Count the number of same elements in an array

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luca
luca on 10 Sep 2019
Edited: Witold Stepien on 14 Aug 2021
Hi given a vector
V = [ 1 2 4 3 4 2 3 5 6 4 5 6 8 4 2 3 5 7 8 5 3 1 3 5 7 8 9 5 3 2 4 6 7 8]
I would like to count how many times the value 1,2,3,4,5,6,7,8,9 are repeated inside V, and obtain a vector that report this values:
C = [2 4 6 5 6 3 3 4 1]
where 1 is repeated 2 times, 2 is repetead 4 times, 3 is repeated 6 times and so on..

Accepted Answer

madhan ravi
madhan ravi on 10 Sep 2019
Edited: madhan ravi on 10 Sep 2019
[~,~,ix] = unique(V);
C = accumarray(ix,1).'
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More Answers (3)

Stephen23
Stephen23 on 10 Sep 2019
Edited: Stephen23 on 10 Sep 2019
Your 1st example:
>> V = [ 1 2 4 3 4 2 3 5 6 4 5 6 8 4 2 3 5 7 8 5 3 1 3 5 7 8 9 5 3 2 4 6 7 8];
>> C = hist(V,1:max(V))
C =
2 4 6 5 6 3 3 4 1
Your 2nd example:
>> V = [2 2 3 4 5 6 7 7 8 8 9 9]
>> C = hist(V,1:max(V))
C =
0 2 1 1 1 1 2 2 2

Witold Stepien
Witold Stepien on 14 Aug 2021
Edited: Witold Stepien on 14 Aug 2021
I found this function extremely useful, and doing exactly what you need:
V = [ 1 2 4 3 4 2 3 5 6 4 5 6 8 4 2 3 5 7 8 5 3 1 3 5 7 8 9 5 3 2 4 6 7 8];
[gc,grps] = groupcounts(V'); % <- need column vector here
grps'
ans = 1×9
1 2 3 4 5 6 7 8 9
gc'
ans = 1×9
2 4 6 5 6 3 3 4 1
Where grps lists the unique values in order, and gc provides the count of each unique values found in v.
This is very similar to madhan ravi's accumarray, but even simpler.
P.S. I turned gc and grps into row vectors only for compactness of the post, it's purely aesthetical. However groupcounts requires a column vector, not a row.

Hugo Diaz
Hugo Diaz on 28 Nov 2020
I use sparse(V(:),V(:), 1) for large arrays with missing indices.

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