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Find group of points in an array of points

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Ndilokelwa Luis
Ndilokelwa Luis 2015-5-26
回答： Walter Roberson 2015-5-26
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.

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Ndilokelwa Luis
Ndilokelwa Luis 2015-5-26
[L, num] = bwlabel(BW, n) solved the problem. Thank you!

更多回答（3 个）

Joseph Cheng
Joseph Cheng 2015-5-26
Do you have access to the image processing toolbox? if so the function bwconncomp() maybe what you're looking for
http://www.mathworks.com/help/images/examples/correcting-nonuniform-illumination.html
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Joseph Cheng
Joseph Cheng 2015-5-26
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.

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Ndilokelwa Luis
Ndilokelwa Luis 2015-5-26
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
Walter Roberson 2015-5-26
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].

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