numpy.where() returns empty array instead of replacing

Question

I have RGB images as numpy arrays and I want to replace to certain pixel value:

import numpy as np
import random

def rgb():
    L = [1]*3
    return [random.randrange(0, 256,1)for l in L]

def makeDummyData():
    #creates list of lists with 3 rgb, which contains values to replace (x)
    L = [1]*10000
    L = [rgb() for l in L]
    x = L[0]
    for i in range(500):
        L[i] = x
    return L, x

L, x = makeDummyData()
a = np.array(L).reshape(-1, 3)
a = np.array(a).reshape(100, 100, 3)
b = np.where(a == (np.array(x), [255,255,255], a))
print(b)

Instead of returning an array, where the target values are replaced, the output is (array([], dtype=int64),)

I think your parenthesis are little off.

Sayandip Dutta
– Sayandip Dutta

2020-01-28 16:04:13 +00:00
Commented Jan 28, 2020 at 16:04 — Sayandip Dutta
– Sayandip Dutta, Commented Jan 28, 2020 at 16:04

Gaurav Agarwal Garg · Accepted Answer · 2020-01-28 16:39:22Z

0

np.where

Bracket seems to be off

b = np.where(a == np.array(x), [255,255,255], a) # After == an extra '(' is used

answered Jan 28, 2020 at 16:39

Gaurav Agarwal Garg

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