # Read the data data=csvread("lemniscate.csv"); X=data(:,1:2); Y=data(:,3); # Set the dimensions of the layers layersDimensions = [2 9 7 1] # Compute the DL network [weights biases costs]=L_Layer_DeepModel(X', Y', layersDimensions, hiddenActivationFunc='relu', learningRate = 0.20, numIterations = 10000); plotCostVsIterations(10000,costs); plotDecisionBoundary(data,weights, biases,hiddenActivationFunc="relu")  ## 4a. Binary Classification using MNIST – Python code Finally I perform a simple classification using the MNIST handwritten digits, which according to Prof Geoffrey Hinton is “the Drosophila of Deep Learning”. The Python code for reading the MNIST data is taken from Alex Kesling’s github link MNIST. In the Python code below, I perform a simple binary classification between the handwritten digit ‘5’ and ‘not 5’ which is all other digits. I will perform the proper classification of all digits using the Softmax classifier some time later. import os import numpy as np import matplotlib.pyplot as plt os.chdir("C:\\software\\DeepLearning-Posts\\part3") execfile("./DLfunctions34.py") execfile("./load_mnist.py") training=list(read(dataset='training',path="./mnist")) test=list(read(dataset='testing',path="./mnist")) lbls=[] pxls=[] print(len(training)) # Select the first 10000 training data and the labels for i in range(10000): l,p=training[i] lbls.append(l) pxls.append(p) labels= np.array(lbls) pixels=np.array(pxls) # Sey y=1 when labels == 5 and 0 otherwise y=(labels==5).reshape(-1,1) X=pixels.reshape(pixels.shape[0],-1) # Create the necessary feature and target variable X1=X.T Y1=y.T # Create the layer dimensions. The number of features are 28 x 28 = 784 since the 28 x 28 # pixels is flattened to single vector of length 784. layersDimensions=[784, 15,9,7,1] # Works very well parameters = L_Layer_DeepModel(X1, Y1, layersDimensions, hiddenActivationFunc='relu', learning_rate = 0.1,num_iterations = 1000, fig="fig7.png") # Test data lbls1=[] pxls1=[] for i in range(800): l,p=test[i] lbls1.append(l) pxls1.append(p) testLabels=np.array(lbls1) testData=np.array(pxls1) ytest=(testLabels==5).reshape(-1,1) Xtest=testData.reshape(testData.shape[0],-1) Xtest1=Xtest.T Ytest1=ytest.T yhat = predict(parameters,Xtest1) from sklearn.metrics import confusion_matrix a=confusion_matrix(Ytest1.T,yhat.T) from sklearn.metrics import accuracy_score, precision_score, recall_score, f1_score print('Accuracy: {:.2f}'.format(accuracy_score(Ytest1.T, yhat.T))) print('Precision: {:.2f}'.format(precision_score(Ytest1.T, yhat.T))) print('Recall: {:.2f}'.format(recall_score(Ytest1.T, yhat.T))) print('F1: {:.2f}'.format(f1_score(Ytest1.T, yhat.T))) probs=predict_proba(parameters,Xtest1) from sklearn.metrics import precision_recall_curve precision, recall, thresholds = precision_recall_curve(Ytest1.T, probs.T) closest_zero = np.argmin(np.abs(thresholds)) closest_zero_p = precision[closest_zero] closest_zero_r = recall[closest_zero] plt.xlim([0.0, 1.01]) plt.ylim([0.0, 1.01]) plt.plot(precision, recall, label='Precision-Recall Curve') plt.plot(closest_zero_p, closest_zero_r, 'o', markersize = 12, fillstyle = 'none', c='r', mew=3) plt.xlabel('Precision', fontsize=16) plt.ylabel('Recall', fontsize=16) plt.savefig("fig8.png",bbox_inches='tight')   ## Accuracy: 0.99 ## Precision: 0.96 ## Recall: 0.89 ## F1: 0.92 In addition to plotting the Cost vs Iterations, I also plot the Precision-Recall curve to show how the Precision and Recall, which are complementary to each other vary with respect to the other. To know more about Precision-Recall, please check my post Practical Machine Learning with R and Python – Part 4. 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($7.99) versions. My book includes implementations of key ML algorithms and associated measures and metrics. The book is ideal for anybody who is familiar with the concepts and would like a quick reference to the different ML algorithms that can be applied to problems and how to select the best model. Pick your copy today!! A physical copy of the book is much better than scrolling down a webpage. Personally, I tend to use my own book quite frequently to refer to R, Python constructs, subsetting, machine Learning function calls and the necessary parameters etc. It is useless to commit any of this to memory, and a physical copy of a book is much easier to thumb through for the relevant code snippet. Pick up your copy today! ## 4b. Binary Classification using MNIST – R code In the R code below the same binary classification of the digit ‘5’ and the ‘not 5’ is performed. The code to read and display the MNIST data is taken from Brendan O’ Connor’s github link at MNIST source("mnist.R") load_mnist() #show_digit(train$x[2,]
layersDimensions=c(784, 7,7,3,1) # Works at 1500
x <- t(train$x) # Choose only 5000 training data x2 <- x[,1:5000] y <-train$y
# Set labels for all digits that are 'not 5' to 0
y[y!=5] <- 0
# Set labels of digit 5 as 1
y[y==5] <- 1
# Set the data
y1 <- as.matrix(y)
y2 <- t(y1)
# Choose the 1st 5000 data
y3 <- y2[,1:5000]

