deeplearning.ai-lecture2-week1-Initialization-homework
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Initialization
Welcome to the first assignment of "Improving Deep Neural Networks".
Training your neural network requires specifying an initial value of the weights. A well chosen initialization method will help learning.
If you completed the previous course of this specialization, you probably followed our instructions for weight initialization, and it has worked out so far. But how do you choose the initialization for a new neural network? In this notebook, you will see how different initializations lead to different results.
A well chosen initialization can:
- Speed up the convergence of gradient descent
- Increase the odds of gradient descent converging to a lower training (and generalization) error
To get started, run the following cell to load the packages and the planar dataset you will try to classify.
import numpy as np
import matplotlib.pyplot as plt
import sklearn
import sklearn.datasets
from init_utils import sigmoid, relu, compute_loss, forward_propagation, backward_propagation
from init_utils import update_parameters, predict, load_dataset, plot_decision_boundary, predict_dec
%matplotlib inline
plt.rcParams['figure.figsize'] = (7.0, 4.0) # set default size of plots
plt.rcParams['image.interpolation'] = 'nearest'
plt.rcParams['image.cmap'] = 'gray'
# load image dataset: blue/red dots in circles
train_X, train_Y, test_X, test_Y = load_dataset()
You would like a classifier to separate the blue dots from the red dots.
1 - Neural Network model
You will use a 3-layer neural network (already implemented for you). Here are the initialization methods you will experiment with:
- Zeros initialization -- setting
initialization = "zeros"
in the input argument. - Random initialization -- setting
initialization = "random"
in the input argument. This initializes the weights to large random values. - He initialization -- setting
initialization = "he"
in the input argument. This initializes the weights to random values scaled according to a paper by He et al., 2015.
Instructions: Please quickly read over the code below, and run it. In the next part you will implement the three initialization methods that this model()
calls.
def model(X, Y, learning_rate = 0.01, num_iterations = 15000, print_cost = True, initialization = "he"):
"""
Implements a three-layer neural network: LINEAR->RELU->LINEAR->RELU->LINEAR->SIGMOID.
Arguments:
X -- input data, of shape (2, number of examples)
Y -- true "label" vector (containing 0 for red dots; 1 for blue dots), of shape (1, number of examples)
learning_rate -- learning rate for gradient descent
num_iterations -- number of iterations to run gradient descent
print_cost -- if True, print the cost every 1000 iterations
initialization -- flag to choose which initialization to use ("zeros","random" or "he")
Returns:
parameters -- parameters learnt by the model
"""
grads = {}
costs = [] # to keep track of the loss
m = X.shape[1] # number of examples
layers_dims = [X.shape[0], 10, 5, 1]
# Initialize parameters dictionary.
if initialization == "zeros":
parameters = initialize_parameters_zeros(layers_dims)
elif initialization == "random":
parameters = initialize_parameters_random(layers_dims)
elif initialization == "he":
parameters = initialize_parameters_he(layers_dims)
# Loop (gradient descent)
for i in range(0, num_iterations):
# Forward propagation: LINEAR -> RELU -> LINEAR -> RELU -> LINEAR -> SIGMOID.
a3, cache = forward_propagation(X, parameters)
# Loss
cost = compute_loss(a3, Y)
# Backward propagation.
grads = backward_propagation(X, Y, cache)
# Update parameters.
parameters = update_parameters(parameters, grads, learning_rate)
# Print the loss every 1000 iterations
if print_cost and i % 1000 == 0:
print("Cost after iteration {}: {}".format(i, cost))
costs.append(cost)
# plot the loss
plt.plot(costs)
plt.ylabel('cost')
plt.xlabel('iterations (per hundreds)')
plt.title("Learning rate =" + str(learning_rate))
plt.show()
return parameters
2 - Zero initialization
There are two types of parameters to initialize in a neural network:
- the weight matrices
(W[1],W[2],W[3],...,W[L−1],W[L]) (W[1],W[2],W[3],...,W[L−1],W[L]) - the bias vectors
(b[1],b[2],b[3],...,b[L−1],b[L]) (b[1],b[2],b[3],...,b[L−1],b[L])
Exercise: Implement the following function to initialize all parameters to zeros. You'll see later that this does not work well since it fails to "break symmetry", but lets try it anyway and see what happens. Use np.zeros((..,..)) with the correct shapes.
# GRADED FUNCTION: initialize_parameters_zeros
def initialize_parameters_zeros(layers_dims):
"""
Arguments:
layer_dims -- python array (list) containing the size of each layer.
