Welcome to your week 3 programming assignment. It's time to build your first neural network, which will have a hidden layer. You will see a big difference between this model and the one you implemented using logistic regression.
You will learn how to:
Let's first import all the packages that you will need during this assignment.
In [1]:
# Package imports
import numpy as np
import matplotlib.pyplot as plt
from testCases_v2 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 consistent
In [2]:
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. Your goal is to build a model to fit this data.
In [3]:
# Visualize the data:
plt.scatter(X[0, :], X[1, :], c=Y, s=40, cmap=plt.cm.Spectral);
You 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.
Exercise: How many training examples do you have? In addition, what is the shape of the variables X and Y?
Hint: How do you get the shape of a numpy array? (help)
In [4]:
### START CODE HERE ### (≈ 3 lines of code)
shape_X = X.shape
shape_Y = Y.shape
m = shape_X[1] # training set size
### END CODE HERE ###
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))
Expected Output:
| **shape of X** | (2, 400) |
| **shape of Y** | (1, 400) |
| **m** | 400 |
In [5]:
# Train the logistic regression classifier
clf = sklearn.linear_model.LogisticRegressionCV();
clf.fit(X.T, Y.T);
You can now plot the decision boundary of these models. Run the code below.
In [6]:
# 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)")
Expected Output:
| **Accuracy** | 47% |
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". You are going to train a Neural Network with a single hidden layer.
Here is our model:
Mathematically:
For one example $x^{(i)}$: $$z^{[1] (i)} = W^{[1]} x^{(i)} + b^{[1] (i)}\tag{1}$$ $$a^{[1] (i)} = \tanh(z^{[1] (i)})\tag{2}$$ $$z^{[2] (i)} = W^{[2]} a^{[1] (i)} + b^{[2] (i)}\tag{3}$$ $$\hat{y}^{(i)} = a^{[2] (i)} = \sigma(z^{ [2] (i)})\tag{4}$$ $$y^{(i)}_{prediction} = \begin{cases} 1 & \mbox{if } a^{[2](i)} > 0.5 \\ 0 & \mbox{otherwise } \end{cases}\tag{5}$$
Given the predictions on all the examples, you can also compute the cost $J$ as follows: $$J = - \frac{1}{m} \sum\limits_{i = 0}^{m} \large\left(\small y^{(i)}\log\left(a^{[2] (i)}\right) + (1-y^{(i)})\log\left(1- a^{[2] (i)}\right) \large \right) \small \tag{6}$$
Reminder: The general methodology to build a Neural Network is to:
1. Define the neural network structure ( # of input units, # of hidden units, etc).
2. Initialize the model's parameters
3. Loop:
- Implement forward propagation
- Compute loss
- Implement backward propagation to get the gradients
- Update parameters (gradient descent)
You often build helper functions to compute steps 1-3 and then merge them into one function we call nn_model(). Once you've built nn_model() and learnt the right parameters, you can make predictions on new data.
In [7]:
# GRADED FUNCTION: layer_sizes
def layer_sizes(X, Y):
"""
Arguments:
X -- input dataset of shape (input size, number of examples)
Y -- labels of shape (output size, number of examples)
Returns:
n_x -- the size of the input layer
n_h -- the size of the hidden layer
n_y -- the size of the output layer
"""
### START CODE HERE ### (≈ 3 lines of code)
n_x = X.shape[0] # size of input layer
n_h = 4
n_y = Y.shape[0] # size of output layer
### END CODE HERE ###
return (n_x, n_h, n_y)
In [8]:
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))
Expected Output (these are not the sizes you will use for your network, they are just used to assess the function you've just coded).
| **n_x** | 5 |
| **n_h** | 4 |
| **n_y** | 2 |
Exercise: Implement the function initialize_parameters().
Instructions:
np.random.randn(a,b) * 0.01 to randomly initialize a matrix of shape (a,b).np.zeros((a,b)) to initialize a matrix of shape (a,b) with zeros.
