Find center of blocks of ones in an 2d array

Question

Lets assume I have the following array

import numpy as np

a = np.array([[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 1. 1. 1. 0. 0. 0.]
 [0. 0. 0. 0. 1. 1. 1. 0. 0. 0.]
 [0. 0. 0. 0. 1. 1. 1. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 1. 1. 1. 0. 0. 0. 0. 0.]
 [0. 0. 1. 1. 1. 0. 0. 0. 0. 0.]
 [0. 0. 1. 1. 1. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]])

and want to get the center of each block of ones, so that I can get the following array:

a_centers = np.array([[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 1. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 1. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]])

I already thought about using means or convolutions, but I couldn't think of a really simple solution.

One in efficient solution that I found is:

a_centers = np.zeros(a.shape)

for i in range(len(a)):
    for j in range(len(a[i])):
        if np.sum(a[i - 1:i + 2, j - 1:j + 2]) == 9:
            a_centers[i, j] = 1

@ScottHunter I added a solution that I found but that seems to be inefficient. — DΦC__WTF, Nov 18 '21 at 13:20
Please read https://stackoverflow.com/questions/70012889/what-is-a-faster-way-to-get-the-center-points-of-objects-represented-in-a-2d-n . This is strangely the 3rd question posted since few hours on this (very?) specific topic. — Jérôme Richard, Nov 18 '21 at 13:26
I want to get rid of the two for loops and iterating over all values — DΦC__WTF, Nov 18 '21 at 14:49

diggusbickus · Answer 1 · 2021-11-18T15:00:40.420

1

if you're sure your blocks are square and don't share cells, then you can skip all the sum stuff. and i do hope that np.where is faster than for loops

x,y=np.where(a==1)
blocks=list(zip(x,y))
for _ in range(x.size//9):
    left_x,left_y=blocks[0]
    a_centers[left_x+1, left_y+1]=1
    for i in range(3):
        for j in range(3):
            blocks.remove((left_x+i, left_y+j))

edited Nov 18 '21 at 15:00

answered Nov 18 '21 at 14:55

diggusbickus

1,537
3
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score 0 · Accepted Answer · answered Nov 18 '21 at 13:47

0

This is a little less inefficient, since no edge cell can be at the center of a 3x3 block:

for i in range(1,len(a)-1):
    for j in range(1,len(a[i])-1):
        if np.sum(a[i - 1:i + 2, j - 1:j + 2]) == 9:
            a_centers[i, j] = 1

answered Nov 18 '21 at 13:47

Scott Hunter

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101

Find center of blocks of ones in an 2d array

2 Answers2