Classification with one hidden layer on Planar data
Building neural network, which will have a hidden layer implementing using logistic regression.
From this project, I have learn how to:
- Implement a 2-class classification neural network with a single hidden layer
- Use units with a non-linear activation function, such as tanh
- Compute the cross entropy loss
- Implement forward and backward propagation
Let's first import all the packages that you will need during this project.
- numpy is the fundamental package for scientific computing with Python.
- sklearn provides simple and efficient tools for data mining and data analysis.
- matplotlib is a library for plotting graphs in Python.
- testCases provides some test examples to assess the correctness of your functions
- planar_utils provide various useful functions used in this project
# Package imports
import numpy as np
import matplotlib.pyplot as plt
from testCases import *
import sklearn
import sklearn.datasets
import sklearn.linear_model
from planar_utils import plot_decision_boundary, sigmoid, load_planar_dataset, load_extra_datasets
%matplotlib inline
np.random.seed(1) # set a seed so that the results are consistentFirst, let's get the dataset that will work on. The following code will load a "flower" 2-class dataset into variables X and Y.
X, Y = load_planar_dataset() Visualize the dataset using matplotlib. The data looks like a "flower" with some red (label y=0) and some blue (y=1) points. My goal is to build a model to fit this data.
# Visualize the data:
plt.scatter(X[0, :], X[1, :], c=Y, s=40, cmap=plt.cm.Spectral);
I have: - a numpy-array (matrix) X that contains your features (x1, x2) - a numpy-array (vector) Y that contains your labels (red:0, blue:1).
Lets first get a better sense of what our data is like.
shape_X = X.shape
shape_Y = Y.shape
m = shape_X[1] # training set size
print ('The shape of X is: ' + str(shape_X))
print ('The shape of Y is: ' + str(shape_Y))
print ('I have m = %d training examples!' % (m))The shape of X is: (2, 400)
The shape of Y is: (1, 400)
I have m = 400 training examples!
Before building a full neural network, lets first see how logistic regression performs on this problem. We can use sklearn's built-in functions to do that. Run the code below to train a logistic regression classifier on the dataset.
# Train the logistic regression classifier
clf = sklearn.linear_model.LogisticRegressionCV();
clf.fit(X.T, Y.T);/Users/rajeshidumalla/opt/anaconda3/lib/python3.8/site-packages/sklearn/utils/validation.py:63: DataConversionWarning: A column-vector y was passed when a 1d array was expected. Please change the shape of y to (n_samples, ), for example using ravel().
return f(*args, **kwargs)
Ploting the decision boundary of these models. Run the code below.
# Plot the decision boundary for logistic regression
plot_decision_boundary(lambda x: clf.predict(x), X, Y)
plt.title("Logistic Regression")
# Print accuracy
LR_predictions = clf.predict(X.T)
print ('Accuracy of logistic regression: %d ' % float((np.dot(Y,LR_predictions) + np.dot(1-Y,1-LR_predictions))/float(Y.size)*100) +
'% ' + "(percentage of correctly labelled datapoints)")Accuracy of logistic regression: 47 % (percentage of correctly labelled datapoints)
Interpretation: The dataset is not linearly separable, so logistic regression doesn't perform well. Hopefully a neural network will do better. Let's try this now!
Logistic regression did not work well on the "flower dataset". I am going to train a Neural Network with a single hidden layer.
Mathematically:
Defining three variables: - n_x: the size of the input layer - n_h: the size of the hidden layer (set this to 4) - n_y: the size of the output layer
I am using the shapes of X and Y to find n_x and n_y. Also, hard code the hidden layer size to be 4.
# GRADED FUNCTION: layer_sizes
def layer_sizes(X, Y):
n_x = X.shape[0] # size of input layer
n_h = 4
n_y = Y.shape[0] # size of output layer
return (n_x, n_h, n_y)X_assess, Y_assess = layer_sizes_test_case()
(n_x, n_h, n_y) = layer_sizes(X_assess, Y_assess)
print("The size of the input layer is: n_x = " + str(n_x))
print("The size of the hidden layer is: n_h = " + str(n_h))
print("The size of the output layer is: n_y = " + str(n_y))The size of the input layer is: n_x = 5
The size of the hidden layer is: n_h = 4
The size of the output layer is: n_y = 2
Implementing the function initialize_parameters().
# GRADED FUNCTION: initialize_parameters
def initialize_parameters(n_x, n_h, n_y):
np.random.seed(2) # I am seting up a seed so that the output matches, although the initialization is random.
