ReLU derivative with NumPy

问题

import numpy as np

def relu(z):
    return np.maximum(0,z)

def d_relu(z):
    z[z>0]=1
    z[z<=0]=0
    return z

x=np.array([5,1,-4,0])
y=relu(x)
z=d_relu(y)
print("y = {}".format(y))
print("z = {}".format(z))

The code above prints out:

y = [1 1 0 0]
z = [1 1 0 0]

instead of

y = [5 1 0 0]
z = [1 1 0 0]

From what I understand the function calls I've used should only be doing passing by value,passing a copy of the variable.

Why is my d_relu function affecting the y variable?

回答1:

Your first mistake is in assuming python passes objects by value... it doesn't - it's pass by assignment (similar to passing by reference, if you're familiar with this concept). However, only mutable objects, as the name suggests, can be modified in-place. This includes, among other things, numpy arrays.

You shouldn't have d_relu modify z inplace, because that's what it's doing right now, through the z[...] = ... syntax. Try instead building a mask using broadcasted comparison and returning that instead.

def d_relu(z):
    return (z > 0).astype(int)

This returns a fresh array instead of modifying z in-place, and your code prints

y = [5 1 0 0]
z = [1 1 0 0]

If you're building a layered architecture, you can leverage the use of a computed mask during the forward pass stage:

class relu:
    def __init__(self):
        self.mask = None

    def forward(self, x):
        self.mask = x > 0
        return x * self.mask

    def backward(self, x):
        return self.mask

Where the derivative is simply 1 if the input during feedforward if > 0, else 0.

来源：https://stackoverflow.com/questions/50121887/relu-derivative-with-numpy