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Deep Learning Performance Improvement 1 - Parameter Initialization
26 minute read Parameter Initialization Importance of Parameter InitializationTo build a machine learning algorithm, we use ideas from defined models such as Linear Regression, Logistic Regression, Support Vector Machine, CNN, RNN, etc. The follwoing is a common training process: Initialize parameters Choose Optimizer Repeat the steps: Forward-Propagation with inputs Compute Loss function after determining which one you use Back-Propagation to Compute Gradients of the loss w.r.t. parameters Update each parameter using the gradients, depending on optimizer.Parameter Initialization can be important to the performance of your model. Initializing all weights with zeros can lead the neurons to learn the same features over and over again during training. A well chosen weight can: Speed up the convergence of gradient descent Increase the odds of gradient descent converging to a lower training (and generalization) error import numpy as np import seaborn as sns; sns.set_style("whitegrid") import matplotlib.pyplot as plt %matplotlib inline import h5py import sklearn import sklearn.datasets def sigmoid(x): """ Compute the sigmoid of x Arguments: x -- A scalar or numpy array of any size. Return: s -- sigmoid(x) """ s = 1/(1+np.exp(-x)) return s def relu(x): """ Compute the relu of x Arguments: x -- A scalar or numpy array of any size. Return: s -- relu(x) """ s = np.maximum(0,x) return s def forward_propagation(X, parameters): """ Implements the forward propagation (and computes the loss) presented in Figure 2. Arguments: X -- input dataset, of shape (input size, number of examples) Y -- true "label" vector (containing 0 if cat, 1 if non-cat) parameters -- python dictionary containing your parameters "W1", "b1", "W2", "b2", "W3", "b3": W1 -- weight matrix of shape () b1 -- bias vector of shape () W2 -- weight matrix of shape () b2 -- bias vector of shape () W3 -- weight matrix of shape () b3 -- bias vector of shape () Returns: loss -- the loss function (vanilla logistic loss) """ # retrieve parameters W1 = parameters["W1"] b1 = parameters["b1"] W2 = parameters["W2"] b2 = parameters["b2"] W3 = parameters["W3"] b3 = parameters["b3"] # gr z1 = np.dot(W1, X) + b1 a1 = relu(z1) z2 = np.dot(W2, a1) + b2 a2 = relu(z2) z3 = np.dot(W3, a2) + b3 a3 = sigmoid(z3) cache = (z1, a1, W1, b1, z2, a2, W2, b2, z3, a3, W3, b3) return a3, cache def backward_propagation(X, Y, cache): """ Implement the backward propagation presented in figure 2. Arguments: X -- input dataset, of shape (input size, number of examples) Y -- true "label" vector (containing 0 if cat, 1 if non-cat) cache -- cache output from forward_propagation() Returns: gradients -- A dictionary with the gradients with respect to each parameter, activation and pre-activation variables """ m = X.shape[1] (z1, a1, W1, b1, z2, a2, W2, b2, z3, a3, W3, b3) = cache dz3 = 1./m * (a3 - Y) dW3 = np.dot(dz3, a2.T) db3 = np.sum(dz3, axis=1, keepdims = True) da2 = np.dot(W3.T, dz3) dz2 = np.multiply(da2, np.int64(a2 > 0)) dW2 = np.dot(dz2, a1.T) db2 = np.sum(dz2, axis=1, keepdims = True) da1 = np.dot(W2.T, dz2) dz1 = np.multiply(da1, np.int64(a1 > 0)) dW1 = np.dot(dz1, X.T) db1 = np.sum(dz1, axis=1, keepdims = True) gradients = {"dz3": dz3, "dW3": dW3, "db3": db3, "da2": da2, "dz2": dz2, "dW2": dW2, "db2": db2, "da1": da1, "dz1": dz1, "dW1": dW1, "db1": db1} return gradients def update_parameters(parameters, grads, learning_rate): """ Update parameters using gradient descent Arguments: parameters -- python dictionary containing your parameters grads -- python dictionary containing your gradients, output of n_model_backward Returns: parameters -- python dictionary containing your updated parameters parameters['W' + str(i)] = ... parameters['b' + str(i)] = ... """ L = len(parameters) // 2 # number of layers in the neural networks # Update rule for each parameter for k in range(L): parameters["W" + str(k+1)] = parameters["W" + str(k+1)] - learning_rate * grads["dW" + str(k+1)] parameters["b" + str(k+1)] = parameters["b" + str(k+1)] - learning_rate * grads["db" + str(k+1)] return parameters def compute_loss(a3, Y): """ Implement the loss function