I could not understand well especially how gradients were computed with regards to matrix transposes. My question is for DW2 but if you want also to discuss about the computation of the other gradients and extend my question I am open to discussion. Mathematically things seem a little bit different but this code is reliable and on github so I trust this code.
from __future__ import print_function
from builtins import range
from builtins import object
import numpy as np
import matplotlib.pyplot as plt
from past.builtins import xrange
class TwoLayerNet(object):
"""
A two-layer fully-connected neural network. The net has an input dimension of
D* (correction), a hidden layer dimension of H, and performs classification over C classes.
We train the network with a softmax loss function and L2 regularization on the
weight matrices. The network uses a ReLU nonlinearity after the first fully
connected layer.
In other words, the network has the following architecture:
input - fully connected layer - ReLU - fully connected layer - softmax
The outputs of the second fully-connected layer are the scores for each class.
"""
def __init__(self, input_size, hidden_size, output_size, std=1e-4):
"""
Initialize the model. Weights are initialized to small random values and
biases are initialized to zero. Weights and biases are stored in the
variable self.params, which is a dictionary with the following keys:
W1: First layer weights; has shape (D, H)
b1: First layer biases; has shape (H,)
W2: Second layer weights; has shape (H, C)
b2: Second layer biases; has shape (C,)
Inputs:
- input_size: The dimension D of the input data.
- hidden_size: The number of neurons H in the hidden layer.
- output_size: The number of classes C.
"""
self.params = {}
self.params['W1'] = std * np.random.randn(input_size, hidden_size)
self.params['b1'] = np.zeros(hidden_size)
self.params['W2'] = std * np.random.randn(hidden_size, output_size)
self.params['b2'] = np.zeros(output_size)
def loss(self, X, y=None, reg=0.0):
"""
Compute the loss and gradients for a two layer fully connected neural
network.
Inputs:
- X: Input data of shape (N, D). Each X[i] is a training sample.
- y: Vector of training labels. y[i] is the label for X[i], and each y[i] is
an integer in the range 0 <= y[i] < C. This parameter is optional; if it
is not passed then we only return scores, and if it is passed then we
instead return the loss and gradients.
- reg: Regularization strength.
Returns:
If y is None, return a matrix scores of shape (N, C) where scores[i, c] is
the score for class c on input X[i].
If y is not None, instead return a tuple of:
- loss: Loss (data loss and regularization loss) for this batch of training
samples.
- grads: Dictionary mapping parameter names to gradients of those parameters
with respect to the loss function; has the same keys as self.params.
"""
# Unpack variables from the params dictionary
W1, b1 = self.params['W1'], self.params['b1']
W2, b2 = self.params['W2'], self.params['b2']
N, D = X.shape
# Compute the forward pass
scores = None
#############################################################################
# TODO: Perform the forward pass, computing the class scores for the input. #
# Store the result in the scores variable, which should be an array of #
# shape (N, C). #
#############################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
# perform the forward pass and compute the class scores for the input
# input - fully connected layer - ReLU - fully connected layer - softmax
# define lamba function for relu
relu = lambda x: np.maximum(0, x)
# a1 = X x W1 = (N x D) x (D x H) = N x H
a1 = relu(X.dot(W1) + b1) # activations of fully connected layer #1
# store the result in the scores variable, which should be an array of
# shape (N, C).
# scores = a1 x W2 = (N x H) x (H x C) = N x C
scores = a1.dot(W2) + b2 # output of softmax
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
# If the targets are not given then jump out, we're done
if y is None:
return scores
# Compute the loss
loss = None
#############################################################################
# TODO: Finish the forward pass, and compute the loss. This should include #
# both the data loss and L2 regularization for W1 and W2. Store the result #
# in the variable loss, which should be a scalar. Use the Softmax #
# classifier loss. #
#############################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
# shift values for 'scores' for numeric reasons (over-flow cautious)
# figure out the max score across all classes
# scores.shape is N x C
scores -= scores.max(axis = 1, keepdims = True)
# probs.shape is N x C
probs = np.exp(scores)/np.sum(np.exp(scores), axis = 1, keepdims = True)
loss = -np.log(probs[np.arange(N), y])
# loss is a single number
loss = np.sum(loss)
