我现在正在学习斯坦福大学的cs231n课程。在完成softmax_loss函数时，我发现将其完全向量化编写并不容易，尤其是处理dw项。以下是我的代码。能否有人优化一下代码，将不胜感激。

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

回答：

以下是一个向量化的实现。但我建议你多花一点时间，自己找到解决方案。思路是构建一个包含所有softmax值的矩阵，并从正确元素中减去-1。

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