W1 = np.random.randn(n_h,n_x) * 0.01
b1 = np.zeros((n_h,1))
W2 = np.random.randn(n_y,n_h) * 0.01
b2 = np.zeros((n_y,1))
assert (W1.shape == (n_h, n_x))
assert (b1.shape == (n_h, 1))
assert (W2.shape == (n_y, n_h))
assert (b2.shape == (n_y, 1))
parameters = {"W1": W1,
"b1": b1,
"W2": W2,
"b2": b2}
return parametersn_x, n_h, n_y = initialize_parameters_test_case()
parameters = initialize_parameters(n_x, n_h, n_y)
print("W1 = " + str(parameters["W1"]))
print("b1 = " + str(parameters["b1"]))
print("W2 = " + str(parameters["W2"]))
print("b2 = " + str(parameters["b2"]))W1 = [[-0.00416758 -0.00056267]
[-0.02136196 0.01640271]
[-0.01793436 -0.00841747]
[ 0.00502881 -0.01245288]]
b1 = [[0.]
[0.]
[0.]
[0.]]
W2 = [[-0.01057952 -0.00909008 0.00551454 0.02292208]]
b2 = [[0.]]
Implementinnng forward_propagation().
# GRADED FUNCTION: forward_propagation
def forward_propagation(X, parameters):
# Retrieve each parameter from the dictionary "parameters"
W1 = parameters["W1"]
b1 = parameters["b1"]
W2 = parameters["W2"]
b2 = parameters["b2"]
# Implement Forward Propagation to calculate A2 (probabilities)
Z1 = np.dot(W1,X) + b1
A1 = np.tanh(Z1)
Z2 = np.dot(W2,A1) + b2
A2 = sigmoid(Z2)
assert(A2.shape == (1, X.shape[1]))
cache = {"Z1": Z1,
"A1": A1,
"Z2": Z2,
"A2": A2}
return A2, cacheX_assess, parameters = forward_propagation_test_case()
A2, cache = forward_propagation(X_assess, parameters)
# Note: we use the mean here just to make sure that your output matches ours.
print(np.mean(cache['Z1']) ,np.mean(cache['A1']),np.mean(cache['Z2']),np.mean(cache['A2']))-0.0004997557777419913 -0.0004969633532317802 0.0004381874509591466 0.500109546852431
Now that I have computed A2"), which contains $a^{2}$ for every example, I can compute the cost function as follows:
Implementinng compute_cost() to compute the value of the cost
# GRADED FUNCTION: compute_cost
def compute_cost(A2, Y, parameters):
m = Y.shape[1] # number of example
# Compute the cross-entropy cost
logprobs = Y*np.log(A2) + (1-Y)* np.log(1-A2)
cost = -1/m * np.sum(logprobs)
cost = np.squeeze(cost) # makes sure cost is the dimension we expect.
# E.g., turns [[17]] into 17
assert(isinstance(cost, float))
return costA2, Y_assess, parameters = compute_cost_test_case()
print("cost = " + str(compute_cost(A2, Y_assess, parameters)))cost = 0.6929198937761265
By using the cache computed during forward propagation, I can now implement backward propagation.
Implementing the function backward_propagation().
# GRADED FUNCTION: backward_propagation
def backward_propagation(parameters, cache, X, Y):
m = X.shape[1]
# First, retrieve W1 and W2 from the dictionary "parameters".
W1 = parameters["W1"]
W2 = parameters["W2"]
# Retrieve also A1 and A2 from dictionary "cache".
A1 = cache["A1"]
A2 = cache["A2"]
# Backward propagation: calculate dW1, db1, dW2, db2.
dZ2= A2 - Y
dW2 = 1 / m * np.dot(dZ2,A1.T)
db2 = 1 / m * np.sum(dZ2,axis=1,keepdims=True)
dZ1 = np.dot(W2.T,dZ2) * (1-np.power(A1,2))
dW1 = 1 / m * np.dot(dZ1,X.T)
db1 = 1 / m * np.sum(dZ1,axis=1,keepdims=True)
grads = {"dW1": dW1,
"db1": db1,
"dW2": dW2,
"db2": db2}
return gradsparameters, cache, X_assess, Y_assess = backward_propagation_test_case()
grads = backward_propagation(parameters, cache, X_assess, Y_assess)
print ("dW1 = "+ str(grads["dW1"]))
print ("db1 = "+ str(grads["db1"]))
print ("dW2 = "+ str(grads["dW2"]))
print ("db2 = "+ str(grads["db2"]))dW1 = [[ 0.01018708 -0.00708701]
[ 0.00873447 -0.0060768 ]
[-0.00530847 0.00369379]
[-0.02206365 0.01535126]]
db1 = [[-0.00069728]
[-0.00060606]
[ 0.000364 ]
[ 0.00151207]]
dW2 = [[ 0.00363613 0.03153604 0.01162914 -0.01318316]]
db2 = [[0.06589489]]
Implementing the update rule using gradient descent. I am using (dW1, db1, dW2, db2) in order to update (W1, b1, W2, b2).