Arguments: a3 -- post-activation, output of forward propagation Y -- "true" labels vector, same shape as a3 Returns: loss - value of the loss function """ m = Y.shape[1] logprobs = np.multiply(-np.log(a3),Y) + np.multiply(-np.log(1 - a3), 1 - Y) loss = 1./m * np.nansum(logprobs) return loss def load_cat_dataset(): train_dataset = h5py.File('datasets/train_catvnoncat.h5', "r") train_set_x_orig = np.array(train_dataset["train_set_x"][:]) # your train set features train_set_y_orig = np.array(train_dataset["train_set_y"][:]) # your train set labels test_dataset = h5py.File('datasets/test_catvnoncat.h5', "r") test_set_x_orig = np.array(test_dataset["test_set_x"][:]) # your test set features test_set_y_orig = np.array(test_dataset["test_set_y"][:]) # your test set labels classes = np.array(test_dataset["list_classes"][:]) # the list of classes train_set_y = train_set_y_orig.reshape((1, train_set_y_orig.shape[0])) test_set_y = test_set_y_orig.reshape((1, test_set_y_orig.shape[0])) train_set_x_orig = train_set_x_orig.reshape(train_set_x_orig.shape[0], -1).T test_set_x_orig = test_set_x_orig.reshape(test_set_x_orig.shape[0], -1).T train_set_x = train_set_x_orig/255 test_set_x = test_set_x_orig/255 return train_set_x, train_set_y, test_set_x, test_set_y, classes def predict(X, y, parameters): """ This function is used to predict the results of a n-layer neural network. Arguments: X -- data set of examples you would like to label parameters -- parameters of the trained model Returns: p -- predictions for the given dataset X """ m = X.shape[1] p = np.zeros((1,m), dtype = np.int) # Forward propagation a3, caches = forward_propagation(X, parameters) # convert probas to 0/1 predictions for i in range(0, a3.shape[1]): if a3[0,i] > 0.5: p[0,i] = 1 else: p[0,i] = 0 # print results print("Accuracy: " + str(np.mean((p[0,:] == y[0,:])))) return p def plot_decision_boundary(model, X, y): # Set min and max values and give it some padding x_min, x_max = X[0, :].min() - 1, X[0, :].max() + 1 y_min, y_max = X[1, :].min() - 1, X[1, :].max() + 1 h = 0.01 # Generate a grid of points with distance h between them xx, yy = np.meshgrid(np.arange(x_min, x_max, h), np.arange(y_min, y_max, h)) # Predict the function value for the whole grid Z = model(np.c_[xx.ravel(), yy.ravel()]) Z = Z.reshape(xx.shape) # Plot the contour and training examples plt.contourf(xx, yy, Z, cmap=plt.cm.coolwarm, alpha=0.3) plt.ylabel('x2') plt.xlabel('x1') plt.scatter(X[0, :], X[1, :], c=np.squeeze(y), cmap=plt.cm.coolwarm, s=10, alpha=0.5) plt.show() def predict_dec(parameters, X): """ Used for plotting decision boundary. Arguments: parameters -- python dictionary containing your parameters X -- input data of size (m, K) Returns predictions -- vector of predictions of our model (red: 0 / blue: 1) """ # Predict using forward propagation and a classification threshold of 0.5 a3, cache = forward_propagation(X, parameters) predictions = (a3>0.5) return predictions def load_dataset(): np.random.seed(1) train_X, train_Y = sklearn.datasets.make_moons(n_samples=300, noise=.05) np.random.seed(2) test_X, test_Y = sklearn.datasets.make_moons(n_samples=100, noise=.05) # Visualize the data plt.scatter(train_X[:, 0], train_X[:, 1], c=train_Y, s=10, cmap=plt.cm.coolwarm, alpha=0.5); train_X = train_X.T train_Y = train_Y.reshape((1, train_Y.shape[0])) test_X = test_X.T test_Y = test_Y.reshape((1, test_Y.shape[0])) return train_X, train_Y, test_X, test_Y plt.subplots(1,1,figsize=(6,4)) # plt.rcParams['figure.figsize'] = (7.0, 4.0) # set default size of plots # plt.rcParams['image.interpolation'] = 'nearest' # plt.rcParams['image.cmap'] = 'viridis' # load image dataset: blue/red dots in circles train_X, train_Y, test_X, test_Y = load_dataset() plt.show()You would like a classifier to separate the blue dots from the red dots. 1 - Neural Network modelYou 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 initializationThere are two types of parameters to initialize in a neural network: the weight matrices \((W^{[1]}, W^{[2]}, W^{[3]}, ..., W^{[L-1]}, W^{[L]})\) the bias vectors \((b^{[1]}, b^{[2]}, b^{[3]}, ..., b^{[L-1]}, b^{[L]})\)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. plt.subplots(1,1,figsize=(6,4)) 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: 