# Find group of points in an array of points

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Ndilokelwa Luis on 26 May 2015
Answered: Walter Roberson on 26 May 2015
I have a array x(i,j) where randomly x(i,j) = 1 or x(i,j) = 0. I need to find groups of x(i,j) adjacent to each other with x(i,j) = 1. Please help.

Ndilokelwa Luis on 26 May 2015
[L, num] = bwlabel(BW, n) solved the problem. Thank you!

Joseph Cheng on 26 May 2015
Do you have access to the image processing toolbox? if so the function bwconncomp() maybe what you're looking for
Joseph Cheng on 26 May 2015
While your response isn't an answer detecting 1's isn't the hard part. Do you have access to the image processing toolbox. bwconncomp() will detect and group adjacent 1's for you. If you do not post what you have created and we can figure out how to stop the infinite loop.

Ndilokelwa Luis on 26 May 2015
Hi Joseph. Actually I found the way to identify x(i,j) = 1. The problem is grouping adjacent x(i,j) = 1. I enter in an infinite loop and can't get out of that.

Walter Roberson on 26 May 2015
B = [A, zeros(size(A,1),1)]';
[Br,Bc] = ind2sub(size(B), strfind(B(:)', [1 1]));
Arc = [Bc(:), Br(:)];
Arc will now be an N x 2 array of row and column indices into A where a [1 1] pair was found. If there are "runs" then there will be one row in Arc for each "left" pixel in the run, and those entries will occur one after another in Arc, so if A(8,17) is the start of a [1 1 1] run, then there will be a row in Arc for [8, 17] and the next row in Arc would be [8, 18].