Numpy 点函数：操作数无法在 MLP 中一起广播

如何解决Numpy 点函数：操作数无法在 MLP 中一起广播

我用 python 写了一个 MLP，我做了这个代码：

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,y):
        output = self.feedForward(X)
       
        self.backward(X,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)

但我遇到了这个问题：

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

我知道问题是由于矩阵问题，numpy不能做点函数，但我不知道如何解决这个问题。

解决方法

请参阅以下与您的错误相同的示例：

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])

doc。好烦。

---

它的作用是将一维数组（形状为 (n,)）转换为二维数组（形状为 (n,1)），然后 numpy 能够处理它。