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How to count the number of consecutive numbers of the same value in an array

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I have an array given as
x = [1 1 1 2 2 1 1 1]
I'd like to know a way I could go through each individual element in the array, and getting a value for how many steps the number stays at the same value. For instance, for this example, the output I would be looking for would be
y = [2 1 0 1 0 2 1 0]
Where the first value of 1 stays constant for another 2 steps, the second stays constant for one more step etc.

Accepted Answer

Jos (10584)
Jos (10584) on 24 Feb 2014
% data
x = [1 1 1 2 2 1 1 1 3 3 3 3 3 5]
% engine
i = find(diff(x))
n = [i numel(x)] - [0 i]
c = arrayfun(@(X) X-1:-1:0, n , 'un',0)
y = cat(2,c{:})

More Answers (2)

Andrei Bobrov
Andrei Bobrov on 25 Feb 2014
c = [1 1 1 2 2 1 1 1];
v = numel(c):-1:1;
ii = [true,diff(c)~=0];
n = v(ii);
t = [n(2:end)+1,1];
out = v - t(cumsum(ii));

Roger Stafford
Roger Stafford on 24 Feb 2014
Here's a slightly different way:
x = [2 2 5 5 5 6 6 6 6 4 7 2 2 2];
n = size(x,2);
f = find([true,diff(x)~=0,true]);
y = zeros(1,n);
y(f(1:end-1)) = diff(f);
y = cumsum(y(1:n))-(1:n);
  1 Comment
Gareth Pritchard
Gareth Pritchard on 25 Feb 2014
This also works great, thank you. Is there any way you could explain line by line what it is doing at all?

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