How to Implement the Relu Function in Numpy

I'm completely revising my original answer because of points raised in the other questions and comments. Here is the new benchmark script:

import time
import numpy as np


def fancy_index_relu(m):
    m[m < 0] = 0


relus = {
    "max": lambda x: np.maximum(x, 0),
    "in-place max": lambda x: np.maximum(x, 0, x),
    "mul": lambda x: x * (x > 0),
    "abs": lambda x: (abs(x) + x) / 2,
    "fancy index": fancy_index_relu,
}

for name, relu in relus.items():
    n_iter = 20
    x = np.random.random((n_iter, 5000, 5000)) - 0.5

    t1 = time.time()
    for i in range(n_iter):
        relu(x[i])
    t2 = time.time()

    print("{:>12s}  {:3.0f} ms".format(name, (t2 - t1) / n_iter * 1000))

It takes care to use a different ndarray for each implementation and iteration. Here are the results:

         max  126 ms
in-place max  107 ms
         mul  136 ms
         abs   86 ms
 fancy index  132 ms

You can do it in much easier way:

def ReLU(x):
    return x * (x > 0)

def dReLU(x):
    return 1. * (x > 0)

EDIT As jirassimok has mentioned below my function will change the data in place, after that it runs a lot faster in timeit. This causes the good results. It's some kind of cheating. Sorry for your inconvenience.

I found a faster method for ReLU with numpy. You can use the fancy index feature of numpy as well.

fancy index:

20.3 ms ± 272 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

>>> x = np.random.random((5,5)) - 0.5 
>>> x
array([[-0.21444316, -0.05676216,  0.43956365, -0.30788116, -0.19952038],
       [-0.43062223,  0.12144647, -0.05698369, -0.32187085,  0.24901568],
       [ 0.06785385, -0.43476031, -0.0735933 ,  0.3736868 ,  0.24832288],
       [ 0.47085262, -0.06379623,  0.46904916, -0.29421609, -0.15091168],
       [ 0.08381359, -0.25068492, -0.25733763, -0.1852205 , -0.42816953]])
>>> x[x<0]=0
>>> x
array([[ 0.        ,  0.        ,  0.43956365,  0.        ,  0.        ],
       [ 0.        ,  0.12144647,  0.        ,  0.        ,  0.24901568],
       [ 0.06785385,  0.        ,  0.        ,  0.3736868 ,  0.24832288],
       [ 0.47085262,  0.        ,  0.46904916,  0.        ,  0.        ],
       [ 0.08381359,  0.        ,  0.        ,  0.        ,  0.        ]])

Here is my benchmark:

import numpy as np
x = np.random.random((5000, 5000)) - 0.5
print("max method:")
%timeit -n10 np.maximum(x, 0)
print("max inplace method:")
%timeit -n10 np.maximum(x, 0,x)
print("multiplication method:")
%timeit -n10 x * (x > 0)
print("abs method:")
%timeit -n10 (abs(x) + x) / 2
print("fancy index:")
%timeit -n10 x[x<0] =0

max method:
241 ms ± 3.53 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
max inplace method:
38.5 ms ± 4 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
multiplication method:
162 ms ± 3.1 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
abs method:
181 ms ± 4.18 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
fancy index:
20.3 ms ± 272 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Richard Möhn's comparison is not fair.
As Andrea Di Biagio's comment, the in-place method np.maximum(x, 0, x) will modify x at the first loop.
So here is my benchmark:

import numpy as np

def baseline():
    x = np.random.random((5000, 5000)) - 0.5
    return x

def relu_mul():
    x = np.random.random((5000, 5000)) - 0.5
    out = x * (x > 0)
    return out

def relu_max():
    x = np.random.random((5000, 5000)) - 0.5
    out = np.maximum(x, 0)
    return out

def relu_max_inplace():
    x = np.random.random((5000, 5000)) - 0.5
    np.maximum(x, 0, x)
    return x 

Timing it:

print("baseline:")
%timeit -n10 baseline()
print("multiplication method:")
%timeit -n10 relu_mul()
print("max method:")
%timeit -n10 relu_max()
print("max inplace method:")
%timeit -n10 relu_max_inplace()

Get the results:

baseline:
10 loops, best of 3: 425 ms per loop
multiplication method:
10 loops, best of 3: 596 ms per loop
max method:
10 loops, best of 3: 682 ms per loop
max inplace method:
10 loops, best of 3: 602 ms per loop

In-place maximum method is only a bit faster than the maximum method, and it may because it omits the variable assignment for 'out'. And it's still slower than the multiplication method.
And since you're implementing the ReLU func. You may have to save the 'x' for backprop through relu. E.g.:

def relu_backward(dout, cache):
    x = cache
    dx = np.where(x > 0, dout, 0)
    return dx

So i recommend you to use multiplication method.

If we have 3 parameters (t0, a0, a1) for Relu, that is we want to implement

if x > t0:
    x = x * a1
else:
    x = x * a0

We can use the following code:

X = X * (X > t0) * a1 +  X * (X < t0) * a0

X there is a matrix.

numpy didn't have the function of relu, but you define it by yourself as follow:

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

for example:

arr = np.array([[-1,2,3],[1,2,3]])

ret = relu(arr)
print(ret) # print [[0 2 3] [1 2 3]]

ReLU(x) also is equal to (x+abs(x))/2