如何用np array数组搭建一个带隐藏层的简单bp网络?

用np array数组搭建一个带隐藏层的简单bp网络,实现异或功能。然后通过pyqt界面让用户输入两个数,并显示预测结果。

希望得到可运行的代码

只需要前向吗?

实现异或功能,功能是明确的。如果只是要进行前向传播的话,可以人工把隐含层的参数确定下来。np实现bp网络,本质就是矩阵相乘。也就是说人工计算确定隐含层的参数即可实现。

可以参考这篇博客的代码,然后自己再加个用户界面


def init_network():
    #init_network()函数会进行权重和偏置的初始化,并将它们保存在字典变量network中。这个字典变量network中保存了每一层所需的参数(权重和偏置)。
    network = {}
    network['W1'] = np.array([[0.1, 0.3, 0.5], [0.2, 0.4, 0.6]])
    network['b1'] = np.array([0.1, 0.2, 0.3])
    network['W2'] = np.array([[0.1, 0.4], [0.2, 0.5], [0.3, 0.6]])
    network['b2'] = np.array([0.1, 0.2])
    network['W3'] = np.array([[0.1, 0.3], [0.2, 0.4]])
    network['b3'] = np.array([0.1, 0.2])
    return network


def forward(network, x):
    #forward()函数中则封装了将输入信号转换为输出信号的处理过程。
    W1, W2, W3 = network['W1'], network['W2'], network['W3']
    b1, b2, b3 = network['b1'], network['b2'], network['b3']
    
    a1 = np.dot(x, W1) + b1
    z1 = sigmoid(a1)
    a2 = np.dot(z1, W2) + b2
    z2 = sigmoid(a2)
    a3 = np.dot(z2, W3) + b3
    y = identity_function(a3)
    return y

network = init_network()
x = np.array([1.0, 0.5])
y = forward(network, x)
print(y) # [ 0.31682708 0.69627909]

import numpy as np

# numpy实现与运算
def AND(x1,x2):
    #判断条件(x1w1+x2w2)>1 retuen 1,否则return 0
    x=np.array([x1,x2])
    w=np.array([0.6,0.5])
    y=np.sum(x*w)
    if y>1:
        return 1
    else:
        return 0
        
# numpy实现非运算
def NOT(x1,x2):
    #判断条件(x1w1+x2w2)<1 retuen 1,否则return 0
    x=np.array([x1,x2])
    w=np.array([0.6,0.5])
    y=np.sum(x*w)
    if y<=1:
        return 1
    else:
        return 0
        
#numpy实现或运算
def OR(x1,x2):
    #权重x1,x2要大于偏执b,判断条件 为真-> b+x1w1+x2w2>=0 为假->  b+x1w1+x2w2<=0
    x=np.array([x1,x2])
    w=np.array([0.2,0.5])
    b=-0.1
    y=np.sum(x*w)+b
    if y<=0:
        return 0
    else:
        return 1

#numpy实现异或运算
def XOR(x1,x2):
    #判断条件将输入值x1,x2进行非not运算和或or运算然后再将其返回值进行与and运算变得到异或xor运算
    m=NOT(x1,x2)
    n=OR(x1,x2)
    k =AND(m,n)
    return k
    
    
if __name__=="__main__":
    data=[1,0]
    print("x1,x2分别为:",data[0],data[1])
    print("与运算:",AND(data[0],data[1]))
    print("非运算:",NOT(data[0],data[1]))
    print("或运算:",OR(data[0],data[1]))
    print("异或运算:",XOR(data[0],data[1]))
    


x1,x2 = np.array([0,0,1,1]), np.array([0,1,0,1])
x = np.array([x1,x2])
y = np.array([0,1,1,0])
nums_input, nums_hidden, nums_output = 2,2,1
lr = 0.01
epochs = 10000
losses = []
bs = x.shape[1]
w1 = np.random.rand(nums_hidden, nums_input)
b1 = np.random.rand(nums_hidden,1)
w2 = np.random.rand(nums_output, nums_hidden)
b2 = np.random.rand(nums_output,1)

进行前项传播计算

def sigmoid(x):
return 1.0 / (1 + np.exp(-x))

前项传播函数

def forward(x,w1,b1,w2,b2):
# 必须将b1变成列向量
z1 = np.dot(w1,x) + b1
a1 = sigmoid(z1)
z2 = np.dot(w2,a1) + b2
a2 = sigmoid(z2)
return z1,a1,z2,a2
def backward(x,y,z1,a1,z2,a2,w1,w2):
bs = x.shape[1]
dz2 = a2 - y
dw2 = np.dot(dz2, a1.T) / bs
db2 = np.sum(dz2, axis=1,keepdims=True) / bs
dz1 = np.dot(w2.T,dz2)a1(1-a1)
dw1 = np.dot(dz1, x.T) / bs
db1 = np.sum(dz1, axis=1,keepdims=True) / bs
# 对dw1和dw2的数据的尺寸进行设置
dw1 = dw1.reshape(w1.shape)
db1 = db1.reshape(b1.shape)
dw2 = dw2.reshape(w2.shape)
db2 = db2.reshape(b2.shape)
return dw1,db1,dw2,db2