import numpy as np
alpha = 0.0251 # as close to true alpha as possible
def nonlinear(x, deriv=False):
if(deriv==True):
return x*(1-x)
return 1/(1+np.e**(-x))
#seed
np.random.seed(1)
#testing sample
test_x = np.array([[251,497,-246],
[299,249,50],
[194,180,14],
[140,148,-8],
[210,140,70]])
#Input Array - This input will be taken directly from a Pong game
X = np.array([[198,200,-2],
[90, 280,-190],
[84, 256,-172],
[140,240,-100],
[114,216,-102],
[72, 95,-23],
[99, 31, 68],
[144, 20, 124],
[640, 216,424],
[32, 464,-432],
[176, 64,112],
[754, 506,248],
[107, 104,3],
[116,101,15]])
#output array - if ball_pos - paddle > 0 move up else move down
Y = np.array([[0,0,0,0,0,0,1,1,1,0,1,1,1,1,]]).T
syn0 = 2*np.random.random((3,14))-1
syn1 = 2*np.random.random((14,14))-1
for j in range(60000):
#forward propagation
l0 = X
l1 = nonlinear(np.dot(l0, syn0))
l2 = nonlinear(np.dot(l1, syn1))
#how much did we miss
l2_error = Y - l2
#multiply how much missed by the slope of sigmoid at the value in l1
l2_delta = l2_error * nonlinear(l2, True)
#how much did l1 contribute to l2 error
#(according to the weights)
l1_error = l2_delta.dot(syn1.T)
#in what direction is the target l1?
# Sure?
l1_delta = l1_error*nonlinear(l1,True)
#update weight
syn1 += alpha * (l1.T.dot(l2_delta))
syn0 += alpha * (l0.T.dot(l1_delta))
# display error
if(j % 10000) == 0:
print("ERROR: " + str(np.mean(np.abs(l2_error))))
#Testing Forward propagation
l0_test = test_x
l1_test = nonlinear(np.dot(l0_test,syn0))
l2_test = nonlinear(np.dot(l1_test,syn1))
#Dress up the array (make it look nice)
l2_test_output = []
for x in range(len(l2_test)):
l2_test_output.append(l2_test[x][0])
print("Test Output")
print(l2_test_output)
#Put all the l2 data in a way I could see it: Just the first probabilites
l2_output = []
for x in range(len(l2)):
l2_output.append(l2[x][0])
print("Output")
print(l2_output)
此代码应该接收一组三个数字 [(value_1),(value_2),(value_1-value_2)] 并在第一个和第二个值之间的差为负或“1”时返回“0”如果差异是积极的。到目前为止，它实际上工作得很好。
这是输出：
ERROR: 0.497132186092
ERROR: 0.105081486632
ERROR: 0.102115299177
ERROR: 0.100813655802
ERROR: 0.100042420179
ERROR: 0.0995185781466
Test Output
[0.0074706006801269686, 0.66687458928464094, 0.66687458928463983, 0.66686236694464551, 0.98341439176739631]
Output
[0.66687459245609326, 0.00083944690766060215, 0.00083946471285455484, 0.0074706634783305243, 0.0074706634765733968, 0.007480987498372226, 0.99646513183073093, 0.99647100131874755, 0.99646513180692531, 0.00083944572383107523, 0.99646513180692531, 0.98324165810211861, 0.66687439729829612, 0.66687459321626519] 错误：0.497132186092
如您所见，给定 alpha = 0.0251（对于梯度下降 - 通过反复试验发现）的误差仅为 9.95 % 左右。
自从我编写了这个程序以来，我了解到leaky RelU 是 Sigmoid 函数的更好替代方案，因为它比 Sigmoid 优化和学习更快。我想在这个程序中使用 numpy 实现泄漏的 RelU 函数，但我不确定从哪里开始，尤其是它的导数是什么。
我怎样才能在这个神经网络中实现leaky RelU？
