我正在观看Stanford CS231:用于视觉识别的卷积神经网络的一些视频,但不太了解如何使用numpy
计算softmax损失函数的分析梯度。
从this stackexchange回答,softmax梯度计算如下:
上面的Python实现是:
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]
有人可以解释上面的代码片段是如何工作的吗? softmax的详细实施也包含在下面。
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
答案 0 :(得分:14)
答案 1 :(得分:1)
我知道这很晚了,但这是我的答案:
我假设您熟悉cs231n Softmax损失函数。 我们知道:
因此,就像我们对SVM损失函数所做的那样,梯度如下:
希望有帮助。
答案 2 :(得分:1)
一个带有小例子的 supplement to this answer。