How to implement the ReLU function in Numpy

Question

I want to make a simple neural network which uses the ReLU function. Can someone give me a clue of how can I implement the function using numpy.

Answer 1

There are a couple of ways.

>>> x = np.random.random((3, 2)) - 0.5
>>> x
array([[-0.00590765,  0.18932873],
       [-0.32396051,  0.25586596],
       [ 0.22358098,  0.02217555]])
>>> np.maximum(x, 0)
array([[ 0.        ,  0.18932873],
       [ 0.        ,  0.25586596],
       [ 0.22358098,  0.02217555]])
>>> x * (x > 0)
array([[-0.        ,  0.18932873],
       [-0.        ,  0.25586596],
       [ 0.22358098,  0.02217555]])
>>> (abs(x) + x) / 2
array([[ 0.        ,  0.18932873],
       [ 0.        ,  0.25586596],
       [ 0.22358098,  0.02217555]])

If timing the results with the following code:

import numpy as np

x = np.random.random((5000, 5000)) - 0.5
print("max method:")
%timeit -n10 np.maximum(x, 0)

print("multiplication method:")
%timeit -n10 x * (x > 0)

print("abs method:")
%timeit -n10 (abs(x) + x) / 2

We get:

max method:
10 loops, best of 3: 239 ms per loop
multiplication method:
10 loops, best of 3: 145 ms per loop
abs method:
10 loops, best of 3: 288 ms per loop

So the multiplication seems to be the fastest.

Answer 2

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.

Answer 3

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

Answer 4

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]]

Answer 5

This is more precise implementation:

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

Answer 6

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)