#Execute the Deep Learning Model
retvals = L_Layer_DeepModel(x2, y3, layersDimensions,
hiddenActivationFunc='tanh',
learningRate = 0.3,
numIterations = 3000, print_cost = True)
# Plot cost vs iteration
costs <- retvals[['costs']]
numIterations = 3000
iterations <- seq(0,numIterations,by=1000)
df <-data.frame(iterations,costs)
ggplot(df,aes(x=iterations,y=costs)) + geom_point() +geom_line(color="blue") +
xlab('No of iterations') + ylab('Cost') + ggtitle("Cost vs No of iterations")

# Compute probability scores
scores <- computeScores(retvals$parameters, x2,hiddenActivationFunc='relu') a=y3==1 b=y3==0 # Compute probabilities of class 0 and class 1 class1=scores[a] class0=scores[b] # Plot ROC curve pr <-pr.curve(scores.class0=class1, scores.class1=class0, curve=T) plot(pr) The AUC curve hugs the top left corner and hence the performance of the classifier is quite good. ## 4c. Binary Classification using MNIST – Octave code This code to load MNIST data was taken from Daniel E blog. Precision recall curves are available in Matlab but are yet to be implemented in Octave’s statistics package.  load('./mnist/mnist.txt.gz'); % load the dataset # Subset the 'not 5' digits a=(trainY != 5); # Subset '5' b=(trainY == 5); #make a copy of trainY #Set 'not 5' as 0 and '5' as 1 y=trainY; y(a)=0; y(b)=1; X=trainX(1:5000,:); Y=y(1:5000); # Set the dimensions of layer layersDimensions=[784, 7,7,3,1]; # Compute the DL network [weights biases costs]=L_Layer_DeepModel(X', Y', layersDimensions, hiddenActivationFunc='relu', learningRate = 0.1, numIterations = 5000);  # Conclusion It was quite a challenge coding a Deep Learning Network in Python, R and Octave. The Deep Learning network implementation, in this post,is the base Deep Learning network, without any of the regularization methods included. Here are some key learning that I got while playing with different multi-layer networks on different problems a. Deep Learning Networks come with many levers, the hyper-parameters, – learning rate – activation unit – number of hidden layers – number of units per hidden layer – number of iterations while performing gradient descent b. Deep Networks are very sensitive. A change in any of the hyper-parameter makes it perform very differently c. Initially I thought adding more hidden layers, or more units per hidden layer will make the DL network better at learning. On the contrary, there is a performance degradation after the optimal DL configuration d. At a sub-optimal number of hidden layers or number of hidden units, gradient descent seems to get stuck at a local minima e. There were occasions when the cost came down, only to increase slowly as the number of iterations were increased. Probably early stopping would have helped. f. I also did come across situations of ‘exploding/vanishing gradient’, cost went to Inf/-Inf. Here I would think inclusion of ‘momentum method’ would have helped I intend to add the additional hyper-parameters of L1, L2 regularization, momentum method, early stopping etc. into the code in my future posts. Feel free to fork/clone the code from Github Deep Learning – Part 3, and take the DL network apart and play around with it. I will be continuing this series with more hyper-parameters to handle vanishing and exploding gradients, early stopping and regularization in the weeks to come. I also intend to add some more activation functions to this basic Multi-Layer Network. Hang around, there are more exciting things to come. Watch this space! To see all posts see Index of posts # Deep Learning from first principles in Python, R and Octave – Part 2 “What does the world outside your head really ‘look’ like? Not only is there no color, there’s also no sound: the compression and expansion of air is picked up by the ears, and turned into electrical signals. The brain then presents these signals to us as mellifluous tones and swishes and clatters and jangles. Reality is also odorless: there’s no such thing as smell outside our brains. Molecules floating through the air bind to receptors in our nose and are interpreted as different smells by our brain. The real world is not full of rich sensory events; instead, our brains light up the world with their own sensuality.” The Brain: The Story of You” by David Eagleman The world is Maya, illusory. The ultimate reality, the Brahman, is all-pervading and all-permeating, which is colourless, odourless, tasteless, nameless and formless Bhagavad Gita ## 1. Introduction This post is a follow-up post to my earlier post Deep Learning from first principles in Python, R and Octave-Part 1. In the first part, I implemented Logistic Regression, in vectorized Python,R and Octave, with a wannabe Neural Network (a Neural Network with no hidden layers). In this second part, I implement a regular, but somewhat primitive Neural Network (a Neural Network with just 1 hidden layer). The 2nd part implements classification of manually created datasets, where the different clusters of the 2 classes are not linearly separable. Neural Network perform really well in learning all sorts of non-linear boundaries between classes. Initially logistic regression is used perform the classification and the decision boundary is plotted. Vanilla logistic regression performs quite poorly. Using SVMs with a radial basis kernel would have performed much better in creating non-linear boundaries. To see R and Python implementations of SVMs take a look at my post Practical Machine Learning with R and Python – Part 4. 