0.6931471805599453 Cost after iteration 1000: 0.6931471805599453 Cost after iteration 2000: 0.6931471805599453 Cost after iteration 3000: 0.6931471805599453 Cost after iteration 4000: 0.6931471805599453 Cost after iteration 5000: 0.6931471805599453 Cost after iteration 6000: 0.6931471805599453 Cost after iteration 7000: 0.6931471805599453 Cost after iteration 8000: 0.6931471805599453 Cost after iteration 9000: 0.6931471805599453 Cost after iteration 10000: 0.6931471805599455 Cost after iteration 11000: 0.6931471805599453 Cost after iteration 12000: 0.6931471805599453 Cost after iteration 13000: 0.6931471805599453 Cost after iteration 14000: 0.6931471805599453 On the train set: Accuracy: 0.5 On the test set: Accuracy: 0.5The 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.subplots(1,1,figsize=(6,4)) 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 \(n^{[l]}=1\) for every layer, and the network is no more powerful than a linear classifier such as logistic regression. **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 as $$W^{[l]}$$ is initialized randomly. 3 - Random initializationTo 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 = [[ 17.88628473 4.36509851 0.96497468] [-18.63492703 -2.77388203 -3.54758979]] b1 = [[0.] [0.]] W2 = [[-0.82741481 -6.27000677]] 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. plt.subplots(1,1,figsize=(6,4)) 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) :147: RuntimeWarning: divide by zero encountered in log logprobs = np.multiply(-np.log(a3),Y) + np.multiply(-np.log(1 - a3), 1 - Y) :147: RuntimeWarning: invalid value encountered in multiply logprobs = np.multiply(-np.log(a3),Y) + np.multiply(-np.log(1 - a3), 1 - Y) Cost after iteration 0: inf :19: RuntimeWarning: overflow encountered in exp s = 1/(1+np.exp(-x)) Cost after iteration 1000: 1.4689728581497818 Cost after iteration 2000: 0.6422096748947888 Cost after iteration 3000: 0.5183352539690133 Cost after iteration 4000: 0.2980509156122723 Cost after iteration 5000: 0.24392073041135126 Cost after iteration 6000: 0.2424874223607829 Cost after iteration 7000: 0.2421717415715247 Cost after iteration 8000: 0.24260135727310358 Cost after iteration 9000: 0.2428797438736504 Cost after iteration 10000: 0.24221439430112404 Cost after iteration 11000: 0.24156059704466773 Cost after iteration 12000: 0.24091176777584963 Cost after iteration 13000: 0.24026236850861002 Cost after iteration 14000: 0.23959997123320914 On the train set: Accuracy: 0.8866666666666667 On the test set: Accuracy: 0.87If 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 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 1 1 0 1 1 0 0 0 0 0 0 1 1 1 1 1 0 1 1 0 0 1 1 1 1 0 1 0 0 1 1 1 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 0 1 0 1 1 0 0 0 1 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 1 0 1 1 0 0 1 1 1 0 1 1 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 0 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 0 1 1 0 0 1 1 0 0 1 0 1 0 0 1 0 1 1 0 1 1 1 1 0 1 1 1 0 0 1 0 1 0 0 0 1 0 1 0 0 1 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 1 0 1 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 1 1 0 0 0 0 1 1 0 0 0 1 1 1 0 1 1 0 0 0 0 1 1 0 1 0]] [[1 0 0 0 0 0 0 0 0 0 1 1 1 0 1 0 0 0 1 1 0 0 1 0 1 0 0 1 1 1 1 1 0 0 0 1 0 1 1 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 0 0 0 0 0 0 1 0 0 0 0 1 0 1 1 0 0 0 1 1 0 0 1 0 0]] plt.subplots(1,1,figsize=(6,4)) 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) :19: RuntimeWarning: overflow encountered in exp s = 1/(1+np.exp(-x))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 \(\log(a^{[3]}) = \log(0)\), 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 initializationFinally, 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 \(W^{[l]}\) of 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 \(\sqrt{\frac{2}{\text{dimension of the previous layer}}}\), which is what He initialization recommends for layers with a ReLU activation. # 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. plt.subplots(1,1,figsize=(6,4)) 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: 0.7342482321384683 Cost after iteration 1000: 0.2956649797057583 Cost after iteration 2000: 