Asked by Giorgos Papakonstantinou
on 21 Jun 2013

I have a vector:

a=[0 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 1 1 1 0 1 ];

in which I would like to count the "areas" of zeros and the "areas" of ones. In the example above I have

4 areas of zeros and 4 areas of ones. Thank you!

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Answer by Andrei Bobrov
on 21 Jun 2013

Accepted answer

a=[0 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 1 1 1 0 1 ];

lc = [true;diff(a(:))~=0];

x = a(lc);

zerosareas = sum(~x); onesareas = sum(x);

Giorgos Papakonstantinou
on 21 Jun 2013

Thank you Andrei

Answer by Jan Simon
on 21 Jun 2013

Another approach which might be faster for large data sets: FEX: RunLength:

a = [0 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 1 1 1 0 1 ]; b = RunLength(a); nZero = sum(b == 0); nOne = length(b) - nZero; % Or: sum(b == 1)

Giorgos Papakonstantinou
on 22 Jun 2013

Thank you Jan!

Jan Simon
on 22 Jun 2013

For 100'000 elements, the RunLength approach is twice as fast as the DIFF method. But as long as both methods need only some milliseconds, this might not really matter.

Answer by Image Analyst
on 23 Jun 2013

I know you already accepted an answer, but if you have the Image Processing Toolbox there is a straightforward one-liner solution to get either of those numbers:

a=[0 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 1 1 1 0 1]; [~, numberOf1Regions] = bwlabel(a) % Get # of "1" regions. [~, numberOf0Regions] = bwlabel(~a) % Get # of "0" regions.

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