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

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Gareth Pritchard on 24 Feb 2014
Commented: Emil Jensen on 10 Mar 2020
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.

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{:})
##### 2 CommentsShowHide 1 older comment
Emil Jensen on 10 Mar 2020
Briliant, tanjs r rot!

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 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);
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?