用Python实现神经网络(周#5 )的成本函数

Question

在计算机学习课程的基础上，我尝试在python中实现神经网络的成本函数。有一个类似于这个的question --有一个被接受的答案--但是答案中的代码是用八度写的。为了不懒惰，我尝试将答案的相关概念适应于我的情况，据我所知，我正在正确地实现这个函数。然而，我输出的成本与预期成本不同，所以我做错了一些事情。

下面是一个可复制的小例子：

下面的链接将导致一个.npz文件，该文件可以加载(如下所示)以获得相关数据。如果您使用"arrays.npz"文件，请重命名它。

1

if __name__ == "__main__":

with np.load("arrays.npz") as data:

    thrLayer = data['thrLayer'] # The final layer post activation; you 
    # can derive this final layer, if verification needed, using weights below

    thetaO = data['thetaO'] # The weight array between layers 1 and 2
    thetaT = data['thetaT'] # The weight array between layers 2 and 3

    Ynew = data['Ynew'] # The output array with a 1 in position i and 0s elsewhere

    #class i is the class that the data described by X[i,:] belongs to

    X = data['X'] #Raw data with 1s appended to the first column
    Y = data['Y'] #One dimensional column vector; entry i contains the class of entry i



import numpy as np

m = len(thrLayer)
k = thrLayer.shape[1]
cost = 0

for i in range(m):
    for j in range(k):
        cost += -Ynew[i,j]*np.log(thrLayer[i,j]) - (1 - Ynew[i,j])*np.log(1 - thrLayer[i,j])
print(cost)
cost /= m

'''
Regularized Cost Component
'''

regCost = 0

for i in range(len(thetaO)):
    for j in range(1,len(thetaO[0])):
        regCost += thetaO[i,j]**2

for i in range(len(thetaT)):
    for j in range(1,len(thetaT[0])):
        regCost += thetaT[i,j]**2

regCost *= lam/(2*m) 


print(cost)
print(regCost)

实际上，cost应该是0.287629，cost + newCost应该是0.383770。

这是上述问题中公布的费用职能，以供参考：

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Answer 1

问题在于您使用的是错误的类标签。在计算成本函数时，您需要使用基本真理或真正的类标签。

我不知道你的新数组是什么，但不是训练输出。因此，我更改了您的代码，使用Y作为类标签的替代Ynew，并得到了正确的成本。

import numpy as np

with np.load("arrays.npz") as data:

    thrLayer = data['thrLayer'] # The final layer post activation; you
    # can derive this final layer, if verification needed, using weights below

    thetaO = data['thetaO'] # The weight array between layers 1 and 2
    thetaT = data['thetaT'] # The weight array between layers 2 and 3

    Ynew = data['Ynew'] # The output array with a 1 in position i and 0s elsewhere

    #class i is the class that the data described by X[i,:] belongs to

    X = data['X'] #Raw data with 1s appended to the first column
    Y = data['Y'] #One dimensional column vector; entry i contains the class of entry i


m = len(thrLayer)
k = thrLayer.shape[1]
cost = 0

Y_arr = np.zeros(Ynew.shape)
for i in xrange(m):
    Y_arr[i,int(Y[i,0])-1] = 1

for i in range(m):
    for j in range(k):
        cost += -Y_arr[i,j]*np.log(thrLayer[i,j]) - (1 - Y_arr[i,j])*np.log(1 - thrLayer[i,j])
cost /= m

'''
Regularized Cost Component
'''

regCost = 0

for i in range(len(thetaO)):
    for j in range(1,len(thetaO[0])):
        regCost += thetaO[i,j]**2

for i in range(len(thetaT)):
    for j in range(1,len(thetaT[0])):
        regCost += thetaT[i,j]**2
lam=1
regCost *= lam/(2.*m)


print(cost)
print(cost + regCost)

这一产出如下：

0.287629165161
0.383769859091

编辑：用regCost *= lam/(2*m)修正了一个整数除法错误，将regCost归零。

Answer 2

您可以尝试此实现。

import scipy.io
mat=scipy.io.loadmat('ex4data1.mat')
X=mat['X']
y=mat['y']

theta=scipy.io.loadmat('ex4weights.mat')
theta1=theta['Theta1']
theta2=theta['Theta2']
theta=[theta1,theta2]



new=np.zeros((10,len(y)))
for i in range(len(y)):
    new[y[i]-1,i]=1

y=new

def sigmoid(x):
    return 1/(1+np.exp(-x))

def reg_cost(theta,X,y,lambda1):
    current=X
    for i in range(len(theta)):
        a= np.append(np.ones((len(current),1)),current,axis=1)
        z=np.matmul(a,theta[i].T)
        z=sigmoid(z)
        current=z
    htheta=current
    ans=np.sum(np.multiply(np.log(htheta),(y).T)) + 
np.sum(np.multiply(np.log(1-htheta),(1-y).T))
    ans=-ans/len(X)
    for i in range(len(theta)):
        new=theta[i][:,1:]
        newsum=np.sum(np.multiply(new,new))
        ans+=newsum*(lambda1)/(2*len(X))

    return ans

print(reg_cost(theta,X,y,1))  

It输出

0.3837698590909236