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I would like to be able to return the count of occurences of each element in a vector.

For example if I have a vector:

x=[10 25 4 10 9 4 4]

I expect the result to be

y=[2 1 3 2 1 3 3].

MathWorks Support Team
on 27 Feb 2020

Edited: MathWorks Support Team
on 27 Feb 2020

As of MATLAB R2019a, you can use the “groupcounts” function to compute the number of times an element appears in a vector as a summary. In other words, the elements of the below output “GC” are the counts of the corresponding element values in “GR” (from the original input vector “x”):

x = [10 25 4 10 9 4 4]';

[GC,GR] = groupcounts(x)

GC =

3

1

2

1

GR =

4

9

10

25

For more information on "groupcounts", please see the documentation link below:

---

As of MATLAB R2018b, you can use the “grouptransform” function if you want to compute the number of times an element appears in a vector and output that count for each corresponding element of the input vector. For example:

x = [10 25 4 10 9 4 4]';

y = grouptransform(x,x,@numel)

y =

2

1

3

2

1

3

3

For more information on "grouptransform", please see the documentation link below:

---

Prior to MATLAB R2018b, while there is no single function to count occurrences of each element, there are a few ways to count elements in a vector:

1. Logical Indexing:

The following code snippet will give the desired output:

y = zeros(size(x));

for i = 1:length(x)

y(i) = sum(x==x(i));

end

For MATLAB R2016b and later, you can use implicit expansion to further simplify the code:

y = sum(x==x')

2. Binning:

You can use the "hist" and "unique" functions as shown here to do the same:

x = [10 25 4 10 9 4 4]

[a,b]=hist(x,unique(x))

3. Third-Party Tools:

For another workaround, see the following file, 'CountMember.m', that was contributed by a MATLAB user to do the same from a single function:

Note that MathWorks does not guarantee or warrant the use or content of submissions to the MATLAB Central File Exchange. Any questions, issues, or complaints should be directed to the contributing author.

Andrei Bobrov
on 14 Aug 2014

[a,b] = histc(x,unique(x));

y = a(b);

SANA
on 27 Apr 2018

Tech Support
on 2 May 2018

Hi,

If you are still experiencing this issue, please contact MathWorks support:

-Justin

Razvan Carbunescu
on 9 May 2019

>> x=[10 25 4 10 9 4 4]';

>> grouptransform(x,x,@numel)

ans =

2

1

3

2

1

3

3

>>[GC,GR]=groupcounts(x)

GC =

3

1

2

1

GR =

4

9

10

25

Razvan Carbunescu
on 6 Jun 2019

Masoud Mirzaei
on 6 Jun 2019

I meant "the same number" in column 1. Here is the expected solution:

(22 , 33) 2

(22 , 44) 2

(33 , 44) 1

and the other pairs have the value of 0.

THe actual problem consists of > 1 million elements. So I am looking for a smaort solution rather than using some loops.

Razvan Carbunescu
on 7 Jun 2019

Julian Hapke
on 1 Jun 2017

Edited: Julian Hapke
on 1 Jun 2017

here is another one:

sum(bsxfun(@eq,x,x'),1)

or if you want the output to be the same orientation as input

sum(bsxfun(@eq,x,x'),(size(x,2)==1)+1)

Josh
on 14 Aug 2014

This is kind of awkward since it requires using the input array within the anonymous function, but:

y = arrayfun(@(t)nnz(x==t), x);

should do the trick, too.

Truong Phan
on 16 Aug 2017

I need help to count the occurrences of each element in a matrix. Thanks

Walter Roberson
on 16 Aug 2017

[uvals, ~, uidx] = unique(YourArray);

output = [uvals, accumarray(uidx, 1)];

This would produce an N x 2 array with the first column being the unique array values and the second column being the associated count.

mittal54
on 16 May 2015

Edited: Walter Roberson
on 16 Aug 2017

numbers=unique(v); %list of elements

count=hist(v,numbers); %provides a count of each element's occurrence

this will give counts. and if you want to have a nice graphical representation then try this

bar(accumarray(v', 1))

Walter Roberson
on 16 May 2015

Ramon Villamangca
on 25 Oct 2018

Edited: Ramon Villamangca
on 25 Oct 2018

This is how I did it:

numOccur = sum(arrayfun(@(x) x == elem,vec))

where 'elem', is the element to search in the given vector 'vec'.

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