Returns:
parameters -- python dictionary containing your parameters "W1", "b1", ..., "WL", "bL":
W1 -- weight matrix of shape (layers_dims[1], layers_dims[0])
b1 -- bias vector of shape (layers_dims[1], 1)
...
WL -- weight matrix of shape (layers_dims[L], layers_dims[L-1])
bL -- bias vector of shape (layers_dims[L], 1)
"""
parameters = {}
L = len(layers_dims) # number of layers in the network
for l in range(1, L):
### START CODE HERE ### (≈ 2 lines of code)
parameters['W' + str(l)] = np.zeros((layers_dims[l],layers_dims[l-1]))
parameters['b' + str(l)] = np.zeros((layers_dims[l],1))
### END CODE HERE ###
return parameters
parameters = initialize_parameters_zeros([3,2,1])
print("W1 = " + str(parameters["W1"]))
print("b1 = " + str(parameters["b1"]))
print("W2 = " + str(parameters["W2"]))
print("b2 = " + str(parameters["b2"]))
W1 = [[ 0. 0. 0.] [ 0. 0. 0.]]b1 = [[ 0.] [ 0.]]W2 = [[ 0. 0.]]b2 = [[ 0.]]
Expected Output:
W1[[ 0. 0. 0.] [ 0. 0. 0.]]b1[[ 0.] [ 0.]]W2[[ 0. 0.]]b2[[ 0.]]Run the following code to train your model on 15,000 iterations using zeros initialization.
parameters = model(train_X, train_Y, initialization = "zeros")
print ("On the train set:")
predictions_train = predict(train_X, train_Y, parameters)
print ("On the test set:")
predictions_test = predict(test_X, test_Y, parameters)
Cost after iteration 0: 0.6931471805599453Cost after iteration 1000: 0.6931471805599453Cost after iteration 2000: 0.6931471805599453Cost after iteration 3000: 0.6931471805599453Cost after iteration 4000: 0.6931471805599453Cost after iteration 5000: 0.6931471805599453Cost after iteration 6000: 0.6931471805599453Cost after iteration 7000: 0.6931471805599453Cost after iteration 8000: 0.6931471805599453Cost after iteration 9000: 0.6931471805599453Cost after iteration 10000: 0.6931471805599455Cost after iteration 11000: 0.6931471805599453Cost after iteration 12000: 0.6931471805599453Cost after iteration 13000: 0.6931471805599453Cost after iteration 14000: 0.6931471805599453
On the train set:Accuracy: 0.5On the test set:Accuracy: 0.5
The performance is really bad, and the cost does not really decrease, and the algorithm performs no better than random guessing. Why? Lets look at the details of the predictions and the decision boundary:
print ("predictions_train = " + str(predictions_train))
print ("predictions_test = " + str(predictions_test))
predictions_train = [[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]]predictions_test = [[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]]
plt.title("Model with Zeros initialization")
axes = plt.gca()
axes.set_xlim([-1.5,1.5])
axes.set_ylim([-1.5,1.5])
plot_decision_boundary(lambda x: predict_dec(parameters, x.T), train_X, train_Y)
The model is predicting 0 for every example.
In general, initializing all the weights to zero results in the network failing to break symmetry. This means that every neuron in each layer will learn the same thing, and you might as well be training a neural network with
What you should remember:
- The weights
W[l] should be initialized randomly to break symmetry. - It is however okay to initialize the biases
b[l] to zeros. Symmetry is still broken so long asW[l] is initialized randomly.
3 - Random initialization
To break symmetry, lets intialize the weights randomly. Following random initialization, each neuron can then proceed to learn a different function of its inputs. In this exercise, you will see what happens if the weights are intialized randomly, but to very large values.
Exercise: Implement the following function to initialize your weights to large random values (scaled by *10) and your biases to zeros. Use np.random.randn(..,..) * 10
for weights and np.zeros((.., ..))
for biases. We are using a fixed np.random.seed(..)
to make sure your "random" weights match ours, so don't worry if running several times your code gives you always the same initial values for the parameters.
# GRADED FUNCTION: initialize_parameters_random
def initialize_parameters_random(layers_dims):
"""
Arguments:
layer_dims -- python array (list) containing the size of each layer.
Returns:
parameters -- python dictionary containing your parameters "W1", "b1", ..., "WL", "bL":
W1 -- weight matrix of shape (layers_dims[1], layers_dims[0])
b1 -- bias vector of shape (layers_dims[1], 1)
...
