numpy dot函数:操作数不能在mlp中一起广播

ufj5ltwl  于 2021-07-14  发布在  Java
关注(0)|答案(1)|浏览(408)

我用python编写了一个mlp,我编写了以下代码:

class NeuralNetwork(object):
    def __init__(self):

        #parameters
        self.inputSize = 13
        self.hidden1Size = 13
        self.hidden2Size = 13
        self.outputSize = 13

        #weights
        self.W1 = np.random.randn(self.inputSize, self.hidden1Size) # weight matrix from input to hidden layer
        self.W2 = np.random.randn(self.hidden1Size, self.hidden2Size) # weight matrix from hidden to hidden2 layer
        self.W3 = np.random.randn(self.hidden2Size, self.outputSize) # weight matrix from hidden to output layer

    def feedForward(self, X):
        self.z = np.dot(X, self.W1)  #Zh1
        self.z2 = self.sigmoid(self.z)  #ah1
        self.z3 = np.dot(self.z2, self.W2) #Zh2
        self.z4 = self.sigmoid(self.z3)  #ah2
        self.z5 = np.dot(self.z3, self.W3) #zout
        output = self.sigmoid(self.z5) #a out
        print(output)
        return output

    def sigmoid(self, s, deriv=False):
        if (deriv == True):
            return s * (1 - s)
        return 1/(1 + np.exp(-s))

    def backward(self, X, y, output):
        #step1
        self.output_error = output - y # error in output

        #step2
        self.output_delta = self.output_error * self.sigmoid(self.z3).T
        self.W3 = self.W3 - self.output_delta

        #step3
        self.z2_error = self.output_delta.dot(self.output_delta)
        self.z2_delta = self.z2_error * self.sigmoid(self.z3)
        self.W2 = self.W2 - self.z2_delta

        #step4
        self.z1_error = self.output_delta.dot(self.W2.T)
        self.z1_delta = self.z2_error * self.sigmoid(self.z2, deriv=True)

        self.W1 = self.W1 - self.z1_delta

    def train(self, X, y):
        output = self.feedForward(X)

        self.backward(X, y, output)

def accuracy (predict , y_test):
    C = 0
    length = len(y_test)
    for i in length:
        if predict[i] == y_test[i]:
            C = C + 1
    return(C/length)

但我有个问题:

ValueError                                Traceback (most recent
 call last) <ipython-input-8-69e28bd743d3> in <module>
       1 NN = NeuralNetwork()
       2 
 ----> 3 NN.train(X_train,y_train)
       4 
       5 predict = NN.feedforward(X_train)

 <ipython-input-7-55e55429732f> in train(self, X, y)
      54         output = self.feedForward(X)
      55 
 ---> 56         self.backward(X, y, output)
      57 
      58 

 <ipython-input-7-55e55429732f> in backward(self, X, y, output)
      31     def backward(self, X, y, output):
      32         #step1
 ---> 33         self.output_error = output - y # error in output
      34 
      35         #step2

 ValueError: operands could not be broadcast together with shapes
 (242,13) (242,)

我知道问题是numpy不能做点函数,因为矩阵的问题,但我不知道如何解决这个问题。

x9ybnkn6

x9ybnkn61#

请参见以下与错误相同的示例:

import numpy as np

a = np.zeros((4, 5))
b = np.zeros(4,)

# c = a - b error!

c = a - b[:, np.newaxis] # no error!

所以你可以试试 self.backward(X, y, output[:, np.newaxis]) 医生。挺烦人的。
它转换一个一维数组(形状) (n,) )二维阵列(形状) (n, 1) )那努比就可以应付了。

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