# GRADED FUNCTION: update_parameters
def update_parameters(parameters, grads, learning_rate = 1.2):
# Retrieve each parameter from the dictionary "parameters"
W1 = parameters["W1"]
b1 = parameters["b1"]
W2 = parameters["W2"]
b2 = parameters["b2"]
# Retrieve each gradient from the dictionary "grads"
dW1 = grads["dW1"]
db1 = grads["db1"]
dW2 = grads["dW2"]
db2 = grads["db2"]
# Update rule for each parameter
W1 = W1 - learning_rate * dW1
b1 = b1 - learning_rate * db1
W2 = W2 - learning_rate * dW2
b2 = b2 - learning_rate * db2
parameters = {"W1": W1,
"b1": b1,
"W2": W2,
"b2": b2}
return parametersparameters, grads = update_parameters_test_case()
parameters = update_parameters(parameters, grads)
print("W1 = " + str(parameters["W1"]))
print("b1 = " + str(parameters["b1"]))
print("W2 = " + str(parameters["W2"]))
print("b2 = " + str(parameters["b2"]))W1 = [[-0.00643025 0.01936718]
[-0.02410458 0.03978052]
[-0.01653973 -0.02096177]
[ 0.01046864 -0.05990141]]
b1 = [[-1.02420756e-06]
[ 1.27373948e-05]
[ 8.32996807e-07]
[-3.20136836e-06]]
W2 = [[-0.01041081 -0.04463285 0.01758031 0.04747113]]
b2 = [[0.00010457]]
# GRADED FUNCTION: nn_model
def nn_model(X, Y, n_h, num_iterations = 10000, print_cost=False):
np.random.seed(3)
n_x = layer_sizes(X, Y)[0]
n_y = layer_sizes(X, Y)[2]
# Initialize parameters, then retrieve W1, b1, W2, b2. Inputs: "n_x, n_h, n_y". Outputs = "W1, b1, W2, b2, parameters".
parameters = initialize_parameters(n_x, n_h, n_y)
W1 = parameters["W1"]
b1 = parameters["b1"]
W2 = parameters["W2"]
b2 = parameters["b2"]
# Loop (gradient descent)
for i in range(0, num_iterations):
# Forward propagation. Inputs: "X, parameters". Outputs: "A2, cache".
A2, cache = forward_propagation(X, parameters)
# Cost function. Inputs: "A2, Y, parameters". Outputs: "cost".
cost = compute_cost(A2, Y, parameters)
# Backpropagation. Inputs: "parameters, cache, X, Y". Outputs: "grads".
grads = backward_propagation(parameters, cache, X, Y)
# Gradient descent parameter update. Inputs: "parameters, grads". Outputs: "parameters".
parameters = update_parameters(parameters, grads)
# Print the cost every 1000 iterations
if print_cost and i % 1000 == 0:
print ("Cost after iteration %i: %f" %(i, cost))
return parametersX_assess, Y_assess = nn_model_test_case()
parameters = nn_model(X_assess, Y_assess, 4, num_iterations=10000, print_cost=False)
print("W1 = " + str(parameters["W1"]))
print("b1 = " + str(parameters["b1"]))
print("W2 = " + str(parameters["W2"]))
print("b2 = " + str(parameters["b2"]))<ipython-input-13-991ce5e9d2fe>:20: RuntimeWarning: divide by zero encountered in log
logprobs = Y*np.log(A2) + (1-Y)* np.log(1-A2)
/Users/rajeshidumalla/planar_utils.py:34: RuntimeWarning: overflow encountered in exp
s = 1/(1+np.exp(-x))
W1 = [[-4.18491249 5.33220652]
[-7.52991891 1.24304481]
[-4.19257906 5.32654053]
[ 7.52989842 -1.24305585]]
b1 = [[ 2.32930709]
[ 3.79453736]
[ 2.33008777]
[-3.79456804]]
W2 = [[-6033.83651818 -6008.12961806 -6033.10073783 6008.06596049]]
b2 = [[-52.66626424]]
Using the model to predict by building predict() by using forward propagation to predict results.
