Welcome to your first (required) programming assignment! You will build a logistic regression classifier to recognize cats. This assignment will step you through how to do this with a Neural Network mindset, and so will also hone your intuitions about deep learning.
Instructions:
You will learn to:
This notebook has been updated over the past few months. The prior version was named "v5", and the current versionis now named '6a'
First, let's run the cell below to import all the packages that you will need during this assignment.
import numpy as np
import matplotlib.pyplot as plt
import h5py
import scipy
from PIL import Image
from scipy import ndimage
from lr_utils import load_dataset
%matplotlib inline
Problem Statement: You are given a dataset ("data.h5") containing:
 a training set of m_train images labeled as cat (y=1) or noncat (y=0)
 a test set of m_test images labeled as cat or noncat
 each image is of shape (num_px, num_px, 3) where 3 is for the 3 channels (RGB). Thus, each image is square (height = num_px) and (width = num_px).
You will build a simple imagerecognition algorithm that can correctly classify pictures as cat or noncat.
Let's get more familiar with the dataset. Load the data by running the following code.
# Loading the data (cat/noncat)
train_set_x_orig, train_set_y, test_set_x_orig, test_set_y, classes = load_dataset()
We added "_orig" at the end of image datasets (train and test) because we are going to preprocess them. After preprocessing, we will end up with train_set_x and test_set_x (the labels train_set_y and test_set_y don't need any preprocessing).
Each line of your train_set_x_orig and test_set_x_orig is an array representing an image. You can visualize an example by running the following code. Feel free also to change the index
value and rerun to see other images.
# Example of a picture
index =10
plt.imshow(train_set_x_orig[index])
print ("y = " + str(train_set_y[0, index]) + ", it's a '" + classes[np.squeeze(train_set_y[:, index])].decode("utf8") + "' picture.")
Many software bugs in deep learning come from having matrix/vector dimensions that don't fit. If you can keep your matrix/vector dimensions straight you will go a long way toward eliminating many bugs.
Exercise: Find the values for:
 m_train (number of training examples)
 m_test (number of test examples)
 num_px (= height = width of a training image)
Remember that train_set_x_orig
is a numpyarray of shape (m_train, num_px, num_px, 3). For instance, you can access m_train
by writing train_set_x_orig.shape[0]
.
### START CODE HERE ### (≈ 3 lines of code)
m_train = train_set_x_orig.shape[0]
m_test = test_set_x_orig.shape[0]
num_px = train_set_x_orig.shape[1]
### END CODE HERE ###
print ("Number of training examples: m_train = " + str(m_train))
print ("Number of testing examples: m_test = " + str(m_test))
print ("Height/Width of each image: num_px = " + str(num_px))
print ("Each image is of size: (" + str(num_px) + ", " + str(num_px) + ", 3)")
print ("train_set_x shape: " + str(train_set_x_orig.shape))
print ("train_set_y shape: " + str(train_set_y.shape))
print ("test_set_x shape: " + str(test_set_x_orig.shape))
print ("test_set_y shape: " + str(test_set_y.shape))
Expected Output for m_train, m_test and num_px:
**m_train**  209 
**m_test**  50 
**num_px**  64 
For convenience, you should now reshape images of shape (num_px, num_px, 3) in a numpyarray of shape (num_px $*$ num_px $*$ 3, 1). After this, our training (and test) dataset is a numpyarray where each column represents a flattened image. There should be m_train (respectively m_test) columns.
Exercise: Reshape the training and test data sets so that images of size (num_px, num_px, 3) are flattened into single vectors of shape (num_px $*$ num_px $*$ 3, 1).
A trick when you want to flatten a matrix X of shape (a,b,c,d) to a matrix X_flatten of shape (b$*$c$*$d, a) is to use:
X_flatten = X.reshape(X.shape[0], 1).T # X.T is the transpose of X
# Reshape the training and test examples
### START CODE HERE ### (≈ 2 lines of code)
train_set_x_flatten = train_set_x_orig.reshape(train_set_x_orig.shape[1]*train_set_x_orig.shape[2]*train_set_x_orig.shape[3],train_set_x_orig.shape[0])
test_set_x_flatten = test_set_x_orig.reshape(test_set_x_orig.shape[1]*test_set_x_orig.shape[2]*test_set_x_orig.shape[3],test_set_x_orig.shape[0])
### END CODE HERE ###
print ("train_set_x_flatten shape: " + str(train_set_x_flatten.shape))
print ("train_set_y shape: " + str(train_set_y.shape))
print ("test_set_x_flatten shape: " + str(test_set_x_flatten.shape))
print ("test_set_y shape: " + str(test_set_y.shape))
print ("sanity check after reshaping: " + str(train_set_x_flatten[0:5,0]))
Expected Output:
**train_set_x_flatten shape**  (12288, 209) 
**train_set_y shape**  (1, 209) 
**test_set_x_flatten shape**  (12288, 50) 
**test_set_y shape**  (1, 50) 
**sanity check after reshaping**  [17 31 56 22 33] 
To represent color images, the red, green and blue channels (RGB) must be specified for each pixel, and so the pixel value is actually a vector of three numbers ranging from 0 to 255.
