Deep Learning from first principles in Python, R and Octave – Part 2

January 11, 2018
By

(This article was first published on R – Giga thoughts …, and kindly contributed to R-bloggers)

“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.

You could also check out my book on Amazon Practical Machine Learning with R and Python – Machine Learning in Stereo,  in which I implement several Machine Learning algorithms on regression and classification, along with other necessary metrics that are used in Machine Learning.

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) - coshx*d/dx(sinhx))/(1-sinhx^{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!!

References
1. Deep Learning Specialization
2. Neural Networks for Machine Learning
3. Deep Learning, Ian Goodfellow, Yoshua Bengio and Aaron Courville
4. Neural Networks: The mechanics of backpropagation
5. Machine Learning

Also see
1. My book ‘Practical Machine Learning with R and Python’ on Amazon
2. GooglyPlus: yorkr analyzes IPL players, teams, matches with plots and tables
3. The 3rd paperback & kindle editions of my books on Cricket, now on Amazon
4. Exploring Quantum Gate operations with QCSimulator
5. Simulating a Web Joint in Android
6. My travels through the realms of Data Science, Machine Learning, Deep Learning and (AI)
7. Presentation on Wireless Technologies – Part 1

To see all posts check Index of posts

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