In [35]:
# GRADED FUNCTION: initialize_parameters
def initialize_parameters(n_x, n_h, n_y):
"""
Argument:
n_x -- size of the input layer
n_h -- size of the hidden layer
n_y -- size of the output layer
Returns:
params -- python dictionary containing your parameters:
W1 -- weight matrix of shape (n_h, n_x)
b1 -- bias vector of shape (n_h, 1)
W2 -- weight matrix of shape (n_y, n_h)
b2 -- bias vector of shape (n_y, 1)
"""
np.random.seed(2) # we set up a seed so that your output matches ours although the initialization is random.
### START CODE HERE ### (≈ 4 lines of code)
W1 = np.random.randn(n_h, n_x) * 0.01
b1 = np.zeros(shape=(n_h, 1))
W2 = np.random.randn(n_y, n_h) * 0.01
b2 = np.zeros((n_y, 1))
### END CODE HERE ###
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 parameters
In [36]:
n_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"]))
Expected Output:
| **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.]] |
Question: Implement forward_propagation().
Instructions:
sigmoid(). It is built-in (imported) in the notebook.np.tanh(). It is part of the numpy library.initialize_parameters()) by using parameters[".."].cache". The cache will be given as an input to the backpropagation function.
In [37]:
# GRADED FUNCTION: forward_propagation
def forward_propagation(X, parameters):
"""
Argument:
X -- input data of size (n_x, m)
parameters -- python dictionary containing your parameters (output of initialization function)
Returns:
A2 -- The sigmoid output of the second activation
cache -- a dictionary containing "Z1", "A1", "Z2" and "A2"
"""
# Retrieve each parameter from the dictionary "parameters"
### START CODE HERE ### (≈ 4 lines of code)
W1 = parameters["W1"]
b1 = parameters["b1"]
W2 = parameters["W2"]
b2 = parameters["b2"]
### END CODE HERE ###
# Implement Forward Propagation to calculate A2 (probabilities)
### START CODE HERE ### (≈ 4 lines of code)
Z1 = np.dot(W1, X) + b1
A1 = np.tanh(Z1)
Z2 = np.dot(W2, A1) + b2
A2 = sigmoid(Z2)
### END CODE HERE ###
assert(A2.shape == (1, X.shape[1]))
cache = {"Z1": Z1,
"A1": A1,
"Z2": Z2,
"A2": A2}
return A2, cache
In [38]:
X_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']))
Expected Output:
| 0.262818640198 0.091999045227 -1.30766601287 0.212877681719 |
Now that you have computed $A^{[2]}$ (in the Python variable "A2"), which contains $a^{[2](i)}$ for every example, you can compute the cost function as follows:
Exercise: Implement compute_cost() to compute the value of the cost $J$.
Instructions:
logprobs = np.multiply(np.log(A2),Y)
cost = - np.sum(logprobs) # no need to use a for loop!
(you can use either np.multiply() and then np.sum() or directly np.dot()).
In [39]:
# GRADED FUNCTION: compute_cost
def compute_cost(A2, Y, parameters):
"""
Computes the cross-entropy cost given in equation (13)
Arguments:
A2 -- The sigmoid output of the second activation, of shape (1, number of examples)
Y -- "true" labels vector of shape (1, number of examples)
parameters -- python dictionary containing your parameters W1, b1, W2 and b2
Returns:
cost -- cross-entropy cost given equation (13)
"""
m = Y.shape[1] # number of example
# Compute the cross-entropy cost
### START CODE HERE ### (≈ 2 lines of code)
logprobs = np.multiply(np.log(A2), Y) + np.multiply((1 - Y), np.log(1 - A2))
cost = - np.sum(logprobs) / m
### END CODE HERE ###
cost = np.squeeze(cost) # makes sure cost is the dimension we expect.
# E.g., turns [[17]] into 17
assert(isinstance(cost, float))
return cost
In [40]:
A2, Y_assess, parameters = compute_cost_test_case()
print("cost = " + str(compute_cost(A2, Y_assess, parameters)))
Expected Output:
| **cost** | 0.693058761... |
Using the cache computed during forward propagation, you can now implement backward propagation.
Question: Implement the function backward_propagation().
