CS231n: How to calculate gradient for Softmax loss function?

Nghia Tran picture Nghia Tran · Jan 15, 2017 · Viewed 29.6k times · Source

I am watching some videos for Stanford CS231: Convolutional Neural Networks for Visual Recognition but do not quite understand how to calculate analytical gradient for softmax loss function using numpy.

From this stackexchange answer, softmax gradient is calculated as:

derivative calculation

Python implementation for above is:

num_classes = W.shape[0]
num_train = X.shape[1]
for i in range(num_train):
  for j in range(num_classes):
    p = np.exp(f_i[j])/sum_i
    dW[j, :] += (p-(j == y[i])) * X[:, i]

Could anyone explain how the above snippet work? Detailed implementation for softmax is also included below.

def softmax_loss_naive(W, X, y, reg):
  """
  Softmax loss function, naive implementation (with loops)
  Inputs:
  - W: C x D array of weights
  - X: D x N array of data. Data are D-dimensional columns
  - y: 1-dimensional array of length N with labels 0...K-1, for K classes
  - reg: (float) regularization strength
  Returns:
  a tuple of:
  - loss as single float
  - gradient with respect to weights W, an array of same size as W
  """
  # Initialize the loss and gradient to zero.
  loss = 0.0
  dW = np.zeros_like(W)

  #############################################################################
  # Compute the softmax loss and its gradient using explicit loops.           #
  # Store the loss in loss and the gradient in dW. If you are not careful     #
  # here, it is easy to run into numeric instability. Don't forget the        #
  # regularization!                                                           #
  #############################################################################

  # Get shapes
  num_classes = W.shape[0]
  num_train = X.shape[1]

  for i in range(num_train):
    # Compute vector of scores
    f_i = W.dot(X[:, i]) # in R^{num_classes}

    # Normalization trick to avoid numerical instability, per http://cs231n.github.io/linear-classify/#softmax
    log_c = np.max(f_i)
    f_i -= log_c

    # Compute loss (and add to it, divided later)
    # L_i = - f(x_i)_{y_i} + log \sum_j e^{f(x_i)_j}
    sum_i = 0.0
    for f_i_j in f_i:
      sum_i += np.exp(f_i_j)
    loss += -f_i[y[i]] + np.log(sum_i)

    # Compute gradient
    # dw_j = 1/num_train * \sum_i[x_i * (p(y_i = j)-Ind{y_i = j} )]
    # Here we are computing the contribution to the inner sum for a given i.
    for j in range(num_classes):
      p = np.exp(f_i[j])/sum_i
      dW[j, :] += (p-(j == y[i])) * X[:, i]

  # Compute average
  loss /= num_train
  dW /= num_train

  # Regularization
  loss += 0.5 * reg * np.sum(W * W)
  dW += reg*W

  return loss, dW

Answer

Ben Barsdell picture Ben Barsdell · Jan 19, 2017

Not sure if this helps, but:

y_i is really the indicator function y_i, as described here. This forms the expression (j == y[i]) in the code.

Also, the gradient of the loss with respect to the weights is:

y_i

where

y_i

which is the origin of the X[:,i] in the code.