WL -- weight matrix of shape (layers_dims[L], layers_dims[L-1])
bL -- bias vector of shape (layers_dims[L], 1)
"""
np.random.seed(3) # This seed makes sure your "random" numbers will be the as ours
parameters = {}
L = len(layers_dims) # integer representing the number of layers
for l in range(1, L):
### START CODE HERE ### (≈ 2 lines of code)
parameters['W' + str(l)] = np.random.randn(layers_dims[l],layers_dims[l-1])*10
parameters['b' + str(l)] = np.zeros((layers_dims[l],1))
### END CODE HERE ###
return parameters
parameters = initialize_parameters_random([3, 2, 1])
print("W1 = " + str(parameters["W1"]))
print("b1 = " + str(parameters["b1"]))
print("W2 = " + str(parameters["W2"]))
print("b2 = " + str(parameters["b2"]))
W1 = [[ 1.78862847 0.43650985 0.09649747] [-1.8634927 -0.2773882 -0.35475898]]b1 = [[ 0.] [ 0.]]W2 = [[-0.08274148 -0.62700068]]b2 = [[ 0.]]
Expected Output:
W1[[ 17.88628473 4.36509851 0.96497468] [-18.63492703 -2.77388203 -3.54758979]]b1[[ 0.] [ 0.]]W2[[-0.82741481 -6.27000677]]b2[[ 0.]]Run the following code to train your model on 15,000 iterations using random initialization.
parameters = model(train_X, train_Y, initialization = "random")
print ("On the train set:")
predictions_train = predict(train_X, train_Y, parameters)
print ("On the test set:")
predictions_test = predict(test_X, test_Y, parameters)
Cost after iteration 0: 1.9698193182646353Cost after iteration 1000: 0.6894749458317239Cost after iteration 2000: 0.675058063210226Cost after iteration 3000: 0.6469210868251528Cost after iteration 4000: 0.5398790761260324Cost after iteration 5000: 0.4062642269764849Cost after iteration 6000: 0.29844708868759456Cost after iteration 7000: 0.2218373466209485Cost after iteration 8000: 0.16926424179038072Cost after iteration 9000: 0.13413308969827087Cost after iteration 10000: 0.10873865543082417Cost after iteration 11000: 0.0916944306812697Cost after iteration 12000: 0.07991173603998075Cost after iteration 13000: 0.07083949901112581Cost after iteration 14000: 0.06370209022580518
On the train set:Accuracy: 0.996666666667On the test set:Accuracy: 0.96
If you see "inf" as the cost after the iteration 0, this is because of numerical roundoff; a more numerically sophisticated implementation would fix this. But this isn't worth worrying about for our purposes.
Anyway, it looks like you have broken symmetry, and this gives better results. than before. The model is no longer outputting all 0s.
print (predictions_train)
print (predictions_test)
[[1 0 1 1 0 0 1 0 1 1 1 0 1 0 0 0 0 1 1 0 0 0 1 0 0 0 1 0 1 1 0 1 1 0 0 0 0 1 0 1 1 1 1 0 0 1 1 1 0 1 0 1 1 1 1 0 0 1 1 1 1 1 0 1 0 1 0 1 0 1 1 0 0 0 0 0 1 0 0 0 1 0 1 0 0 1 1 1 1 1 1 0 0 0 1 1 0 1 1 0 1 0 0 1 0 1 1 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 1 0 0 0 0 1 0 1 1 0 1 1 1 1 0 1 1 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 1 0 1 0 0 1 0 1 0 0 0 1 0 1 1 1 0 0 1 1 0 0 1 0 1 0 0 1 0 1 1 0 1 1 1 1 0 1 0 1 0 0 1 0 1 0 0 0 1 1 1 1 0 1 0 0 1 1 0 0 1 0 1 0 0 1 0 0 1 1 0 0 0 0 1 1 0 1 0 1 1 0 1 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 1 0 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 0 0 0 0 1 1 0 1 0]][[1 0 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 1 1 1 0 1 0 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 1 0 1 0 1 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 1 1 0 1 0 0 0 0 1 0 1 1 0 0 0 1 1 0 0 0 0 0]]
plt.title("Model with large random initialization")
axes = plt.gca()
axes.set_xlim([-1.5,1.5])
axes.set_ylim([-1.5,1.5])
plot_decision_boundary(lambda x: predict_dec(parameters, x.T), train_X, train_Y)
Observations:
- The cost starts very high. This is because with large random-valued weights, the last activation (sigmoid) outputs results that are very close to 0 or 1 for some examples, and when it gets that example wrong it incurs a very high loss for that example. Indeed, when
log(a[3])=log(0) log(a[3])=log(0), the loss goes to infinity. - Poor initialization can lead to vanishing/exploding gradients, which also slows down the optimization algorithm.
- If you train this network longer you will see better results, but initializing with overly large random numbers slows down the optimization.
In summary:
- Initializing weights to very large random values does not work well.