Instructions: Backpropagation is usually the hardest (most mathematical) part in deep learning. To help you, here again is the slide from the lecture on backpropagation. You'll want to use the six equations on the right of this slide, since you are building a vectorized implementation.
(1 - np.power(A1, 2)).
In [41]:
# GRADED FUNCTION: backward_propagation
def backward_propagation(parameters, cache, X, Y):
"""
Implement the backward propagation using the instructions above.
Arguments:
parameters -- python dictionary containing our parameters
cache -- a dictionary containing "Z1", "A1", "Z2" and "A2".
X -- input data of shape (2, number of examples)
Y -- "true" labels vector of shape (1, number of examples)
Returns:
grads -- python dictionary containing your gradients with respect to different parameters
"""
m = X.shape[1]
# First, retrieve W1 and W2 from the dictionary "parameters".
### START CODE HERE ### (≈ 2 lines of code)
W1 = parameters["W1"]
W2 = parameters["W2"]
### END CODE HERE ###
# Retrieve also A1 and A2 from dictionary "cache".
### START CODE HERE ### (≈ 2 lines of code)
A1 = cache["A1"]
A2 = cache["A2"]
### END CODE HERE ###
# Backward propagation: calculate dW1, db1, dW2, db2.
### START CODE HERE ### (≈ 6 lines of code, corresponding to 6 equations on slide above)
dZ2 = A2 - Y
dW2 = (1 / m) * np.dot(dZ2, A1.T)
db2 = (1 / m) * np.sum(dZ2, axis=1, keepdims=True)
dZ1 = np.multiply(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)
### END CODE HERE ###
grads = {"dW1": dW1,
"db1": db1,
"dW2": dW2,
"db2": db2}
return grads
In [42]:
parameters, 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"]))
Expected output:
| **dW1** | [[ 0.00301023 -0.00747267] [ 0.00257968 -0.00641288] [-0.00156892 0.003893 ] [-0.00652037 0.01618243]] |
| **db1** | [[ 0.00176201] [ 0.00150995] [-0.00091736] [-0.00381422]] |
| **dW2** | [[ 0.00078841 0.01765429 -0.00084166 -0.01022527]] |
| **db2** | [[-0.16655712]] |
Question: Implement the update rule. Use gradient descent. You have to use (dW1, db1, dW2, db2) in order to update (W1, b1, W2, b2).
General gradient descent rule: $ \theta = \theta - \alpha \frac{\partial J }{ \partial \theta }$ where $\alpha$ is the learning rate and $\theta$ represents a parameter.
Illustration: The gradient descent algorithm with a good learning rate (converging) and a bad learning rate (diverging). Images courtesy of Adam Harley.
In [43]:
# GRADED FUNCTION: update_parameters
def update_parameters(parameters, grads, learning_rate = 1.2):
"""
Updates parameters using the gradient descent update rule given above
Arguments:
parameters -- python dictionary containing your parameters
grads -- python dictionary containing your gradients
Returns:
parameters -- python dictionary containing your updated parameters
"""
# Retrieve each parameter from the dictionary "parameters"
### START CODE HERE ### (≈ 4 lines of code)
W1 = parameters["W1"]
b1 = parameters["b1"]
W2 = parameters["W2"]
b2 = parameters["b2"]
### END CODE HERE ###
# Retrieve each gradient from the dictionary "grads"
### START CODE HERE ### (≈ 4 lines of code)
dW1 = grads["dW1"]
db1 = grads["db1"]
dW2 = grads["dW2"]
db2 = grads["db2"]
## END CODE HERE ###
# Update rule for each parameter
### START CODE HERE ### (≈ 4 lines of code)
W1 = W1 - learning_rate * dW1
b1 = b1 - learning_rate * db1
W2 = W2 - learning_rate * dW2
b2 = b2 - learning_rate * db2
### END CODE HERE ###
parameters = {"W1": W1,
"b1": b1,
"W2": W2,
"b2": b2}
return parameters
In [44]:
parameters, 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"]))
Expected Output:
| **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]] |
In [45]:
# GRADED FUNCTION: nn_model
def nn_model(X, Y, n_h, num_iterations = 10000, print_cost=False):
"""
Arguments:
X -- dataset of shape (2, number of examples)
Y -- labels of shape (1, number of examples)
n_h -- size of the hidden layer
num_iterations -- Number of iterations in gradient descent loop
print_cost -- if True, print the cost every 1000 iterations
Returns:
parameters -- parameters learnt by the model. They can then be used to predict.
