# How to implement the ReLU function in Numpy

Hello Guys, How are you all? Hope You all Are Fine. Today We Are Going To learn about How to implement the ReLU function in Numpy in Python. So Here I am Explain to you all the possible Methods here.

## How to implement the ReLU function in Numpy?

1. 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.

2. implement the ReLU function in Numpy

I'm completely revising my original answer because of points raised in the other questions and comments.

## Method 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.

## Method 2

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

## Conclusion

It’s all About this issue. Hope all Methods helped you a lot. Comment below Your thoughts and your queries. Also, Comment below which Method worked for you? Thank You.