- Hopefully intializing with small random values does better. The important question is: how small should be these random values be? Lets find out in the next part!
4 - He initialization
Finally, try "He Initialization"; this is named for the first author of He et al., 2015. (If you have heard of "Xavier initialization", this is similar except Xavier initialization uses a scaling factor for the weights sqrt(1./layers_dims[l-1])
where He initialization would use sqrt(2./layers_dims[l-1])
.)
Exercise: Implement the following function to initialize your parameters with He initialization.
Hint: This function is similar to the previous initialize_parameters_random(...)
. The only difference is that instead of multiplying np.random.randn(..,..)
by 10, you will multiply it by
# GRADED FUNCTION: initialize_parameters_he
def initialize_parameters_he(layers_dims):
"""
Arguments:
layer_dims -- python array (list) containing the size of each layer.
Returns:
parameters -- python dictionary containing your parameters "W1", "b1", ..., "WL", "bL":
W1 -- weight matrix of shape (layers_dims[1], layers_dims[0])
b1 -- bias vector of shape (layers_dims[1], 1)
...
WL -- weight matrix of shape (layers_dims[L], layers_dims[L-1])
bL -- bias vector of shape (layers_dims[L], 1)
"""
np.random.seed(3)
parameters = {}
L = len(layers_dims) - 1 # integer representing the number of layers
for l in range(1, L + 1):
### START CODE HERE ### (≈ 2 lines of code)
parameters['W' + str(l)] = np.random.randn(layers_dims[l],layers_dims[l-1])*np.sqrt(2./layers_dims[l-1])
parameters['b' + str(l)] = np.zeros((layers_dims[l],1))
### END CODE HERE ###
return parameters
parameters = initialize_parameters_he([2, 4, 1])
print("W1 = " + str(parameters["W1"]))
print("b1 = " + str(parameters["b1"]))
print("W2 = " + str(parameters["W2"]))
print("b2 = " + str(parameters["b2"]))
W1 = [[ 1.78862847 0.43650985] [ 0.09649747 -1.8634927 ] [-0.2773882 -0.35475898] [-0.08274148 -0.62700068]]b1 = [[ 0.] [ 0.] [ 0.] [ 0.]]W2 = [[-0.03098412 -0.33744411 -0.92904268 0.62552248]]b2 = [[ 0.]]
Expected Output:
W1[[ 1.78862847 0.43650985] [ 0.09649747 -1.8634927 ] [-0.2773882 -0.35475898] [-0.08274148 -0.62700068]]b1[[ 0.] [ 0.] [ 0.] [ 0.]]W2[[-0.03098412 -0.33744411 -0.92904268 0.62552248]]b2[[ 0.]]Run the following code to train your model on 15,000 iterations using He initialization.
parameters = model(train_X, train_Y, initialization = "he")
print ("On the train set:")
predictions_train = predict(train_X, train_Y, parameters)
print ("On the test set:")
predictions_test = predict(test_X, test_Y, parameters)
Cost after iteration 0: 0.8830537463419761Cost after iteration 1000: 0.6879825919728063Cost after iteration 2000: 0.6751286264523371Cost after iteration 3000: 0.6526117768893807Cost after iteration 4000: 0.6082958970572938Cost after iteration 5000: 0.5304944491717495Cost after iteration 6000: 0.4138645817071794Cost after iteration 7000: 0.3117803464844441Cost after iteration 8000: 0.23696215330322562Cost after iteration 9000: 0.18597287209206836Cost after iteration 10000: 0.1501555628037182Cost after iteration 11000: 0.12325079292273548Cost after iteration 12000: 0.09917746546525937Cost after iteration 13000: 0.0845705595402428Cost after iteration 14000: 0.07357895962677366
On the train set:Accuracy: 0.993333333333On the test set:Accuracy: 0.96
plt.title("Model with He initialization")
axes = plt.gca()
axes.set_xlim([-1.5,1.5])
axes.set_ylim([-1.5,1.5])
plot_decision_boundary(lambda x: predict_dec(parameters, x.T), train_X, train_Y)
Observations:
- The model with He initialization separates the blue and the red dots very well in a small number of iterations.
5 - Conclusions
You have seen three different types of initializations. For the same number of iterations and same hyperparameters the comparison is:
ModelTrain accuracyProblem/Comment3-layer NN with zeros initialization50%fails to break symmetry3-layer NN with large random initialization83%too large weights3-layer NN with He initialization99%recommended methodWhat you should remember from this notebook:
- Different initializations lead to different results
- Random initialization is used to break symmetry and make sure different hidden units can learn different things
- Don't intialize to values that are too large
- He initialization works well for networks with ReLU activations.
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