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). You may also like my companion 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/Rs449) versions. This book is ideal for a quick reference of the various ML functions and associated measurements in both R and Python which are essential to delve deep into Deep Learning. Take a look at my video presentation which discusses the below derivation step-by- step Elements of Neural Networks and Deep Learning – Part 3 You can clone and fork this R Markdown file along with the vectorized implementations of the 3 layer Neural Network for Python, R and Octave from Github DeepLearning-Part2 ### 2. The 3 layer Neural Network A simple representation of a 3 layer Neural Network (NN) with 1 hidden layer is shown below. In the above Neural Network, there are 2 input features at the input layer, 3 hidden units at the hidden layer and 1 output layer as it deals with binary classification. The activation unit at the hidden layer can be a tanh, sigmoid, relu etc. At the output layer the activation is a sigmoid to handle binary classification # Superscript indicates layer 1 $z_{11} = w_{11}^{1}x_{1} + w_{21}^{1}x_{2} + b_{1}$ $z_{12} = w_{12}^{1}x_{1} + w_{22}^{1}x_{2} + b_{1}$ $z_{13} = w_{13}^{1}x_{1} + w_{23}^{1}x_{2} + b_{1}$ Also $a_{11} = tanh(z_{11})$ $a_{12} = tanh(z_{12})$ $a_{13} = tanh(z_{13})$ # Superscript indicates layer 2 $z_{21} = w_{11}^{2}a_{11} + w_{21}^{2}a_{12} + w_{31}^{2}a_{13} + b_{2}$ $a_{21} = sigmoid(z21)$ Hence $Z1= \begin{pmatrix} z11\\ z12\\ z13 \end{pmatrix} =\begin{pmatrix} w_{11}^{1} & w_{21}^{1} \\ w_{12}^{1} & w_{22}^{1} \\ w_{13}^{1} & w_{23}^{1} \end{pmatrix} * \begin{pmatrix} x1\\ x2 \end{pmatrix} + b_{1}$ And $A1= \begin{pmatrix} a11\\ a12\\ a13 \end{pmatrix} = \begin{pmatrix} tanh(z11)\\ tanh(z12)\\ tanh(z13) \end{pmatrix}$ Similarly $Z2= z_{21} = \begin{pmatrix} w_{11}^{2} & w_{21}^{2} & w_{31}^{2} \end{pmatrix} *\begin{pmatrix} z_{11}\\ z_{12}\\ z_{13} \end{pmatrix} +b_{2}$ and $A2 = a_{21} = sigmoid(z_{21})$ These equations can be written as $Z1 = W1 * X + b1$ $A1 = tanh(Z1)$ $Z2 = W2 * A1 + b2$ $A2 = sigmoid(Z2)$ I) Some important results (a memory refresher!) $d/dx(e^{x}) = e^{x}$ and $d/dx(e^{-x}) = -e^{-x}$ -(a) and $sinhx = (e^{x} - e^{-x})/2$ and $coshx = (e^{x} + e^{-x})/2$ Using (a) we can shown that $d/dx(sinhx) = coshx$ and $d/dx(coshx) = sinhx$ (b) Now $d/dx(f(x)/g(x)) = (g(x)*d/dx(f(x)) - f(x)*d/dx(g(x)))/g(x)^{2}$ -(c) Since $tanhx =z= sinhx/coshx$ and using (b) we get $tanhx = (coshx*d/dx(sinhx) - sinhx*d/dx(coshx))/(cosh^{2})$ Using the values of the derivatives of sinhx and coshx from (b) above we get $d/dx(tanhx) = (coshx^{2} - sinhx{2})/coshx{2} = 1 - tanhx^{2}$ Since $tanhx =z$ $d/dx(tanhx) = 1 - tanhx^{2}= 1 - z^{2}$ -(d) II) Derivatives $L=-(Ylog(A2) + (1-Y)log(1-A2))$ $dL/dA2 = -(Y/A2 + (1-Y)/(1-A2))$ Since $A2 = sigmoid(Z2)$ therefore $dA2/dZ2 = A2(1-A2)$ see Part1 $Z2 = W2A1 +b2$ $dZ2/dW2 = A1$ $dZ2/db2 = 1$ $A1 = tanh(Z1)$ and $dA1/dZ1 = 1 - A1^{2}$ $Z1 = W1X + b1$ $dZ1/dW1 = X$ $dZ1/db1 = 1$ III) Back propagation Using the derivatives from II) we can derive the following results using Chain Rule $\partial L/\partial Z2 = \partial L/\partial A2 * \partial A2/\partial Z2$ $= -(Y/A2 + (1-Y)/(1-A2)) * A2(1-A2) = A2 - Y$ $\partial L/\partial W2 = \partial L/\partial A2 * \partial A2/\partial Z2 * \partial Z2/\partial W2$ $= (A2-Y) *A1$ -(A) $\partial L/\partial b2 = \partial L/\partial A2 * \partial A2/\partial Z2 * \partial Z2/\partial b2 = (A2-Y)$ -(B) $\partial L/\partial Z1 = \partial L/\partial A2 * \partial A2/\partial Z2 * \partial Z2/\partial A1 *\partial A1/\partial Z1 = (A2-Y) * W2 * (1-A1^{2})$ $\partial L/\partial W1 = \partial L/\partial A2 * \partial A2/\partial Z2 * \partial Z2/\partial A1 *\partial A1/\partial Z1 *\partial Z1/\partial W1$ $=(A2-Y) * W2 * (1-A1^{2}) * X$ -(C) $\partial L/\partial b1 = \partial L/\partial A2 * \partial A2/\partial Z2 * \partial Z2/\partial A1 *dA1/dZ1 *dZ1/db1$ $= (A2-Y) * W2 * (1-A1^{2})$ -(D) IV) Gradient Descent The key computations in the backward cycle are $W1 = W1-learningRate * \partial L/\partial W1$ – From (C) $b1 = b1-learningRate * \partial L/\partial b1$ – From (D) $W2 = W2-learningRate * \partial L/\partial W2$ – From (A) $b2 = b2-learningRate * \partial L/\partial b2$ – From (B) The weights and biases (W1,b1,W2,b2) are updated for each iteration thus minimizing the loss/cost. These derivations can be represented pictorially using the computation graph (from the book Deep Learning by Ian Goodfellow, Joshua Bengio and Aaron Courville) ### 3. Manually create a data set that is not lineary separable Initially I create a dataset with 2 classes which has around 9 clusters that cannot be separated by linear boundaries. Note: This data set is saved as data.csv and is used for the R and Octave Neural networks to see how they perform on the same dataset. import numpy as np import matplotlib.pyplot as plt import matplotlib.colors import sklearn.linear_model from sklearn.model_selection import train_test_split from sklearn.datasets import make_classification, make_blobs from matplotlib.colors import ListedColormap import sklearn import sklearn.datasets colors=['black','gold'] cmap = matplotlib.colors.ListedColormap(colors) X, y = make_blobs(n_samples = 400, n_features = 2, centers = 7, cluster_std = 1.3, random_state = 4) #Create 2 classes y=y.reshape(400,1) y = y % 2 #Plot the figure plt.figure() plt.title('Non-linearly separable classes') plt.scatter(X[:,0], X[:,1], c=y, marker= 'o', s=50,cmap=cmap) plt.savefig('fig1.png', bbox_inches='tight') ### 4. Logistic Regression On the above created dataset, classification with logistic regression is performed, and the decision boundary is plotted. It can be seen that logistic regression performs quite poorly import numpy as np import matplotlib.pyplot as plt import matplotlib.colors import sklearn.linear_model from sklearn.model_selection import train_test_split from sklearn.datasets import make_classification, make_blobs from matplotlib.colors import ListedColormap import sklearn import