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

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...

Answer 1

[Sean de Wolski]

idx = find(diff([~a(1) a]))';
pcs = spdiags(rot90([idx-1 idx]));
pcs(end) = numel(a)

[Ilya]

this is great!! Thanks!

[Image Analyst]

Wow - never would have guessed that.

[Ilya]

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.

Answer 2

[Image Analyst]

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

[Salaheddin Hosseinzadeh]

Interesting to see such different perspective!