Count the number of same elements in an array

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

[~,~,ix] = unique(V);
C = accumarray(ix,1).'

5 Comments

may you explain your answer?
there's a precision that I have to put.
If there is no 1 element like
V = [2 2 3 4 5 6 7 7 8 8 9 9]
I would like to obtain
C= [0 2 1 1 1 1 2 2 2]
Where the entities 1 is not present so the first element is 0.
Do you know how to do it?
C=accumarray(V(:),1).'
%or
C=sum(V(:)==(1:max(V)))
As per your latter comment Stephen’s approach below will do what you want.
@madhan ravi : this is really quite neat:
accumarray(V(:),1)
Simple idea which works well:
>> V = [2 2 3 4 5 6 7 7 8 8 9 9]
>> accumarray(V(:),1)
ans =
0
2
1
1
1
1
2
2
2
Thank you Stephen :) !

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

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Release

R2019a

Asked:

on 10 Sep 2019

Edited:

on 14 Aug 2021

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