sklearn.datasets #from DLfunctions import plot_decision_boundary execfile("./DLfunctions.py") # Since import does not work in Rmd!!! colors=['black','gold'] cmap = matplotlib.colors.ListedColormap(colors) X, y = make_blobs(n_samples = 400, n_features = 2, centers = 7, cluster_std = 1.3, random_state = 4) #Create 2 classes y=y.reshape(400,1) y = y % 2 # Train the logistic regression classifier clf = sklearn.linear_model.LogisticRegressionCV(); clf.fit(X, y); # Plot the decision boundary for logistic regression plot_decision_boundary_n(lambda x: clf.predict(x), X.T, y.T,"fig2.png")  ### 5. The 3 layer Neural Network in Python (vectorized) The vectorized implementation is included below. Note that in the case of Python a learning rate of 0.5 and 3 hidden units performs very well. ## Random data set with 9 clusters import numpy as np import matplotlib import matplotlib.pyplot as plt import sklearn.linear_model import pandas as pd from sklearn.datasets import make_classification, make_blobs execfile("./DLfunctions.py") # Since import does not work in Rmd!!! X1, Y1 = make_blobs(n_samples = 400, n_features = 2, centers = 9, cluster_std = 1.3, random_state = 4) #Create 2 classes Y1=Y1.reshape(400,1) Y1 = Y1 % 2 X2=X1.T Y2=Y1.T #Perform gradient descent parameters,costs = computeNN(X2, Y2, numHidden = 4, learningRate=0.5, numIterations = 10000) plot_decision_boundary(lambda x: predict(parameters, x.T), X2, Y2,str(4),str(0.5),"fig3.png") ## Cost after iteration 0: 0.692669 ## Cost after iteration 1000: 0.246650 ## Cost after iteration 2000: 0.227801 ## Cost after iteration 3000: 0.226809 ## Cost after iteration 4000: 0.226518 ## Cost after iteration 5000: 0.226331 ## Cost after iteration 6000: 0.226194 ## Cost after iteration 7000: 0.226085 ## Cost after iteration 8000: 0.225994 ## Cost after iteration 9000: 0.225915 ### 6. The 3 layer Neural Network in R (vectorized) For this the dataset created by Python is saved to see how R performs on the same dataset. The vectorized implementation of a Neural Network was just a little more interesting as R does not have a similar package like ‘numpy’. While numpy handles broadcasting implicitly, in R I had to use the ‘sweep’ command to broadcast. The implementaion is included below. Note that since the initialization with random weights is slightly different, R performs best with a learning rate of 0.1 and with 6 hidden units source("DLfunctions2_1.R") z <- as.matrix(read.csv("data.csv",header=FALSE)) # x <- z[,1:2] y <- z[,3] x1 <- t(x) y1 <- t(y) #Perform gradient descent nn <-computeNN(x1, y1, 6, learningRate=0.1,numIterations=10000) # Good ## [1] 0.7075341 ## [1] 0.2606695 ## [1] 0.2198039 ## [1] 0.2091238 ## [1] 0.211146 ## [1] 0.2108461 ## [1] 0.2105351 ## [1] 0.210211 ## [1] 0.2099104 ## [1] 0.2096437 ## [1] 0.209409 plotDecisionBoundary(z,nn,6,0.1) ### 7. The 3 layer Neural Network in Octave (vectorized) This uses the same dataset that was generated using Python code. source("DL-function2.m") data=csvread("data.csv"); X=data(:,1:2); Y=data(:,3); # Make sure that the model parameters are correct. Take the transpose of X & Y #Perform gradient descent [W1,b1,W2,b2,costs]= computeNN(X', Y',4, learningRate=0.5, numIterations = 10000); ### 8a. Performance for different learning rates (Python) import numpy as np import matplotlib import matplotlib.pyplot as plt import sklearn.linear_model import pandas as pd from sklearn.datasets import make_classification, make_blobs execfile("./DLfunctions.py") # Since import does not work in Rmd!!! # Create data X1, Y1 = make_blobs(n_samples = 400, n_features = 2, centers = 9, cluster_std = 1.3, random_state = 4) #Create 2 classes Y1=Y1.reshape(400,1) Y1 = Y1 % 2 X2=X1.T Y2=Y1.T # Create a list of learning rates learningRate=[0.5,1.2,3.0] df=pd.DataFrame() #Compute costs for each learning rate for lr in learningRate: parameters,costs = computeNN(X2, Y2, numHidden = 4, learningRate=lr, numIterations = 10000) print(costs) df1=pd.DataFrame(costs) df=pd.concat([df,df1],axis=1) #Set the iterations iterations=[0,1000,2000,3000,4000,5000,6000,7000,8000,9000] #Create data frame #Set index df1=df.set_index([iterations]) df1.columns=[0.5,1.2,3.0] fig=df1.plot() fig=plt.title("Cost vs No of Iterations for different learning rates") plt.savefig('fig4.png', bbox_inches='tight') ### 8b. Performance for different hidden units (Python) import numpy as np import matplotlib import matplotlib.pyplot as plt import sklearn.linear_model import pandas as pd from sklearn.datasets import make_classification, make_blobs execfile("./DLfunctions.py") # Since import does not work in Rmd!!! #Create data set X1, Y1 = make_blobs(n_samples = 400, n_features = 2, centers = 9, cluster_std = 1.3, random_state = 4) #Create 2 classes Y1=Y1.reshape(400,1) Y1 = Y1 % 2 X2=X1.T Y2=Y1.T # Make a list of hidden unis numHidden=[3,5,7] df=pd.DataFrame() #Compute costs for different hidden units for numHid in numHidden: parameters,costs = computeNN(X2, Y2, numHidden = numHid, learningRate=1.2, numIterations = 10000) print(costs) df1=pd.DataFrame(costs) df=pd.concat([df,df1],axis=1) #Set the iterations iterations=[0,1000,2000,3000,4000,5000,6000,7000,8000,9000] #Set index df1=df.set_index([iterations]) df1.columns=[3,5,7] #Plot fig=df1.plot() fig=plt.title("Cost vs No of Iterations for different no of hidden units") plt.savefig('fig5.png', bbox_inches='tight') ### 9a. Performance for different learning rates (R) source("DLfunctions2_1.R") # Read data z <- as.matrix(read.csv("data.csv",header=FALSE)) # x <- z[,1:2] y <- z[,3] x1 <- t(x) y1 <- t(y) #Loop through learning rates and compute costs learningRate <-c(0.1,1.2,3.0) df <- NULL for(i in seq_along(learningRate)){ nn <- computeNN(x1, y1, 6, learningRate=learningRate[i],numIterations=10000) cost <- nn$costs
df <- cbind(df,cost)