# Right now the loss is a sum over all training examples, but we want it
# to be an average instead so we divide by N.
loss /= N
# Add regularization to the loss.
loss += reg * (np.sum(W1 * W1) + np.sum(W2 * W2))
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
# Backward pass: compute gradients
grads = {}
#############################################################################
# TODO: Compute the backward pass, computing the derivatives of the weights #
# and biases. Store the results in the grads dictionary. For example, #
# grads['W1'] should store the gradient on W1, and be a matrix of same size #
#############################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
# since dL(i)/df(k) = p(k) - 1 (if k = y[i]), where f is a vector of scores for the given example
# i is the training sample and k is the class
dscores = probs.reshape(N, -1) # dscores is (N x C)
dscores[np.arange(N), y] -= 1
# since scores = a1.dot(W2), we get dW2 by multiplying a1.T and dscores
# W2 is H x C so dW2 should also match those dimensions
# a1.T x dscores = (H x N) x (N x C) = H x C
dW2 = np.dot(a1.T, dscores)
# Right now the gradient is a sum over all training examples, but we want it
# to be an average instead so we divide by N.
dW2 /= N
# b2 gradient: sum dscores over all N and C
db2 = dscores.sum(axis = 0)/N
# since a1 = X.dot(W1), we get dW1 by multiplying X.T and da1
# W1 is D x H so dW1 should also match those dimensions
# X.T x da1 = (D x N) x (N x H) = D x H
# first get da1 using scores = a1.dot(W2)
# a1 is N x H so da1 should also match those dimensions
# dscores x W2.T = (N x C) x (C x H) = N x H
da1 = dscores.dot(W2.T)
da1[a1 == 0] = 0 # set gradient of units that did not activate to 0
dW1 = X.T.dot(da1)
# Right now the gradient is a sum over all training examples, but we want it
# to be an average instead so we divide by N.
dW1 /= N
# b1 gradient: sum da1 over all N and H
db1 = da1.sum(axis = 0)/N
# Add regularization loss to the gradient
dW1 += 2 * reg * W1
dW2 += 2 * reg * W2
grads = {'W1': dW1, 'b1': db1, 'W2': dW2, 'b2': db2}
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
return loss, grads
def train(self, X, y, X_val, y_val,
learning_rate=1e-3, learning_rate_decay=0.95,
reg=5e-6, num_iters=100,
batch_size=200, verbose=False):
"""
Train this neural network using stochastic gradient descent.
Inputs:
- X: A numpy array of shape (N, D) giving training data.
- y: A numpy array f shape (N,) giving training labels; y[i] = c means that
X[i] has label c, where 0 <= c < C.
- X_val: A numpy array of shape (N_val, D) giving validation data.
- y_val: A numpy array of shape (N_val,) giving validation labels.
- learning_rate: Scalar giving learning rate for optimization.
- learning_rate_decay: Scalar giving factor used to decay the learning rate
after each epoch.
- reg: Scalar giving regularization strength.
- num_iters: Number of steps to take when optimizing.
- batch_size: Number of training examples to use per step.
- verbose: boolean; if true print progress during optimization.
"""
num_train = X.shape[0]
iterations_per_epoch = max(num_train / batch_size, 1)
# Use SGD to optimize the parameters in self.model
loss_history = []
train_acc_history = []
val_acc_history = []
for it in range(num_iters):
X_batch = None
y_batch = None
#########################################################################
# TODO: Create a random minibatch of training data and labels, storing #
# them in X_batch and y_batch respectively. #
#########################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
# generate random indices
indices = np.random.choice(num_train, batch_size)
X_batch, y_batch = X[indices], y[indices]
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
# Compute loss and gradients using the current minibatch
loss, grads = self.loss(X_batch, y=y_batch, reg=reg)
loss_history.append(loss)
#########################################################################
# TODO: Use the gradients in the grads dictionary to update the #
# parameters of the network (stored in the dictionary self.params) #
# using stochastic gradient descent. You'll need to use the gradients #
# stored in the grads dictionary defined above. #
#########################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
self.params['W1'] -= learning_rate * grads['W1']
self.params['W2'] -= learning_rate * grads['W2']
self.params['b1'] -= learning_rate * grads['b1']
self.params['b2'] -= learning_rate * grads['b2']
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
if verbose and it % 100 == 0:
print('iteration %d / %d: loss %f' % (it, num_iters, loss))
# Every epoch, check train and val accuracy and decay learning rate.
if it % iterations_per_epoch == 0:
# Check accuracy
train_acc = (self.predict(X_batch) == y_batch).mean()
val_acc = (self.predict(X_val) == y_val).mean()
train_acc_history.append(train_acc)
val_acc_history.append(val_acc)
# Decay learning rate
learning_rate *= learning_rate_decay
return {
'loss_history': loss_history,
'train_acc_history': train_acc_history,
'val_acc_history': val_acc_history,
}
def predict(self, X):
"""
Use the trained weights of this two-layer network to predict labels for
data points. For each data point we predict scores for each of the C
classes, and assign each data point to the class with the highest score.
Inputs:
- X: A numpy array of shape (N, D) giving N D-dimensional data points to
classify.
Returns:
- y_pred: A numpy array of shape (N,) giving predicted labels for each of
the elements of X. For all i, y_pred[i] = c means that X[i] is predicted
to have class c, where 0 <= c < C.
"""
y_pred = None
###########################################################################
# TODO: Implement this function; it should be VERY simple! #
###########################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
# define lamba function for relu
relu = lambda x: np.maximum(0, x)
# activations of fully connected layer #1
a1 = relu(X.dot(self.params['W1']) + self.params['b1'])
# output of softmax
# scores = a1 x W2 = (N x H) x (H x C) = N x C
scores = a1.dot(self.params['W2']) + self.params['b2']
y_pred = np.argmax(scores, axis = 1)
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
return y_pred
With regards to above code, I could not understand how DW2 was computed well. I took picture of the point I need to clarify and need an explanation for the difference.enter image description here
My ideas
Question
In CS231 Computing the Analytic Gradient with Backpropagation which is first implementing a Softmax Classifier, the gradient from (softmax + log loss) is divided by the batch size (number of data being used in a cycle of forward cost calculation and backward propagation in the training).
Please help me understand why it needs to be divided by the batch size.
The chain rule to get the gradient should be below. Where should I incorporate the division?
Derivative of Softmax loss function
Code
N = 100 # number of points per class
D = 2 # dimensionality
K = 3 # number of classes
X = np.zeros((N*K,D)) # data matrix (each row = single example)
y = np.zeros(N*K, dtype='uint8') # class labels
#Train a Linear Classifier
# initialize parameters randomly
W = 0.01 * np.random.randn(D,K)
b = np.zeros((1,K))
# some hyperparameters
step_size = 1e-0
reg = 1e-3 # regularization strength
# gradient descent loop
num_examples = X.shape[0]
for i in range(200):
# evaluate class scores, [N x K]
scores = np.dot(X, W) + b
# compute the class probabilities
exp_scores = np.exp(scores)
probs = exp_scores / np.sum(exp_scores, axis=1, keepdims=True) # [N x K]
# compute the loss: average cross-entropy loss and regularization
correct_logprobs = -np.log(probs[range(num_examples),y])
data_loss = np.sum(correct_logprobs)/num_examples
reg_loss = 0.5*reg*np.sum(W*W)
loss = data_loss + reg_loss
if i % 10 == 0:
print "iteration %d: loss %f" % (i, loss)
# compute the gradient on scores
dscores = probs
dscores[range(num_examples),y] -= 1
dscores /= num_examples # <---------------------- Why?
# backpropate the gradient to the parameters (W,b)
dW = np.dot(X.T, dscores)
db = np.sum(dscores, axis=0, keepdims=True)
dW += reg*W # regularization gradient
# perform a parameter update
W += -step_size * dW
b += -step_size * db
It's because you are averaging the gradients instead of taking directly the sum of all the gradients.
You could of course not divide for that size, but this division has a lot of advantages. The main reason is that it's a sort of regularization (to avoid overfitting). With smaller gradients the weights cannot grow out of proportions.
And this normalization allows comparison between different configuration of batch sizes in different experiments (How can I compare two batch performances if they are dependent to the batch size?)
If you divide for that size the gradients sum it could be useful to work with greater learning rates to make the training faster.
This answer in the crossvalidated community is quite useful.
Came to notice that the dot in dW = np.dot(X.T, dscores) for the gradient at W is Σ over the num_sample instances. Since the dscore, which is probability (softmax output), was divided by the num_samples, did not understand that it was normalization for dot and sum part later in the code. Now understood divide by num_sample is required (may still work without normalization if the learning rate is trained though).
I believe the code below explains better.
# compute the gradient on scores
dscores = probs
dscores[range(num_examples),y] -= 1
# backpropate the gradient to the parameters (W,b)
dW = np.dot(X.T, dscores) / num_examples
db = np.sum(dscores, axis=0, keepdims=True) / num_examples
This is not a question for a specific problem I am trying to solve. I am just trying to understand why a gradient is calculated by multiplying the layers (matrices) in a mostly backward fashion. I also didn't know subtracting y from the prediction could also give you something called a gradient.
grad_y_pred = 2.0 * (y_pred - y)
grad_w2 = h_relu.T.dot(grad_y_pred)
I don't know what I thought Pytorch was doing finding the gradients. I figured it was some kind of algorithm that did the power rule and followed other derivative rules somehow.
import numpy as np
# N is batch size; D_in is input dimension;
# H is hidden dimension; D_out is output dimension.
N, D_in, H, D_out = 64, 1000, 100, 10
# Create random input and output data
x = np.random.randn(N, D_in)
y = np.random.randn(N, D_out)
# Randomly initialize weights
w1 = np.random.randn(D_in, H)
w2 = np.random.randn(H, D_out)
learning_rate = 1e-6
for t in range(500):
# Forward pass: compute predicted y
h = x.dot(w1)
h_relu = np.maximum(h, 0)
y_pred = h_relu.dot(w2)
# Compute and print loss
loss = np.square(y_pred - y).sum()
print(t, loss)
# Backprop to compute gradients of w1 and w2 with respect to loss
grad_y_pred = 2.0 * (y_pred - y)
grad_w2 = h_relu.T.dot(grad_y_pred)
grad_h_relu = grad_y_pred.dot(w2.T)
grad_h = grad_h_relu.copy()
grad_h[h < 0] = 0
grad_w1 = x.T.dot(grad_h)
# Update weights
w1 -= learning_rate * grad_w1
w2 -= learning_rate * grad_w2
I am trying to compute a function that calculates the gradient descent in python. I know how to compute it without vectors for example:
def gradient_descent(x,y):
m_curr = b_curr = 0
iterations = 10000
n = len(x)
learning_rate = 0.08
for i in range(iterations):
y_predicted = m_curr * x + b_curr
cost = (1/n) * sum([val**2 for val in (y-y_predicted)])
md = -(2/n)*sum(x*(y-y_predicted))
bd = -(2/n)*sum(y-y_predicted)
m_curr = m_curr - learning_rate * md
b_curr = b_curr - learning_rate * bd
However, I'm having trouble when the parameters are vectors. Any help would be appreciated. I'm new to python
# computeMSEBatchGradient:
# weights - vector of weights (univariate linear = 2 weights)
# features - vector (or matrix) of feature values
# targets - vector of target values, same length as features
#
# returns average gradient over the batch of features
def computeMSEBatchGradient(weights,features,targets):
# insert calculation of gradient here
#return the gradient as a vector
return gradient
I am now learning the stanford cs231n course. When completing the softmax_loss function, I found it is not easy to write in a full-vectorized type, especially dealing with the dw term. Below is my code. Can somebody optimize the code. Would be appreciated.
def softmax_loss_vectorized(W, X, y, reg):
loss = 0.0
dW = np.zeros_like(W)
num_train = X.shape[0]
num_classes = W.shape[1]
scores = X.dot(W)
scores -= np.max(scores, axis = 1)[:, np.newaxis]
exp_scores = np.exp(scores)
sum_exp_scores = np.sum(exp_scores, axis = 1)
correct_class_score = scores[range(num_train), y]
loss = np.sum(np.log(sum_exp_scores)) - np.sum(correct_class_score)
exp_scores = exp_scores / sum_exp_scores[:,np.newaxis]
# **maybe here can be rewroten into matrix operations**
for i in xrange(num_train):
dW += exp_scores[i] * X[i][:,np.newaxis]
dW[:, y[i]] -= X[i]
loss /= num_train
loss += 0.5 * reg * np.sum( W*W )
dW /= num_train
dW += reg * W
return loss, dW
Here's a vectorized implementation below. But I suggest you try to spend a little bit more time and get to the solution yourself. The idea is to construct a matrix with all softmax values and subtract -1 from the correct elements.
def softmax_loss_vectorized(W, X, y, reg):
num_train = X.shape[0]
scores = X.dot(W)
scores -= np.max(scores)
correct_scores = scores[np.arange(num_train), y]
# Compute the softmax per correct scores in bulk, and sum over its logs.
exponents = np.exp(scores)
sums_per_row = np.sum(exponents, axis=1)
softmax_array = np.exp(correct_scores) / sums_per_row
information_array = -np.log(softmax_array)
loss = np.mean(information_array)
# Compute the softmax per whole scores matrix, which gives the matrix for X rows coefficients.
# Their linear combination is algebraically dot product X transpose.
all_softmax_matrix = (exponents.T / sums_per_row).T
grad_coeff = np.zeros_like(scores)
grad_coeff[np.arange(num_train), y] = -1
grad_coeff += all_softmax_matrix
dW = np.dot(X.T, grad_coeff) / num_train
# Regularization
loss += 0.5 * reg * np.sum(W * W)
dW += reg * W
return loss, dW