One common preprocessing step in machine learning is to center and standardize your dataset, meaning that you substract the mean of the whole numpy array from each example, and then divide each example by the standard deviation of the whole numpy array. But for picture datasets, it is simpler and more convenient and works almost as well to just divide every row of the dataset by 255 (the maximum value of a pixel channel).
Let's standardize our dataset.
train_set_x = train_set_x_flatten/255.
test_set_x = test_set_x_flatten/255.
What you need to remember:
Common steps for preprocessing a new dataset are:
It's time to design a simple algorithm to distinguish cat images from noncat images.
You will build a Logistic Regression, using a Neural Network mindset. The following Figure explains why Logistic Regression is actually a very simple Neural Network!
Mathematical expression of the algorithm:
For one example $x^{(i)}$: $$z^{(i)} = w^T x^{(i)} + b \tag{1}$$ $$\hat{y}^{(i)} = a^{(i)} = sigmoid(z^{(i)})\tag{2}$$ $$ \mathcal{L}(a^{(i)}, y^{(i)}) =  y^{(i)} \log(a^{(i)})  (1y^{(i)} ) \log(1a^{(i)})\tag{3}$$
The cost is then computed by summing over all training examples: $$ J = \frac{1}{m} \sum_{i=1}^m \mathcal{L}(a^{(i)}, y^{(i)})\tag{6}$$
Key steps: In this exercise, you will carry out the following steps:
 Initialize the parameters of the model
 Learn the parameters for the model by minimizing the cost
 Use the learned parameters to make predictions (on the test set)
 Analyse the results and conclude
The main steps for building a Neural Network are:
You often build 13 separately and integrate them into one function we call model()
.
Exercise: Using your code from "Python Basics", implement sigmoid()
. As you've seen in the figure above, you need to compute $sigmoid( w^T x + b) = \frac{1}{1 + e^{(w^T x + b)}}$ to make predictions. Use np.exp().
# GRADED FUNCTION: sigmoid
def sigmoid(z):
"""
Compute the sigmoid of z
Arguments:
z  A scalar or numpy array of any size.
Return:
s  sigmoid(z)
"""
### START CODE HERE ### (≈ 1 line of code)
s = 1/(1+np.exp(z))
### END CODE HERE ###
return s
print ("sigmoid([0, 2]) = " + str(sigmoid(np.array([0,2]))))
Expected Output:
**sigmoid([0, 2])**  [ 0.5 0.88079708] 
Exercise: Implement parameter initialization in the cell below. You have to initialize w as a vector of zeros. If you don't know what numpy function to use, look up np.zeros() in the Numpy library's documentation.
# GRADED FUNCTION: initialize_with_zeros
def initialize_with_zeros(dim):
"""
This function creates a vector of zeros of shape (dim, 1) for w and initializes b to 0.
Argument:
dim  size of the w vector we want (or number of parameters in this case)
Returns:
w  initialized vector of shape (dim, 1)
b  initialized scalar (corresponds to the bias)
"""
### START CODE HERE ### (≈ 1 line of code)
w = np.zeros((dim,1))
b = 0
### END CODE HERE ###
assert(w.shape == (dim, 1))
assert(isinstance(b, float) or isinstance(b, int))
return w, b
dim = 2
w, b = initialize_with_zeros(dim)
print ("w = " + str(w))
print ("b = " + str(b))
Expected Output:
** w **  [[ 0.] [ 0.]] 
** b **  0 
For image inputs, w will be of shape (num_px $\times$ num_px $\times$ 3, 1).
Now that your parameters are initialized, you can do the "forward" and "backward" propagation steps for learning the parameters.
Exercise: Implement a function propagate()
that computes the cost function and its gradient.
