【发布时间】:2017-08-21 18:21:16
【问题描述】:
我有这个代码:
import numpy as np
def sigmoid(x):
"""
Calculate sigmoid
"""
return 1 / (1 + np.exp(-x))
x = np.array([0.5, 0.1, -0.2])
target = 0.6
learnrate = 0.5
weights_input_hidden = np.array([[0.5, -0.6],
[0.1, -0.2],
[0.1, 0.7]])
weights_hidden_output = np.array([0.1, -0.3])
## Forward pass
hidden_layer_input = np.dot(x, weights_input_hidden)
hidden_layer_output = sigmoid(hidden_layer_input)
output_layer_in = np.dot(hidden_layer_output, weights_hidden_output)
output = sigmoid(output_layer_in)
## Backwards pass
## TODO: Calculate error
error = target - output
# TODO: Calculate error gradient for output layer
del_err_output = error * output * (1 - output)
print("del_err_output", del_err_output)
# TODO: Calculate error gradient for hidden layer
del_err_hidden = np.dot(del_err_output, weights_hidden_output) * hidden_layer_output * (1 - hidden_layer_output)
print("del_err_hidden", del_err_hidden)
print("del_err_hidden.shape", del_err_hidden.shape)
print("x", x)
print("x.shape", x.shape)
print("x[:,None]")
print(x[:,None])
print("x[:,None].shape", x[:,None].shape)
print("del_err_hidden * x[:, None]")
print(del_err_hidden * x[:, None])
生成此输出:
del_err_output 0.0287306695435
del_err_hidden [ 0.00070802 -0.00204471]
del_err_hidden.shape (2,)
x [ 0.5 0.1 -0.2]
x.shape (3,)
x[:,None]
[[ 0.5]
[ 0.1]
[-0.2]]
x[:,None].shape (3, 1)
del_err_hidden * x[:, None]
[[ 3.54011093e-04 -1.02235701e-03]
[ 7.08022187e-05 -2.04471402e-04]
[ -1.41604437e-04 4.08942805e-04]]
我的问题是这个操作:del_err_hidden * x[:, None]
*是哪种操作?
其次,如果 del_err_hidden.shape 是 (2,) 而 x[:,None].shape 是 (3, 1),为什么我可以将它们相乘?
有人告诉我它与元素和广播有关,但我不明白这些术语。因为要进行元素乘法,两个矩阵必须具有相同的大小,而这里它们没有。
【问题讨论】:
-
Numpy 自动做
del_err_hidden[None,:],所以乘法是(1,2) * (3,1) => (3,2) -
@hpaulj 不是反过来吗?
(3, 1) * (1, 2) => (3, 2)在数学上是正确的 -
广播元素智能乘法是可交换的。
标签: python numpy neural-network deep-learning backpropagation