CODEWITHSUNDEEP
numpyactivation-functionrelu
Liu Hantao
Liu Hantao

Reputation: 636

How do I implement leaky relu using Numpy functions

I am trying to implement leaky Relu, the problem is I have to do 4 for loops for a 4 dimensional array of input.

Is there a way that I can do leaky relu only using Numpy functions?

Upvotes: 8

Views: 15910

Answers (5)

boblareleve
boblareleve

Reputation: 11

Just to add, this approche :

def leaky_relu(x):
    return np.maximum(0.01*x, x)

is 2 time faster than this one :

leaky_way1 = np.where(x > 0, x, x * 0.01)

Upvotes: 1

user17260419
user17260419

Reputation: 1

def leaky_relu_forward(x, alpha):
  out = x                                                
  out[out <= 0]=out[out <= 0]* alpha
  return out

Upvotes: 0

vikanksh nath
vikanksh nath

Reputation: 419

import numpy as np




def leaky_relu(arr):
    alpha = 0.1
    
    return np.maximum(alpha*arr, arr)

Upvotes: 2

Amir
Amir

Reputation: 16587

Here are two approaches to implement leaky_relu:

import numpy as np                                                 

x = np.random.normal(size=[1, 5])

# first approach                           
leaky_way1 = np.where(x > 0, x, x * 0.01)                          

# second approach                                                                   
y1 = ((x > 0) * x)                                                 
y2 = ((x <= 0) * x * 0.01)                                         
leaky_way2 = y1 + y2

Upvotes: 13

Lzkatz
Lzkatz

Reputation: 173

Going off the wikipedia entry for leaky relu, should be able to do this with a simple masking function.

output = np.where(arr > 0, arr, arr * 0.01)

Anywhere you are above 0, you keep the value, everywhere else, you replace it with arr * 0.01.

Upvotes: 1

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