}      

#Create dataframe
df <- data.frame(df)
iterations=seq(0,10000,by=1000)
df <- cbind(iterations,df)
names(df) <- c("iterations","0.5","1.2","3.0")
library(reshape2)
df1 <- melt(df,id="iterations")  # Melt the data
#Plot
ggplot(df1) + geom_line(aes(x=iterations,y=value,colour=variable),size=1)  +
xlab("Iterations") +
ylab('Cost') + ggtitle("Cost vs No iterations for  different learning rates")

### 9b. Performance  for different hidden units (R)

source("DLfunctions2_1.R")
# Loop through Num hidden units
numHidden <-c(4,6,9)
df <- NULL
for(i in seq_along(numHidden)){
nn <-  computeNN(x1, y1, numHidden[i], learningRate=0.1,numIterations=10000)
cost <- nn$costs df <- cbind(df,cost) }  df <- data.frame(df) iterations=seq(0,10000,by=1000) df <- cbind(iterations,df) names(df) <- c("iterations","4","6","9") library(reshape2) # Melt df1 <- melt(df,id="iterations") # Plot ggplot(df1) + geom_line(aes(x=iterations,y=value,colour=variable),size=1) + xlab("Iterations") + ylab('Cost') + ggtitle("Cost vs No iterations for different number of hidden units") ## 10a. Performance of the Neural Network for different learning rates (Octave) source("DL-function2.m") plotLRCostVsIterations() print -djph figa.jpg ## 10b. Performance of the Neural Network for different number of hidden units (Octave) source("DL-function2.m") plotHiddenCostVsIterations() print -djph figa.jpg ## 11. Turning the heat on the Neural Network In this 2nd part I create a a central region of positives and and the outside region as negatives. The points are generated using the equation of a circle (x – a)^{2} + (y -b) ^{2} = R^{2} . How does the 3 layer Neural Network perform on this? Here’s a look! Note: The same dataset is also used for R and Octave Neural Network constructions ## 12. Manually creating a circular central region import numpy as np import matplotlib.pyplot as plt import matplotlib.colors import sklearn.linear_model from sklearn.model_selection import train_test_split from sklearn.datasets import make_classification, make_blobs from matplotlib.colors import ListedColormap import sklearn import sklearn.datasets colors=['black','gold'] cmap = matplotlib.colors.ListedColormap(colors) x1=np.random.uniform(0,10,800).reshape(800,1) x2=np.random.uniform(0,10,800).reshape(800,1) X=np.append(x1,x2,axis=1) X.shape # Create (x-a)^2 + (y-b)^2 = R^2 # Create a subset of values where squared is <0,4. Perform ravel() to flatten this vector a=(np.power(X[:,0]-5,2) + np.power(X[:,1]-5,2) <= 6).ravel() Y=a.reshape(800,1) cmap = matplotlib.colors.ListedColormap(colors) plt.figure() plt.title('Non-linearly separable classes') plt.scatter(X[:,0], X[:,1], c=Y, marker= 'o', s=15,cmap=cmap) plt.savefig('fig6.png', bbox_inches='tight') ### 13a. Decision boundary with hidden units=4 and learning rate = 2.2 (Python) With the above hyper parameters the decision boundary is triangular import numpy as np import matplotlib.pyplot as plt import matplotlib.colors import sklearn.linear_model execfile("./DLfunctions.py") x1=np.random.uniform(0,10,800).reshape(800,1) x2=np.random.uniform(0,10,800).reshape(800,1) X=np.append(x1,x2,axis=1) X.shape # Create a subset of values where squared is <0,4. Perform ravel() to flatten this vector a=(np.power(X[:,0]-5,2) + np.power(X[:,1]-5,2) <= 6).ravel() Y=a.reshape(800,1) X2=X.T Y2=Y.T parameters,costs = computeNN(X2, Y2, numHidden = 4, learningRate=2.2, numIterations = 10000) plot_decision_boundary(lambda x: predict(parameters, x.T), X2, Y2,str(4),str(2.2),"fig7.png")  ## Cost after iteration 0: 0.692836 ## Cost after iteration 1000: 0.331052 ## Cost after iteration 2000: 0.326428 ## Cost after iteration 3000: 0.474887 ## Cost after iteration 4000: 0.247989 ## Cost after iteration 5000: 0.218009 ## Cost after iteration 6000: 0.201034 ## Cost after iteration 7000: 0.197030 ## Cost after iteration 8000: 0.193507 ## Cost after