0.25184038318406815 Cost after iteration 3000: 0.22284974489831486 Cost after iteration 4000: 0.197785000749911 Cost after iteration 5000: 0.17049197570926786 Cost after iteration 6000: 0.13351846788644978 Cost after iteration 7000: 0.09373065428140327 Cost after iteration 8000: 0.06003854894423607 Cost after iteration 9000: 0.0367254088164847 Cost after iteration 10000: 0.023649094476709844 Cost after iteration 11000: 0.016344535872923487 Cost after iteration 12000: 0.011999996046588598 Cost after iteration 13000: 0.009296262994410533 Cost after iteration 14000: 0.00755073011114619 On the train set: Accuracy: 1.0 On the test set: Accuracy: 1.0 plt.subplots(1,1,figsize=(6,4)) 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 - ConclusionsYou have seen three different types of initializations. For the same number of iterations and same hyperparameters the comparison is: comparison is: Model Accuracy Problem 3-layer NN with zeros initialization 50% Fails to break symmetry 3-layer NN with large random initialization 83% Unstable variance 3-layer NN with He initialization 99% Normalized variance with 0 mean **What 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. 6. Looking at the weight initialization problem from the distribution of activation valuesObserving the distribution of the activation value of the hidden layer, we can get a lot of inspiration. Here, let’s do a simple experiment to see how the initial value of the weight affects the distribution of the activation value of the hidden layer. Pass in randomly generated input data to a 5-layer neural network (the activation function uses the sigmoid function), and use a histogram to plot the data distribution of the activation values of each layer. input_data = np.random.randn(1000, 100) # 1000 data node_num = 100 # The number of nodes (neurons) in each hidden layer hidden_layer_size = 5 # There are 5 hidden layers activations = {} # The result of activation value is saved here x = input_data for i in range(hidden_layer_size): if i != 0: x = activations[i-1] # Change the initial value to experiment! w = np.random.randn(node_num, node_num) * 1 # w = np.random.randn(node_num, node_num) * 0.01 # w = np.random.randn(node_num, node_num) * np.sqrt(1.0 / node_num) # w = np.random.randn(node_num, node_num) * np.sqrt(2.0 / node_num) a = np.dot(x, w) # Change the type of activation function to experiment! z = sigmoid(a) # z = ReLU(a) # z = tanh(a) activations[i] = z # Draw a histogram for i, a in activations.items(): plt.subplot(1, len(activations), i+1) plt.title(str(i+1) + "-layer") if i != 0: plt.yticks([], []) plt.hist(a.flatten(), 30, range=(0,1)) plt.show()Here it is assumed that the neural network has 5 layers and each layer has 100 units. Then, use Gaussian distribution to randomly generate 1000 data as input data, and pass them to the 5-layer neural network. The activation values of each layer are distributed towards 0 and 1. The sigmoid function used here is an sigmoid function. As the output continuously approaches 0 (or approaches 1), the value of its derivative gradually approaches 0. Therefore, the data distribution biased towards 0 and 1 will cause the value of the gradient in the backpropagation to become smaller and eventually disappear. This is the problem of gradient disappearance. In deep learning with deeper levels, the problem of gradient disappearance may be more serious. Next, set the standard deviation of the weight to 0.01 and implement it again. The code is modified as follows: for i in range(hidden_layer_size): if i != 0: x = activations[i-1] # Change the initial value to experiment! # w = np.random.randn(node_num, node_num) * 1 w = np.random.randn(node_num, node_num) * 0.01 # w = np.random.randn(node_num, node_num) * np.sqrt(1.0 / node_num) # w = np.random.randn(node_num, node_num) * np.sqrt(2.0 / node_num) a = np.dot(x, w) # Change the type of activation function to experiment! z = sigmoid(a) # z = ReLU(a) # z = tanh(a) activations[i] = z # Draw a histogram for i, a in activations.items(): plt.subplot(1, len(activations), i+1) plt.title(str(i+1) + "-layer") if i != 0: plt.yticks([], []) plt.hist(a.flatten(), 30, range=(0,1)) plt.show()When using a Gaussian distribution with a standard deviation of 0.01, the