# Count the number of same elements in an array

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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..

madhan ravi on 10 Sep 2019
Edited: madhan ravi on 10 Sep 2019
[~,~,ix] = unique(V);
C = accumarray(ix,1).'
madhan ravi on 10 Sep 2019
Thank you Stephen :) !

Stephen23 on 10 Sep 2019
Edited: Stephen23 on 10 Sep 2019
>> 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
>> 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
luca on 10 Sep 2019
thanks Stephen

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

R2019a

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