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