CODEWITHSUNDEEP
pythonimage-processingmatrixdisjoint-sets
Vahid S. Bokharaie
Vahid S. Bokharaie

Reputation: 997

Find all disjoint subsets of a binray matrix in Pyhton

I have a binary matrix, and I want to find all disjoint subsets that exist in this matrix. To clarify the problem, the matrix is a collection of image masks, masks of irregular shapes, and each disjoint subset of 1s representing a separate mask. In other words, if I have a collection of N disjoint subsets in a matrix of size d1xd2, I want to end up with N matrices of size d1xd2, each having only one of the disjoint subsets of 1s. Any help on that is highly appreciated.

Upvotes: 0

Views: 116

Answers (1)

mimocha
mimocha

Reputation: 1121

In image processing, what you are trying to do is also called Connected Component Labelling (CCL). Woodford's suggestion of Flood fill algorithm is one algorithm for performing CCL.

A variation of your question is already answered in this other thread, which uses the function scipy.ndimage.label.
Here's a slight modification of that answer, which should give you exactly what you are looking for:

import numpy as np
from scipy.ndimage import label

# Define your matrix here
X = 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, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 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, 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, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 1, 1, 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, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 1, 1, 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, 1, 0, 0, 0, 0, 0, 0, 0],
    [0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 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, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
], dtype=int)

# How do you define "connected" or "disjointed"? 4-neighbors vs 8-neighbors.
# https://en.wikipedia.org/wiki/Connected-component_labeling
structure = np.ones((3, 3), dtype=int)

# Perform connected component labelling
labeled, ncomponents = label(X, structure)

# Iterate through each labelled blob
for ii in range(1, ncomponents+1):
    output = np.zeros(X.shape)
    output[labeled == ii] = 1

    # Do something to each resulting blob...
    print(output)

Upvotes: 1

Related Questions