Hints:
Forward Propagation:
Here are the two formulas you will be using:
$$ \frac{\partial J}{\partial w} = \frac{1}{m}X(AY)^T\tag{7}$$$$ \frac{\partial J}{\partial b} = \frac{1}{m} \sum_{i=1}^m (a^{(i)}y^{(i)})\tag{8}$$# GRADED FUNCTION: propagate
def propagate(w, b, X, Y):
"""
Implement the cost function and its gradient for the propagation explained above
Arguments:
w  weights, a numpy array of size (num_px * num_px * 3, 1)
b  bias, a scalar
X  data of size (num_px * num_px * 3, number of examples)
Y  true "label" vector (containing 0 if noncat, 1 if cat) of size (1, number of examples)
Return:
cost  negative loglikelihood cost for logistic regression
dw  gradient of the loss with respect to w, thus same shape as w
db  gradient of the loss with respect to b, thus same shape as b
Tips:
 Write your code step by step for the propagation. np.log(), np.dot()
"""
m = X.shape[1]
# FORWARD PROPAGATION (FROM X TO COST)
### START CODE HERE ### (≈ 2 lines of code)
A = sigmoid(np.dot(w.T,X) + b) # compute activation
cost = np.sum((( np.log(A))*Y + (np.log(1A))*(1Y)))/m # compute cost
### END CODE HERE ###
# BACKWARD PROPAGATION (TO FIND GRAD)
### START CODE HERE ### (≈ 2 lines of code)
dw = (np.dot(X,(AY).T))/m
db = (np.sum(AY))/m
### END CODE HERE ###
assert(dw.shape == w.shape)
assert(db.dtype == float)
cost = np.squeeze(cost)
assert(cost.shape == ())
grads = {"dw": dw,
"db": db}
return grads, cost
w, b, X, Y = np.array([[1.],[2.]]), 2., np.array([[1.,2.,1.],[3.,4.,3.2]]), np.array([[1,0,1]])
grads, cost = propagate(w, b, X, Y)
print ("dw = " + str(grads["dw"]))
print ("db = " + str(grads["db"]))
print ("cost = " + str(cost))
Expected Output:
** dw **  [[ 0.99845601] [ 2.39507239]] 
** db **  0.00145557813678 
** cost **  5.801545319394553 
Exercise: Write down the optimization function. The goal is to learn $w$ and $b$ by minimizing the cost function $J$. For a parameter $\theta$, the update rule is $ \theta = \theta  \alpha \text{ } d\theta$, where $\alpha$ is the learning rate.
# GRADED FUNCTION: optimize
def optimize(w, b, X, Y, num_iterations, learning_rate, print_cost = False):
"""
This function optimizes w and b by running a gradient descent algorithm
Arguments:
w  weights, a numpy array of size (num_px * num_px * 3, 1)
b  bias, a scalar
X  data of shape (num_px * num_px * 3, number of examples)
Y  true "label" vector (containing 0 if noncat, 1 if cat), of shape (1, number of examples)
num_iterations  number of iterations of the optimization loop
learning_rate  learning rate of the gradient descent update rule
print_cost  True to print the loss every 100 steps
Returns:
params  dictionary containing the weights w and bias b
grads  dictionary containing the gradients of the weights and bias with respect to the cost function
costs  list of all the costs computed during the optimization, this will be used to plot the learning curve.
Tips:
You basically need to write down two steps and iterate through them:
1) Calculate the cost and the gradient for the current parameters. Use propagate().
2) Update the parameters using gradient descent rule for w and b.
"""
costs = []
for i in range(num_iterations):
# Cost and gradient calculation (≈ 14 lines of code)
### START CODE HERE ###
grads, cost = propagate(w, b, X, Y)
### END CODE HERE ###
# Retrieve derivatives from grads
dw = grads["dw"]
db = grads["db"]
# update rule (≈ 2 lines of code)
### START CODE HERE ###
w = w  (learning_rate*dw)
b = b  (learning_rate*db)
### END CODE HERE ###
# Record the costs
if i % 100 == 0:
costs.append(cost)
# Print the cost every 100 training iterations
if print_cost and i % 100 == 0:
print ("Cost after iteration %i: %f" %(i, cost))
params = {"w": w,
"b": b}
grads = {"dw": dw,
"db": db}
return params, grads, costs
params, grads, costs = optimize(w, b, X, Y, num_iterations= 100, learning_rate = 0.009, print_cost = False)
print ("w = " + str(params["w"]))
print ("b = " + str(params["b"]))
print ("dw = " + str(grads["dw"]))
print ("db = " + str(grads["db"]))
import matplotlib.pyplot as plt
plt.plot(costs)
Expected Output:
**w**  [[ 0.19033591] [ 0.12259159]] 
**b**  1.92535983008 
**dw**  [[ 0.67752042] [ 1.41625495]] 
**db**  0.219194504541 
Exercise: The previous function will output the learned w and b. We are able to use w and b to predict the labels for a dataset X. Implement the predict()
function. There are two steps to computing predictions:
Calculate $\hat{Y} = A = \sigma(w^T X + b)$
Convert the entries of a into 0 (if activation <= 0.5) or 1 (if activation > 0.5), stores the predictions in a vector Y_prediction
. If you wish, you can use an if
/else
statement in a for
loop (though there is also a way to vectorize this).