iteration 9000: 0.191949 ### 13b. Decision boundary with hidden units=12 and learning rate = 2.2 (Python) With the above hyper parameters the decision boundary is triangular import numpy as np import matplotlib.pyplot as plt import matplotlib.colors import sklearn.linear_model execfile("./DLfunctions.py") x1=np.random.uniform(0,10,800).reshape(800,1) x2=np.random.uniform(0,10,800).reshape(800,1) X=np.append(x1,x2,axis=1) X.shape # Create a subset of values where squared is <0,4. Perform ravel() to flatten this vector a=(np.power(X[:,0]-5,2) + np.power(X[:,1]-5,2) <= 6).ravel() Y=a.reshape(800,1) X2=X.T Y2=Y.T parameters,costs = computeNN(X2, Y2, numHidden = 12, learningRate=2.2, numIterations = 10000) plot_decision_boundary(lambda x: predict(parameters, x.T), X2, Y2,str(12),str(2.2),"fig8.png")  ## Cost after iteration 0: 0.693291 ## Cost after iteration 1000: 0.383318 ## Cost after iteration 2000: 0.298807 ## Cost after iteration 3000: 0.251735 ## Cost after iteration 4000: 0.177843 ## Cost after iteration 5000: 0.130414 ## Cost after iteration 6000: 0.152400 ## Cost after iteration 7000: 0.065359 ## Cost after iteration 8000: 0.050921 ## Cost after iteration 9000: 0.039719 ### 14a. Decision boundary with hidden units=9 and learning rate = 0.5 (R) When the number of hidden units is 6 and the learning rate is 0,1, is also a triangular shape in R source("DLfunctions2_1.R") z <- as.matrix(read.csv("data1.csv",header=FALSE)) # N x <- z[,1:2] y <- z[,3] x1 <- t(x) y1 <- t(y) nn <-computeNN(x1, y1, 9, learningRate=0.5,numIterations=10000) # Triangular ## [1] 0.8398838 ## [1] 0.3303621 ## [1] 0.3127731 ## [1] 0.3012791 ## [1] 0.3305543 ## [1] 0.3303964 ## [1] 0.2334615 ## [1] 0.1920771 ## [1] 0.2341225 ## [1] 0.2188118 ## [1] 0.2082687 plotDecisionBoundary(z,nn,6,0.1) ### 14b. Decision boundary with hidden units=8 and learning rate = 0.1 (R) source("DLfunctions2_1.R") z <- as.matrix(read.csv("data1.csv",header=FALSE)) # N x <- z[,1:2] y <- z[,3] x1 <- t(x) y1 <- t(y) nn <-computeNN(x1, y1, 8, learningRate=0.1,numIterations=10000) # Hemisphere ## [1] 0.7273279 ## [1] 0.3169335 ## [1] 0.2378464 ## [1] 0.1688635 ## [1] 0.1368466 ## [1] 0.120664 ## [1] 0.111211 ## [1] 0.1043362 ## [1] 0.09800573 ## [1] 0.09126161 ## [1] 0.0840379 plotDecisionBoundary(z,nn,8,0.1) ### 15a. Decision boundary with hidden units=12 and learning rate = 1.5 (Octave) source("DL-function2.m") data=csvread("data1.csv"); X=data(:,1:2); Y=data(:,3); # Make sure that the model parameters are correct. Take the transpose of X & Y [W1,b1,W2,b2,costs]= computeNN(X', Y',12, learningRate=1.5, numIterations = 10000); plotDecisionBoundary(data, W1,b1,W2,b2) print -djpg fige.jpg Conclusion: This post implemented a 3 layer Neural Network to create non-linear boundaries while performing classification. Clearly the Neural Network performs very well when the number of hidden units and learning rate are varied. To be continued… Watch this space!! To see all posts check Index of posts # Neural Networks: The mechanics of backpropagation The initial work in the ‘Backpropagation Algorithm’ started in the 1980’s and led to an explosion of interest in Neural Networks and the application of backpropagation The ‘Backpropagation’ algorithm computes the minimum of an error function with respect to the weights in the Neural Network. It uses the method of gradient descent. The combination of weights in a multi-layered neural network, which minimizes the error/cost function is considered to be a solution of the learning problem. In the Neural Network above $out_{o1} =\sum_{i} w_{i}*x_{i}$ $E = 1/2(target - out)^{2}$ $\partial E/\partial out= 1/2*2*(target - out) *-1 = -(target - out)$ $\partial E/\partial w_{i} =\partial E/\partial y* \partial y/\partial w_{i}$ $\partial E/\partial w_{i} = -(target - out) * x_{i}$ 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).

Perceptrons and single layered neural networks can only classify, if the sample space is linearly separable. For non-linear decision boundaries, a multi layered neural network with  backpropagation is required to generate more complex boundaries.The backpropagation algorithm, computes the minimum of the error function in weight space using the method of gradient descent. This computation of the gradient, requires the activation function to be both differentiable and continuous. Hence the sigmoid or logistic function is typically chosen as the activation function at every layer.

This post looks at a 3 layer neural network with 1 input, 1 hidden and 1 output. To a large extent this post is based on Matt Mazur’s detailed “A step by step backpropagation example“, and Prof Hinton’s “Neural Networks for Machine Learning” at Coursera and a few other sources.

While Matt Mazur’s post uses example values, I generate the formulas for the gradient derivatives for each weight in the hidden and input layers. I intend to implement a vector version of backpropagation in Octave, R and Python. So this post is a prequel to that.

The 3 layer neural network is as below

Some basic derivations which are used in backpropagation

Chain rule of differentiation
Let y=f(u)
and u=g(x) then
$\partial y/\partial x = \partial y/\partial u * \partial u/\partial x$

An important result
$y=1/(1+e^{-z})$
Let $x= 1 + e^{-z}$  then
$y = 1/x$
$\partial y/\partial x = -1/x^{2}$
$\partial x/\partial z = -e^{-z}$

Using the chain rule of differentiation we get
$\partial y/\partial z = \partial y/\partial x * \partial x/\partial z$
$=-1/(1+e^{-z})^{2}* -e^{-z} = e^{-z}/(1+e^{-z})^{2}$
Therefore $\partial y/\partial z = y(1-y)$                                   -(A)

1) Feed forward network
The net output at the 1st hidden layer
$in_{h1} = w_{1}i_{1} + w_{2}i_{2} + b_{1}$
$in_{h2} = w_{3}i_{1} + w_{4}i_{2} + b_{1}$

The sigmoid/logistic function function is used to generate the activation outputs for each hidden layer. The sigmoid is chosen because it is continuous and also has a continuous derivative

$out_{h1} = 1/1+e^{-in_{h1}}$
$out_{h2} = 1/1+e^{-in_{h2}}$

The net output at the output layer
$in_{o1} = w_{5}out_{h_{1}} + w_{6}out_{h_{2}} + b_{2}$
$in_{o2} = w_{7}out_{h_{1}} + w_{8}out_{h_{2}} + b_{2}$

Total error
$E_{total} = 1/2\sum (target - output)^{2}$
$E_{total} = E_{o1} + E_{o2}$
$E_{total} = 1/2(target_{o_{1}} - out_{o_{1}})^{2} + 1/2(target_{o_{2}} - out_{o_{2}})^{2}$

2)The backwards pass
In the backward pass we need to compute how the squared error changes with changing weight. i.e we compute $\partial E_{total}/\partial w_{i}$ for each weight $w_{i}$. This is shown below

A squared error is assumed

Error gradient  with $w_{5}$

$\partial E_{total}/\partial w_{5} = \partial E_{total}/\partial out_{o_{1}} * \partial out_{o_{1}}/\partial in_{o_{1}} * \partial in_{o_{1}}/ \partial w_{5}$                -(B)

Since
$E_{total} = 1/2\sum (target - output)^{2}$
$E_{total} = 1/2(target_{o_{1}} - out_{o_{1}})^{2} + 1/2(target_{o_{2}} - out_{o_{2}})^{2}$
$\partial E _{total}/\partial out_{o1} = \partial E_{o1}/\partial out_{o1} + \partial E_{o2}/\partial out_{o1}$
$\partial E _{total}/\partial out_{o1} = \partial /\partial _{out_{o1}}[1/2(target_{01}-out_{01})^{2}- 1/2(target_{02}-out_{02})^{2}]$
$\partial E _{total}/\partial out_{o1} = 2 * 1/2*(target_{01} - out_{01}) *-1 + 0$

Now considering the 2nd term in (B)
$\partial out_{o1}/\partial in_{o1} = \partial/\partial in_{o1} [1/(1+e^{-in_{o1}})]$

Using result (A)
$\partial out_{o1}/\partial in_{o1} = \partial/\partial in_{o1} [1/(1+e^{-in_{o1}})] = out_{o1}(1-out_{o1})$

The 3rd term in (B)
$\partial in_{o1}/\partial w_{5} = \partial/\partial w_{5} [w_{5}*out_{h1} + w_{6}*out_{h2}] = out_{h1}$
$\partial E_{total}/\partial w_{5}=-(target_{o1} - out_{o1}) * out_{o1} *(1-out_{o1}) * out_{h1}$

Having computed $\partial E_{total}/\partial w_{5}$, we now perform gradient descent, by computing a new weight, assuming a learning rate $\alpha$
$w_{5}^{+} = w_{5} - \alpha * \partial E_{total}/\partial w_{5}$

If we do this for $\partial E_{total}/\partial w_{6}$ we would get
$\partial E_{total}/\partial w_{6}=-(target_{02} - out_{02}) * out_{02} *(1-out_{02}) * out_{h2}$

3)Hidden layer

We now compute how the total error changes for a change in weight $w_{1}$
$\partial E_{total}/\partial w_{1}= \partial E_{total}/\partial out_{h1}* \partial out_{h1}/\partial in_{h1} * \partial in_{h1}/\partial w_{1}$ – (C)

Using
$E_{total} = E_{o1} + E_{o2}$ we get
$\partial E_{total}/\partial w_{1}= (\partial E_{o1}/\partial out_{h1}+ \partial E_{o2}/\partial out_{h1}) * \partial out_{h1}/\partial in_{h1} * \partial in_{h1}/\partial w_{1}$
$\partial E_{total}/\partial w_{1}=(\partial E_{o1}/\partial out_{h1}+ \partial E_{o2}/\partial out_{h1}) * out_{h1}*(1-out_{h1})*i_{1}$     -(D)

Considering the 1st term in (C)
$\partial E_{total}/\partial out_{h1}= \partial E_{o1}/\partial out_{h1}+ \partial E_{o2}/\partial out_{h1}$

Now
$\partial E_{o1}/\partial out_{h1} = \partial E_{o1}/\partial out_{o1} *\partial out_{o1}/\partial in_{01} * \partial in_{o1}/\partial out_{h1}$
$\partial E_{o2}/\partial out_{h1} = \partial E_{o2}/\partial out_{o2} *\partial out_{o2}/\partial in_{02} * \partial in_{o2}/\partial out_{h1}$

which gives the following
$\partial E_{o1}/\partial out_{o1} *\partial out_{o1}/\partial in_{o1} * \partial in_{o1}/\partial out_{h1} =-(target_{o1}-out_{o1}) *out_{o1}(1-out_{o1})*w_{5}$ – (E)
$\partial E_{o2}/\partial out_{o2} *\partial out_{o2}/\partial in_{02} * \partial in_{o2}/\partial out_{h1} =-(target_{o2}-out_{o2}) *out_{o2}(1-out_{o2})*w_{6}$ – (F)

Combining (D), (E) & (F) we get
$\partial E_{total}/\partial w_{1} = -[(target_{o1}-out_{o1}) *out_{o1}(1-out_{o1})*w_{5} + (target_{o2}-out_{o2}) *out_{o2}(1-out_{o2})*w_{6}]*out_{h1}*(1-out_{h1})*i_{1}$

This can be represented as
$\partial E_{total}/\partial w_{1} = -\sum_{i}[(target_{oi}-out_{oi}) *out_{oi}(1-out_{oi})*w_{j}]*out_{h1}*(1-out_{h1})*i_{1}$

With this derivative a new value of $w_{1}$ is computed
$w_{1}^{+} = w_{1} - \alpha * \partial E_{total}/\partial w_{1}$

Hence there are 2 important results
At the output layer we have
a) $\partial E_{total}/\partial w_{j}=-(target_{oi} - out_{oi}) * out_{oi} *(1-out_{oi}) * out_{hi}$
At each hidden layer we compute
b) $\partial E_{total}/\partial w_{k} = -\sum_{i}[(target_{oi}-out_{oi}) *out_{oi}(1-out_{oi})*w_{j}]*out_{hk}*(1-out_{hk})*i_{k}$

Backpropagation, was very successful in the early years,  but the algorithm does have its problems for e.g the issue of the ‘vanishing’ and ‘exploding’ gradient. Yet it is a very key development in Neural Networks, and  the issues with the backprop gradients have been addressed through techniques such as the  momentum method and adaptive learning rate etc.

In this post. I derive the weights at the output layer and the hidden layer. As I already mentioned above, I intend to implement a vector version of the backpropagation algorithm in Octave, R and Python in the days to come.

Watch this space! I’ll be back

P.S. If you find any typos/errors, do let me know!

References
1. Neural Networks for Machine Learning by Prof Geoffrey Hinton
2. A Step by Step Backpropagation Example by Matt Mazur
3. The Backpropagation algorithm by R Rojas
4. Backpropagation Learning Artificial Neural Networks David S Touretzky
5. Artificial Intelligence, Prof Sudeshna Sarkar, NPTEL

To see all my posts go to ‘Index of Posts