"""
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".
### START CODE HERE ### (≈ 5 lines of code)
parameters = initialize_parameters(n_x, n_h, n_y)
W1 = parameters["W1"]
b1 = parameters["b1"]
W2 = parameters["W2"]
b2 = parameters["b2"]
### END CODE HERE ###
# Loop (gradient descent)
for i in range(0, num_iterations):
### START CODE HERE ### (≈ 4 lines of code)
# 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)
### END CODE HERE ###
# Print the cost every 1000 iterations
if print_cost and i % 1000 == 0:
print ("Cost after iteration %i: %f" %(i, cost))
return parameters
In [46]:
X_assess, Y_assess = nn_model_test_case()
parameters = nn_model(X_assess, Y_assess, 4, num_iterations=10000, print_cost=True)
print("W1 = " + str(parameters["W1"]))
print("b1 = " + str(parameters["b1"]))
print("W2 = " + str(parameters["W2"]))
print("b2 = " + str(parameters["b2"]))
Expected Output:
| **cost after iteration 0** | 0.692739 |
|
|
|
| **W1** | [[-0.65848169 1.21866811] [-0.76204273 1.39377573] [ 0.5792005 -1.10397703] [ 0.76773391 -1.41477129]] |
| **b1** | [[ 0.287592 ] [ 0.3511264 ] [-0.2431246 ] [-0.35772805]] |
| **W2** | [[-2.45566237 -3.27042274 2.00784958 3.36773273]] |
| **b2** | [[ 0.20459656]] |
Question: Use your model to predict by building predict(). Use forward propagation to predict results.
Reminder: predictions = $y_{prediction} = \mathbb 1 \text{{activation > 0.5}} = \begin{cases} 1 & \text{if}\ activation > 0.5 \\ 0 & \text{otherwise} \end{cases}$
As an example, if you would like to set the entries of a matrix X to 0 and 1 based on a threshold you would do: X_new = (X > threshold)
In [47]:
# GRADED FUNCTION: predict
def predict(parameters, X):
"""
Using the learned parameters, predicts a class for each example in X
Arguments:
parameters -- python dictionary containing your parameters
X -- input data of size (n_x, m)
Returns
predictions -- vector of predictions of our model (red: 0 / blue: 1)
"""
# Computes probabilities using forward propagation, and classifies to 0/1 using 0.5 as the threshold.
### START CODE HERE ### (≈ 2 lines of code)
A2, cache = forward_propagation(X, parameters)
predictions = np.round(A2)
### END CODE HERE ###
return predictions
In [48]:
parameters, X_assess = predict_test_case()
predictions = predict(parameters, X_assess)
print("predictions mean = " + str(np.mean(predictions)))
Expected Output:
| **predictions mean** | 0.666666666667 |
It is time to run the model and see how it performs on a planar dataset. Run the following code to test your model with a single hidden layer of $n_h$ hidden units.
In [49]:
# 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))
Out[49]:
Expected Output:
| **Cost after iteration 9000** | 0.218607 |
In [50]:
# 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) + '%')
Expected Output:
| **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.
In [51]:
# This may take about 2 minutes to run
plt.figure(figsize=(16, 32))
hidden_layer_sizes = [1, 2, 3, 4, 5, 20, 50]
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))
Interpretation:
Optional questions:
Note: Remember to submit the assignment but clicking the blue "Submit Assignment" button at the upper-right.
Some optional/ungraded questions that you can explore if you wish:
You've learnt to:
Nice work!
If you want, you can rerun the whole notebook (minus the dataset part) for each of the following datasets.
In [52]:
# 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}
### START CODE HERE ### (choose your dataset)
dataset = "noisy_moons"
### END CODE HERE ###
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);
Congrats on finishing this Programming Assignment!
Reference: