fastest way too fill a rows that contain all zeros

Question

what is the fastest way to fill every row in 2d array that contains all zeros with different value.

Only the rows that all contain Zero..

the only idea I have is a loop and using np.all() to check every row

array([[2, 6, 9, 7, 0],
       [0, 0, 0, 0, 0],
       [3, 8, 5, 4, 7]])

array([[2, 6, 9, 7, 0],
       [1, 1, 1, 1, 1],
       [3, 8, 5, 4, 7]])

Asked By: sten

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Source

Answer 1

import numpy as np

a = np.array([[2, 6, 9, 7, 0],
       [0, 0, 0, 0, 0],
       [3, 8, 5, 4, 7]])

rowi_0 = (a==0).all(axis=1)
 
for idx, val in enumerate(rowi_0):
    if val == True:
        a[idx] = np.random.randint(10, size=a.shape[1])

print(a)

I think this is the easiest and fast way, rewriting migth take little bit of time tho, but loops work fine for this scenario

Answered By: Ogabek

Answer 2

Well, if this is sufficient:

a = np.array([[2, 6, 9, 7, 0],
              [0, 0, 0, 0, 0],
              [3, 8, 5, 4, 7]])

a[~a.any(axis=1)] = 1

Slight note: I use ~np.any(...) instead of np.all(a==0) because the a array is larger than the boolean array returned by all or any. The a==0 creates a 2D boolean array. Negating the 1D boolean array created by any requires less work and memory.

Answered By: Homer512

fastest way too fill a rows that contain all zeros

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