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

Question

Gareth Pritchard 2014-2-24

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https://ww2.mathworks.cn/matlabcentral/answers/118828-how-to-count-the-number-of-consecutive-numbers-of-the-same-value-in-an-array

评论： Emil Jensen 2020-3-10

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.

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Answer 1

Jos (10584) 2014-2-24

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https://ww2.mathworks.cn/matlabcentral/answers/118828-how-to-count-the-number-of-consecutive-numbers-of-the-same-value-in-an-array#answer_125900

在 MATLAB Online 中打开

% 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{:})

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Gareth Pritchard 2014-2-24

Brilliant, thanks a lot!

Emil Jensen 2020-3-10

Briliant, tanjs r rot!

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Answer 2

Andrei Bobrov 2014-2-25

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在 MATLAB Online 中打开

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));

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Answer 3

Roger Stafford 2014-2-24

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在 MATLAB Online 中打开

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);