# GRADED FUNCTION: predict
def predict(parameters, X):
# Computes probabilities using forward propagation, and classifies to 0/1 using 0.5 as the threshold.
A2, cache = forward_propagation(X, parameters)
predictions = np.round(A2)
return predictionsparameters, X_assess = predict_test_case()
predictions = predict(parameters, X_assess)
print("predictions mean = " + str(np.mean(predictions)))predictions mean = 0.6666666666666666
It is time to run the model and see how it performs on a planar dataset. Running the following code to test the model with a single hidden layer of
# Build a model with a n_h-dimensional hidden layer
parameters = nn_model(X, Y, n_h = 4, num_iterations = 10000, print_cost=True)
# Plot the decision boundary
plot_decision_boundary(lambda x: predict(parameters, x.T), X, Y)
plt.title("Decision Boundary for hidden layer size " + str(4))Cost after iteration 0: 0.693048
Cost after iteration 1000: 0.288083
Cost after iteration 2000: 0.254385
Cost after iteration 3000: 0.233864
Cost after iteration 4000: 0.226792
Cost after iteration 5000: 0.222644
Cost after iteration 6000: 0.219731
Cost after iteration 7000: 0.217504
Cost after iteration 8000: 0.219456
Cost after iteration 9000: 0.218558
Text(0.5, 1.0, 'Decision Boundary for hidden layer size 4')
# Print accuracy
predictions = predict(parameters, X)
print ('Accuracy: %d' % float((np.dot(Y,predictions.T) + np.dot(1-Y,1-predictions.T))/float(Y.size)*100) + '%')Accuracy: 90%
Accuracy is really high compared to Logistic Regression. The model has learnt the leaf patterns of the flower! Neural networks are able to learn even highly non-linear decision boundaries, unlike logistic regression.
Now, let's try out several hidden layer sizes.
4.6 - Tuning hidden layer size (optional/ungraded exercise)
By running the following code, we will observe different behaviors of the model for various hidden layer sizes.
# This may take about 2 minutes to run
plt.figure(figsize=(16, 32))
hidden_layer_sizes = [1, 2, 3, 4, 5, 10, 20]
for i, n_h in enumerate(hidden_layer_sizes):
plt.subplot(5, 2, i+1)
plt.title('Hidden Layer of size %d' % n_h)
parameters = nn_model(X, Y, n_h, num_iterations = 5000)
plot_decision_boundary(lambda x: predict(parameters, x.T), X, Y)
predictions = predict(parameters, X)
accuracy = float((np.dot(Y,predictions.T) + np.dot(1-Y,1-predictions.T))/float(Y.size)*100)
print ("Accuracy for {} hidden units: {} %".format(n_h, accuracy))Accuracy for 1 hidden units: 67.5 %
Accuracy for 2 hidden units: 67.25 %
Accuracy for 3 hidden units: 90.75 %
Accuracy for 4 hidden units: 90.5 %
Accuracy for 5 hidden units: 91.25 %
Accuracy for 10 hidden units: 90.25 %
Accuracy for 20 hidden units: 90.5 %
Interpretation:
- The larger models (with more hidden units) are able to fit the training set better, until eventually the largest models overfit the data.
- The best hidden layer size seems to be around n_h = 5. Indeed, a value around here seems to fits the data well without also incurring noticable overfitting.
- We also do some regularization, which lets you use very large models (such as n_h = 50) without much overfitting.
If you want, you can rerun the whole notebook (minus the dataset part) for each of the following datasets.
# Datasets
noisy_circles, noisy_moons, blobs, gaussian_quantiles, no_structure = load_extra_datasets()
datasets = {"noisy_circles": noisy_circles,
"noisy_moons": noisy_moons,
"blobs": blobs,
"gaussian_quantiles": gaussian_quantiles}
dataset = "gaussian_quantiles"
X, Y = datasets[dataset]
X, Y = X.T, Y.reshape(1, Y.shape[0])
# make blobs binary
if dataset == "blobs":
Y = Y%2
# Visualize the data
plt.scatter(X[0, :], X[1, :], c=Y, s=40, cmap=plt.cm.Spectral);