我正在尝试通过使用2感知器网络来制作XOR门但由于某种原因网络没有学习,当我在图中绘制误差的变化时,误差变为静态水平并在该区域中振荡。
目前我没有对网络添加任何偏见。
import numpy as np
def S(x):
return 1/(1+np.exp(-x))
win = np.random.randn(2,2)
wout = np.random.randn(2,1)
eta = 0.15
# win = [[1,1], [2,2]]
# wout = [[1],[2]]
obj = [[0,0],[1,0],[0,1],[1,1]]
target = [0,1,1,0]
epoch = int(10000)
emajor = ""
for r in range(0,epoch):
for xy in range(len(target)):
tar = target[xy]
fdata = obj[xy]
fdata = S(np.dot(1,fdata))
hnw = np.dot(fdata,win)
hnw = S(np.dot(fdata,win))
out = np.dot(hnw,wout)
out = S(out)
diff = tar-out
E = 0.5 * np.power(diff,2)
emajor += str(E[0]) + ",\n"
delta_out = (out-tar)*(out*(1-out))
nindelta_out = delta_out * eta
wout_change = np.dot(nindelta_out[0], hnw)
for x in range(len(wout_change)):
change = wout_change[x]
wout[x] -= change
delta_in = np.dot(hnw,(1-hnw)) * np.dot(delta_out[0], wout)
nindelta_in = eta * delta_in
for x in range(len(nindelta_in)):
midway = np.dot(nindelta_in[x][0], fdata)
for y in range(len(win)):
win[y][x] -= midway[y]
f = open('xor.csv','w')
f.write(emajor) # python will convert \n to os.linesep
f.close() # you can omit in most cases as the destructor will call it
这是因学习轮数而改变的错误。它是否正确?红色线是我期望错误应该如何变化的线。
我在代码中做错了什么?因为我似乎无法弄清楚导致错误的原因。非常感谢。
提前致谢
答案 0 :(得分:1)
这是一个带有反向传播的隐藏层网络,可以自定义运行relu,sigmoid和其他激活的实验。经过几次实验后得出的结论是,网络表现更好,并且更快地达到收敛,而在sigmoid下,损失值波动。这是因为," https://stackoverflow.com/a/25110513"。
import numpy as np
import matplotlib.pyplot as plt
from operator import xor
class neuralNetwork():
def __init__(self):
# Define hyperparameters
self.noOfInputLayers = 2
self.noOfOutputLayers = 1
self.noOfHiddenLayerNeurons = 2
# Define weights
self.W1 = np.random.rand(self.noOfInputLayers,self.noOfHiddenLayerNeurons)
self.W2 = np.random.rand(self.noOfHiddenLayerNeurons,self.noOfOutputLayers)
def relu(self,z):
return np.maximum(0,z)
def sigmoid(self,z):
return 1/(1+np.exp(-z))
def forward (self,X):
self.z2 = np.dot(X,self.W1)
self.a2 = self.relu(self.z2)
self.z3 = np.dot(self.a2,self.W2)
yHat = self.relu(self.z3)
return yHat
def costFunction(self, X, y):
#Compute cost for given X,y, use weights already stored in class.
self.yHat = self.forward(X)
J = 0.5*sum((y-self.yHat)**2)
return J
def costFunctionPrime(self,X,y):
# Compute derivative with respect to W1 and W2
delta3 = np.multiply(-(y-self.yHat),self.sigmoid(self.z3))
djw2 = np.dot(self.a2.T, delta3)
delta2 = np.dot(delta3,self.W2.T)*self.sigmoid(self.z2)
djw1 = np.dot(X.T,delta2)
return djw1,djw2
if __name__ == "__main__":
EPOCHS = 6000
SCALAR = 0.01
nn= neuralNetwork()
COST_LIST = []
inputs = [ np.array([[0,0]]), np.array([[0,1]]), np.array([[1,0]]), np.array([[1,1]])]
for epoch in xrange(1,EPOCHS):
cost = 0
for i in inputs:
X = i #inputs
y = xor(X[0][0],X[0][1])
cost += nn.costFunction(X,y)[0]
djw1,djw2 = nn.costFunctionPrime(X,y)
nn.W1 = nn.W1 - SCALAR*djw1
nn.W2 = nn.W2 - SCALAR*djw2
COST_LIST.append(cost)
plt.plot(np.arange(1,EPOCHS),COST_LIST)
plt.ylim(0,1)
plt.xlabel('Epochs')
plt.ylabel('Loss')
plt.title(str('Epochs: '+str(EPOCHS)+', Scalar: '+str(SCALAR)))
plt.show()
inputs = [ np.array([[0,0]]), np.array([[0,1]]), np.array([[1,0]]), np.array([[1,1]])]
print "X\ty\ty_hat"
for inp in inputs:
print (inp[0][0],inp[0][1]),"\t",xor(inp[0][0],inp[0][1]),"\t",round(nn.forward(inp)[0][0],4)
结束结果:
the gradient of sigmoids becomes increasingly small as the absolute value of x increases
X y y_hat
(0, 0) 0 0.0
(0, 1) 1 0.9997
(1, 0) 1 0.9997
(1, 1) 0 0.0005
训练后获得的体重是:
nn.w1
[ [-0.81781753 0.71323677]
[ 0.48803631 -0.71286155] ]
nn.w2
[ [ 2.04849235]
[ 1.40170791] ]
我知道的只有很少,也可以在这个答案中解释。如果你想更好地理解神经网络,那么我建议你通过以下链接:Neural networks demystified
答案 1 :(得分:0)
每个时期计算的误差应该是所有平方误差的总和(即每个目标的误差)
import numpy as np
def S(x):
return 1/(1+np.exp(-x))
win = np.random.randn(2,2)
wout = np.random.randn(2,1)
eta = 0.15
# win = [[1,1], [2,2]]
# wout = [[1],[2]]
obj = [[0,0],[1,0],[0,1],[1,1]]
target = [0,1,1,0]
epoch = int(10000)
emajor = ""
for r in range(0,epoch):
# ***** initialize final error *****
finalError = 0
for xy in range(len(target)):
tar = target[xy]
fdata = obj[xy]
fdata = S(np.dot(1,fdata))
hnw = np.dot(fdata,win)
hnw = S(np.dot(fdata,win))
out = np.dot(hnw,wout)
out = S(out)
diff = tar-out
E = 0.5 * np.power(diff,2)
# ***** sum all errors *****
finalError += E
delta_out = (out-tar)*(out*(1-out))
nindelta_out = delta_out * eta
wout_change = np.dot(nindelta_out[0], hnw)
for x in range(len(wout_change)):
change = wout_change[x]
wout[x] -= change
delta_in = np.dot(hnw,(1-hnw)) * np.dot(delta_out[0], wout)
nindelta_in = eta * delta_in
for x in range(len(nindelta_in)):
midway = np.dot(nindelta_in[x][0], fdata)
for y in range(len(win)):
win[y][x] -= midway[y]
# ***** Save final error *****
emajor += str(finalError[0]) + ",\n"
f = open('xor.csv','w')
f.write(emajor) # python will convert \n to os.linesep
f.close() # you can omit in most cases as the destructor will call it