我试图运行多个嵌套循环,然后根据循环中的条件检索一个值数组。我有14个alpha值需要测试,每个on都需要测试1,2,3,4,5,6的历元。当我测试所有6个历元的每个alpha值时,我想记录误差小于0.05的次数。最后,我想要一个2D数组,其中的行表示14个不同的alpha值,每列表示一个不同的历元值
我想知道是否有更好的方法,然后使用带numpy的张量。当我尝试扩展这个项目时,使用这种方法会给我带来很多问题
对于感兴趣的人来说,这只是一个2输入、单隐层、输出的神经网络,用于自学反向传播。我提交的代码是针对2个神经元的,但我现在正试图将其扩展到4个,最终是n个神经元。存储错误值并计算在2D数组中输出的成功率,在那里我可以看到什么alpha和epoch对产生最好的结果,这将非常有帮助
在使用此代码之前,我已完成此任务:
import numpy as np
for l in range(0,14):
alpha = [0.001, 0.002, 0.003, 0.004, 0.005, 0.006, 0.007, 0.008, 0.009, 0.01, 0.02, 0.03, 0.04, 0.05]
nEpoch=1
for n in range(0,6):
nSuccess = 0
w11f = np.zeros(nEpoch*nTrain)
w12f = np.zeros(nEpoch*nTrain)
for j in range(0,50):
w11 = 0.5 - np.random.rand();
w12 = 0.5 - np.random.rand();
w21 = 0.5 - np.random.rand();
w22 = 0.5 - np.random.rand();
w31 = 0.5 - np.random.rand();
w32 = 0.5 - np.random.rand();
w41 = 0.5 - np.random.rand();
w42 = 0.5 - np.random.rand();
b4 = 0.5 - np.random.rand();
b3 = 0.5 - np.random.rand();
b2 = 0.5 - np.random.rand();
b1 = 0.5 - np.random.rand();
ww1 = 0.5 - np.random.rand();
ww2 = 0.5 - np.random.rand();
ww3 = 0.5 - np.random.rand();
ww4 = 0.5 - np.random.rand();
bb = 0.5 - np.random.rand();
sp = random.sample(a,nTrain + nTest)
p = 0
for epoch in range(0,nEpoch):
for i in range(0,nTrain):
y1 = b1 + w11*x[sp[i],0] + w12*x[sp[i],1]
y2 = b2 + w21*x[sp[i],0] + w22*x[sp[i],1]
y3 = b3 + w31*x[sp[i],0] + w32*x[sp[i],1]
y4 = b4 + w41*x[sp[i],0] + w42*x[sp[i],1]
dxx1 = y1 > 0
xx1 = y1*dxx1
dxx2 = y2 > 0
xx2 = y2*dxx2
dxx3 = y3 > 0
xx3 = y3*dxx3
dxx4 = y4 > 0
xx4 = y4*dxx4
yy = bb + ww1*xx1 + ww2*xx2 + ww3*xx3 + ww4*xx4
yy = yy > 0
e = t[sp[i]] - yy
#Updating parameters
ww1 = ww1 + alpha[l]*e*xx1
ww2 = ww2 + alpha[l]*e*xx2
ww3 = ww3 + alpha[l]*e*xx3
ww4 = ww4 + alpha[l]*e*xx4
bb = bb + alpha[l]*e
w11 = w11 + alpha[l]*e*ww1*dxx1*x[sp[i],0]
w12 = w12 + alpha[l]*e*ww1*dxx1*x[sp[i],1]
w21 = w21 + alpha[l]*e*ww2*dxx2*x[sp[i],0]
w22 = w22 + alpha[l]*e*ww2*dxx2*x[sp[i],1]
w31 = w31 + alpha[l]*e*ww3*dxx3*x[sp[i],0]
w32 = w32 + alpha[l]*e*ww3*dxx3*x[sp[i],1]
w41 = w41 + alpha[l]*e*ww4*dxx4*x[sp[i],0]
w42 = w42 + alpha[l]*e*ww4*dxx4*x[sp[i],1]
b1 = b1 + alpha[l]*e*ww1*dxx1
b2 = b2 + alpha[l]*e*ww2*dxx2
b3 = b3 + alpha[l]*e*ww3*dxx3
b4 = b4 + alpha[l]*e*ww4*dxx4
w11f[p] = w11
w12f[p] = w12
p = p + 1
er = 0
for k in range(nTrain,nTrain + nTest):
y1 = b1 + w11*x[sp[i],0] + w12*x[sp[i],1]
y2 = b2 + w21*x[sp[i],0] + w22*x[sp[i],1]
y3 = b3 + w31*x[sp[i],0] + w32*x[sp[i],1]
y4 = b4 + w41*x[sp[i],0] + w42*x[sp[i],1]
dxx1 = y1 > 0
xx1 = y1*dxx1
dxx2 = y2 > 0
xx2 = y2*dxx2
dxx3 = y3 > 0
xx3 = y3*dxx3
dxx4 = y4 > 0
xx4 = y4*dxx4
yy = bb + ww1*xx1 + ww2*xx2 + ww3*xx3 + ww4*xx4
yy = yy > 0
e = abs(t[sp[k]] - yy)
er = er + e #Accumulates error
er = er/nTest #Calculates average error
er_List[l,j,n] = er
if er_List[l,j,n] < 0.05:
nSuccess = nSuccess + 1
#Part C - Creating an Array that contains the success values of each
#alpha and epoch value pair
nSuccess_Array[l,n] = nSuccess #Array that contains the success
if nEpoch < 6:
nEpoch += 1
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