神经网络返回相同的错误

时间:2017-10-31 07:57:17

标签: python neural-network

我是机器学习的初学者,并尝试按照本网站提供的教程自行构建神经网络 http://iamtrask.github.io/2015/07/12/basic-python-network/

在本教程的第3部分中,有一个输入图层,一个输出图层和一个隐藏图层。

但是,当我尝试运行代码时,它会打印相同的错误。所以错误并没有像预期的那样变小。这是代码:

import numpy as np;

def nonlin(x,deriv=False):
    if (deriv==True):
        return x * 1-x

    return 1/(1+np.exp(-x))

x = np.array([  [0,0,1],
                [0,1,1],
                [1,0,1],
                [1,1,1]  ])

y = np.array([[0],[1],[1],[0]])

np.random.seed(1)

#randomly initialize our weights with mean 0
syn0 = 2*np.random.random((3,4)) - 1
syn1 = 2*np.random.random((4,1)) - 1

for j in range (60000):

    #feed forward through layers 0, 1, and 2
    l0 = x
    l1 = nonlin(np.dot(l0,syn0))
    l2 = nonlin(np.dot(l1,syn1))

    # how much did we miss the target value?
    l2_error = y - l2

    if (j%10000) == 0:
        print ("Error:" + str (np.mean(np.abs(l2_error))))

    #in what direction is the target value?
    # were we really sure? if so, dont change too much.
    l2_delta = l2_error*nonlin(l2,deriv=True)

    # how much did each L1 value contribute to the l2 error
    #(according to the weights)?
    l1_error = l2_delta.dot(syn1.T)

    # in what direction is the target L1?
    # were we really sure? if so dont change too much.
    l1_delta = l1_error * nonlin(l1,deriv=True)

    syn1 += l1.T.dot(l2_delta)
    syn0 += l0.T.dot(l1_delta)

感谢您的反馈

P.S:我正在使用python 3.5.2,windows 7

1 个答案:

答案 0 :(得分:1)

你需要注意BODMAS(https://www.skillsyouneed.com/num/bodmas.html)。您从def nonlin(x,deriv=False): if (deriv==True): return x * 1-x return 1/(1+np.exp(-x)) 函数返回零:

return x*1-x = x-x = 0

基本上是return x*(1-x) 。你应该:

{{1}}