# GRADED FUNCTION: predict
def predict(w, b, X):
'''
Predict whether the label is 0 or 1 using learned logistic regression parameters (w, b)
Arguments:
w  weights, a numpy array of size (num_px * num_px * 3, 1)
b  bias, a scalar
X  data of size (num_px * num_px * 3, number of examples)
Returns:
Y_prediction  a numpy array (vector) containing all predictions (0/1) for the examples in X
'''
m = X.shape[1]
Y_prediction = np.zeros((1,m))
w = w.reshape(X.shape[0], 1)
# Compute vector "A" predicting the probabilities of a cat being present in the picture
### START CODE HERE ### (≈ 1 line of code)
A = sigmoid(np.dot(w.T,X) + b) # Dimentions = (1, m)
### END CODE HERE ###
#### WORKING SOLUTION 1: USING IF ELSE ####
#for i in range(A.shape[1]):
## Convert probabilities A[0,i] to actual predictions p[0,i]
### START CODE HERE ### (≈ 4 lines of code)
#if (A[0,i] >= 0.5):
# Y_prediction[0, i] = 1
#else:
# Y_prediction[0, i] = 0
### END CODE HERE ###
#### WORKING SOLUTION 2: ONE LINE ####
#for i in range(A.shape[1]):
## Convert probabilities A[0,i] to actual predictions p[0,i]
### START CODE HERE ### (≈ 4 lines of code)
#Y_prediction[0, i] = 1 if A[0,i] >=0.5 else 0
### END CODE HERE ###
#### WORKING SOLUTION 3: VECTORISED IMPLEMENTATION ####
Y_prediction = (A >= 0.5) * 1.0
assert(Y_prediction.shape == (1, m))
return Y_prediction
w = np.array([[0.1124579],[0.23106775]])
b = 0.3
X = np.array([[1.,1.1,3.2],[1.2,2.,0.1]])
print ("predictions = " + str(predict(w, b, X)))
Expected Output:
**predictions**  [[ 1. 1. 0.]] 
What to remember: You've implemented several functions that:
You will now see how the overall model is structured by putting together all the building blocks (functions implemented in the previous parts) together, in the right order.
Exercise: Implement the model function. Use the following notation:
 Y_prediction_test for your predictions on the test set
 Y_prediction_train for your predictions on the train set
 w, costs, grads for the outputs of optimize()
# GRADED FUNCTION: model
def model(X_train, Y_train, X_test, Y_test, num_iterations = 2000, learning_rate = 0.5, print_cost = False):
"""
Builds the logistic regression model by calling the function you've implemented previously
Arguments:
X_train  training set represented by a numpy array of shape (num_px * num_px * 3, m_train)
Y_train  training labels represented by a numpy array (vector) of shape (1, m_train)
X_test  test set represented by a numpy array of shape (num_px * num_px * 3, m_test)
Y_test  test labels represented by a numpy array (vector) of shape (1, m_test)
num_iterations  hyperparameter representing the number of iterations to optimize the parameters
learning_rate  hyperparameter representing the learning rate used in the update rule of optimize()
print_cost  Set to true to print the cost every 100 iterations
Returns:
d  dictionary containing information about the model.
"""
### START CODE HERE ###
# initialize parameters with zeros (≈ 1 line of code)
w, b = initialize_with_zeros(X_train.shape[0])
# Gradient descent (≈ 1 line of code)
parameters, grads, costs = optimize(w, b, X_train, Y_train, num_iterations, learning_rate, print_cost)
# Retrieve parameters w and b from dictionary "parameters"
w = parameters["w"]
b = parameters["b"]
# Predict test/train set examples (≈ 2 lines of code)
Y_prediction_test = predict(w, b, X_test)
Y_prediction_train = predict(w, b, X_train)
### END CODE HERE ###
# Print train/test Errors
print("train accuracy: {} %".format(100  np.mean(np.abs(Y_prediction_train  Y_train)) * 100))
print("test accuracy: {} %".format(100  np.mean(np.abs(Y_prediction_test  Y_test)) * 100))
d = {"costs": costs,
"Y_prediction_test": Y_prediction_test,
"Y_prediction_train" : Y_prediction_train,
"w" : w,
"b" : b,
"learning_rate" : learning_rate,
"num_iterations": num_iterations}
return d
Run the following cell to train your model.
