# Detect parts of a 0/1 (binary) vector consisting of only 0 or 1

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Ilya on 3 Apr 2014
Edited: Ilya on 6 Apr 2014
Let's say I have a vector of 0 and 1:
a=[1 1 1 0 0 0 1 0 1 0 0 1 1 0];
It's needed to find a way to extract the vector's indices so that each group of extracted indices contains a part of a vector consisting of only zeros or only ones. For the given example it should be like:
[1 3] [4 6] [7] [8] [9] [10 11] [12 13] [14]
I thought about diff(), but some post-processing is definitely required after it...

Sean de Wolski on 3 Apr 2014
idx = find(diff([~a(1) a]))';
pcs = spdiags(rot90([idx-1 idx]));
pcs(end) = numel(a)

Ilya on 3 Apr 2014
this is great!! Thanks!
Image Analyst on 3 Apr 2014
Wow - never would have guessed that.
Ilya on 6 Apr 2014
A more robust solution (the same idea):
idx1 = find(diff([~a(1) a]));
idx2 = circshift(idx1-1,[0 -1]);
idx2(end) = numel(a);
pcs = [idx1; idx2];
The initial one may be wrong i.e. in case a=[1 1 1 1 0], due to the unpredictable spdiags() behavior.

Image Analyst on 3 Apr 2014
If you have the Image Processing Toolbox, the 1's are given by the PixelIdxList property. You can do it in one line with a call to regionprops:
a=[1 1 1 0 0 0 1 0 1 0 0 1 1 0];
% Find 1's.
locations = regionprops(logical(a), 'PixelIdxList');
% Print out to command line:
for k = 1 : length(locations)
locations(k).PixelIdxList
end
% Find 0's.
locations = regionprops(logical(~a), 'PixelIdxList');
% Print out to command line:
for k = 1 : length(locations)
locations(k).PixelIdxList
end

#### 1 Comment

Salaheddin Hosseinzadeh on 3 Apr 2014
Interesting to see such different perspective!