activation value distribution of each layer is shown in the figure above. This time concentrate on the distribution around 0.5. Because it is not biased towards 0 and 1 as in the previous example, the problem of vanishing gradient will not occur. However, the distribution of activation values is biased, indicating that there will be a big problem in expressiveness, that is, there is a symmetry problem. Why do you say that? Because if there are multiple neurons that all output almost the same value, then they have no meaning. For example, if 100 neurons all output almost the same value, then one neuron can also express basically the same thing. Therefore, if the activation value is biased in the distribution, the problem of “restricted expressiveness” will occur. for i in range(hidden_layer_size): if i != 0: x = activations[i-1] # Change the initial value to experiment! # w = np.random.randn(node_num, node_num) * 1 # w = np.random.randn(node_num, node_num) * 0.01 w = np.random.randn(node_num, node_num) * np.sqrt(1.0 / node_num) # w = np.random.randn(node_num, node_num) * np.sqrt(2.0 / node_num) a = np.dot(x, w) # Change the type of activation function to experiment! z = sigmoid(a) # z = ReLU(a) # z = tanh(a) activations[i] = z # Draw a histogram for i, a in activations.items(): plt.subplot(1, len(activations), i+1) plt.title(str(i+1) + "-layer") if i != 0: plt.yticks([], []) plt.hist(a.flatten(), 30, range=(0,1)) plt.show()The result after using Xavier’s initial value is shown in the figure above. It can be seen from the image that the more behind the layer, the more skewed the image becomes, but it presents a wider distribution than before. Because the data passed between the layers has an appropriate breadth, the expressive power of the sigmoid function is not limited, and it is expected to be efficient learning. Because the sigmoid function and the tanh function are symmetrical, the initial value of Xavier is suitable. But when the activation function uses ReLU, it is generally recommended to use the initial value of He. Now let’s look at the distribution of activation values when the activation function uses ReLU. We have given three reals, which are the results when the initial value of the weight is a Gaussian distribution with a standard deviation of 0.01, when the initial value is the initial value of Xavier, and the initial value is the initial value of He dedicated to ReLU. Also modify the drawing parameters below to see more clearly: for i in range(hidden_layer_size): if i != 0: x = activations[i-1] # Change the initial value to experiment! # w = np.random.randn(node_num, node_num) * 1 # w = np.random.randn(node_num, node_num) * 0.01 # w = np.random.randn(node_num, node_num) * np.sqrt(1.0 / node_num) w = np.random.randn(node_num, node_num) * np.sqrt(2.0 / node_num) a = np.dot(x, w) # Change the type of activation function to experiment! # z = sigmoid(a) z = relu(a) # z = tanh(a) activations[i] = z # Draw a histogram for i, a in activations.items(): plt.subplot(1, len(activations), i+1) plt.title(str(i+1) + "-layer") if i != 0: plt.yticks([], []) plt.hist(a.flatten(), 30, range=(0,1)) plt.show()Observing the above figure, we can see that when the standard deviation is a Gaussian distribution of 0.01, the activation value of each layer is very small. The value passed on the neural network is very small, indicating that the gradient of the weight is also very small during back propagation. This is a very serious problem. In fact, there is basically no progress in learning. Next is the result when the initial value is the initial value of Xavier. In this case, as the layer deepens, the bias becomes a little bit larger. In fact, after the layer is deepened, the bias of the activation value becomes larger, and the problem of the gradient disappears during learning. When the initial value is the initial value of He, the breadth of distribution in each layer is the same. Since the breadth of the data can remain the same even if the layer is deepened, the appropriate value will be transmitted during the reverse propagation. Categories: machine-learning-concept Updated: November 01, 2020 Share on Twitter Facebook LinkedIn |
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