python - 如何实现 ReLU 代替 Sigmoid 函数

Question

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)

此代码应该接受一组三个数字 [(value_1),(value_2),(value_1-value_2)] 并在第一个和第二个值之间的差为负数时返回“0”或如果差异为正，则为“1”。到目前为止，它实际上运行良好。

这是输出: 错误:0.497132186092 错误:0.105081486632 错误:0.102115299177 错误:0.100813655802 错误:0.100042420179 错误:0.0995185781466 测试输出 [0.0074706006801269686、0.66687458928464094、0.66687458928463983、0.66686236694464551、0.98341439176739631] 输出 [0.66687459245609326, 0.00083944690766060215, 0.00083946471285455484, 0.0074706634783305243, 0.0074706634765733968, 0.00724986378, 0.00724986378 6513183073093, 0.99647100131874755, 0.99646513180692531, 0.00083944572383107523, 0.99646513180692531, 0.98324165810211861629, 0.98324165810211861629, 7,46618, 0.98324165810211861629 .66687459321626519] 错误:0.497132186092

如您所见，给定 alpha = 0.0251 的误差(对于梯度下降 - 通过反复试验发现)仅为 9.95% 左右。

自从我制作这个程序后，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。我想在此程序中使用 numpy 实现泄漏的 RelU 函数，但我不确定从哪里开始，更具体地说，它的导数是什么。

如何在这个神经网络中实现 leaky RelU？

Answer 1

我想在这里补充一点，实际上有很多类似 ReLu 的激活函数可以用来代替标准的 ReLu activation。 :

• 您自己提到过 Leaky ReLu(由 alpha 参数化)。
• Parametric Rectified Linear Unit (PReLU)。该公式与 Leaky ReLu 相同，但允许学习系数 alpha。另见 this discussion .
• Exponential linear unit (ELU)，它试图使平均激活值接近于零，从而加速学习:

• Scaled exponential linear unit (SELU) 最近发布。它是 ELU 的扩展，具有特定的参数选择，具有额外的归一化效果并有助于更快地学习。

Here's the list所有激活及其导数。