d = model(train_set_x, train_set_y, test_set_x, test_set_y, num_iterations = 2000, learning_rate = 0.005, print_cost = False)
Expected Output:
**Cost after iteration 0 **  0.693147 


**Train Accuracy**  99.04306220095694 % 
**Test Accuracy**  70.0 % 
Comment: Training accuracy is close to 100%. This is a good sanity check: your model is working and has high enough capacity to fit the training data. Test accuracy is 68%. It is actually not bad for this simple model, given the small dataset we used and that logistic regression is a linear classifier. But no worries, you'll build an even better classifier next week!
Also, you see that the model is clearly overfitting the training data. Later in this specialization you will learn how to reduce overfitting, for example by using regularization. Using the code below (and changing the index
variable) you can look at predictions on pictures of the test set.
# Example of a picture that was wrongly classified.
index = 25
plt.imshow(test_set_x[:,index].reshape((num_px, num_px, 3)))
print ("y = " + str(test_set_y[0,index]) + ", you predicted that it is a \"" + classes[d["Y_prediction_test"][0,index]].decode("utf8") + "\" picture.")
Let's also plot the cost function and the gradients.
# Plot learning curve (with costs)
costs = np.squeeze(d['costs'])
plt.plot(costs)
plt.ylabel('cost')
plt.xlabel('iterations (per hundreds)')
plt.title("Learning rate =" + str(d["learning_rate"]))
plt.show()
Interpretation: You can see the cost decreasing. It shows that the parameters are being learned. However, you see that you could train the model even more on the training set. Try to increase the number of iterations in the cell above and rerun the cells. You might see that the training set accuracy goes up, but the test set accuracy goes down. This is called overfitting.
Congratulations on building your first image classification model. Let's analyze it further, and examine possible choices for the learning rate $\alpha$.
Reminder: In order for Gradient Descent to work you must choose the learning rate wisely. The learning rate $\alpha$ determines how rapidly we update the parameters. If the learning rate is too large we may "overshoot" the optimal value. Similarly, if it is too small we will need too many iterations to converge to the best values. That's why it is crucial to use a welltuned learning rate.
Let's compare the learning curve of our model with several choices of learning rates. Run the cell below. This should take about 1 minute. Feel free also to try different values than the three we have initialized the learning_rates
variable to contain, and see what happens.
learning_rates = [0.01, 0.001, 0.0001]
models = {}
for i in learning_rates:
print ("learning rate is: " + str(i))
models[str(i)] = model(train_set_x, train_set_y, test_set_x, test_set_y, num_iterations = 1500, learning_rate = i, print_cost = False)
print ('\n' + "" + '\n')
for i in learning_rates:
plt.plot(np.squeeze(models[str(i)]["costs"]), label= str(models[str(i)]["learning_rate"]))
plt.ylabel('cost')
plt.xlabel('iterations (hundreds)')
legend = plt.legend(loc='upper center', shadow=True)
frame = legend.get_frame()
frame.set_facecolor('0.90')
plt.show()
Interpretation:
Congratulations on finishing this assignment. You can use your own image and see the output of your model. To do that:
1. Click on "File" in the upper bar of this notebook, then click "Open" to go on your Coursera Hub.
2. Add your image to this Jupyter Notebook's directory, in the "images" folder
3. Change your image's name in the following code
4. Run the code and check if the algorithm is right (1 = cat, 0 = noncat)!
## START CODE HERE ## (PUT YOUR IMAGE NAME)
my_image = "my_image.jpg" # change this to the name of your image file
## END CODE HERE ##
# We preprocess the image to fit your algorithm.
fname = "images/" + my_image
image = np.array(ndimage.imread(fname, flatten=False))
image = image/255.
my_image = scipy.misc.imresize(image, size=(num_px,num_px)).reshape((1, num_px*num_px*3)).T
my_predicted_image = predict(d["w"], d["b"], my_image)
plt.imshow(image)
print("y = " + str(np.squeeze(my_predicted_image)) + ", your algorithm predicts a \"" + classes[int(np.squeeze(my_predicted_image)),].decode("utf8") + "\" picture.")
What to remember from this assignment:
Finally, if you'd like, we invite you to try different things on this Notebook. Make sure you submit before trying anything. Once you submit, things you can play with include:
 Play with the learning rate and the number of iterations
 Try different initialization methods and compare the results
 Test other preprocessings (center the data, or divide each row by its standard deviation)