Count the number of same elements in an array

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luca
luca 2019-9-10
编辑： Vitek Stepien 2021-8-14
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..

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采纳的回答

madhan ravi
madhan ravi 2019-9-10
编辑：madhan ravi 2019-9-10
[~,~,ix] = unique(V);
C = accumarray(ix,1).'
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Stephen23
Stephen23 2019-9-10
编辑：Stephen23 2019-9-10
@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

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更多回答（3 个）

Stephen23
Stephen23 2019-9-10
编辑：Stephen23 2019-9-10
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
Vitek Stepien
Vitek Stepien 2021-8-14
编辑：Vitek Stepien 2021-8-14
Ran in:
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 2020-11-28
I use sparse(V(:),V(:), 1) for large arrays with missing indices.

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