python – 为什么将误差乘以神经网络中sigmoid的导数？

import numpy as np

# sigmoid function
def nonlin(x,deriv=False):
    if(deriv==True):
        return x*(1-x)
    return 1/(1+np.exp(-x))

# input dataset
X = np.array([  [0,1],[0,1,[1,1] ])

# output dataset            
y = np.array([[0,1]]).T

# seed random numbers to make calculation
# deterministic (just a good practice)
np.random.seed(1)

# initialize weights randomly with mean 0
syn0 = 2*np.random.random((3,1)) - 1

for iter in xrange(10000):

    # forward propagation
    l0 = X
    l1 = nonlin(np.dot(l0,syn0))

    # how much did we miss?
    l1_error = y - l1

    # multiply how much we missed by the 
    # slope of the sigmoid at the values in l1
    l1_delta = l1_error * nonlin(l1,True)

    # update weights
    syn0 += np.dot(l0.T,l1_delta)

print "Output After Training:"
print l1

这是网站：http://iamtrask.github.io/2015/07/12/basic-python-network/

代码的第36行,l1误差乘以用权重点缀的输入的导数.我不知道为什么这样做,并花了好几个小时试图解决它.我刚刚得出结论这是错误的,但有些事情告诉我,考虑到有多少人推荐并使用本教程作为学习神经网络的起点,这可能是不对的.

在文章中,他们说

Look at the sigmoid picture again! If the slope was really shallow
(close to 0),then the network either had a very high value,or a very
low value. This means that the network was quite confident one way or
the other. However,if the network guessed something close to (x=0,
y=0.5) then it isn’t very confident.

我似乎无法理解为什么输入sigmoid函数的高低有与置信度有关.当然,它有多高并不重要,因为如果预测的输出很低,那么它将真的不自信,不像他们所说的那样应该有信心因为它很高.

如果你想强调错误,那么将l1_error复制一下会更好吗？

这是一个真正的失望考虑到这一点,它最终看起来我找到了一个有前途的方式来真正直观地开始学习神经网络,但我又错了.如果你有一个很好的地方开始学习我可以很容易理解的地方,我们将不胜感激.