# How can I count the occurrences of each element in a vector in MATLAB?

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MathWorks Support Team on 11 Jan 2012
Commented: Walter Roberson on 19 May 2020 at 17:08
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
---
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
---
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.

Show 5 older comments
ABDALHADI ABU ZEYNEH on 16 May 2020 at 15:10
how to calculate the occurence of all possibilities in which a comes before b
for example the occurence of a a a b b =6
also abbaab=5
Walter Roberson on 19 May 2020 at 16:38
Why is abbab 5 ?
If we number the positions 1:5 then you have (1,2), (1,3), (1,5), (4,5) which is 4 combinations.
Walter Roberson on 19 May 2020 at 17:08
S = 'aaabb'
nnz(triu((S'=='a') & S=='b'))

### More Answers (8)

Andrei Bobrov on 14 Aug 2014
[a,b] = histc(x,unique(x));
y = a(b);

Show 2 older comments
Thanks Andrei, that works!
SANA on 27 Apr 2018
I want to count the different number of elements in an array, whether unique or repeating but different, for example I have an array A = [1 2 2 3 1 2 3 3 3 4 1 5 7 9 9 2 8] I want answer to be; ans = 8 How to do this plz help
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
There is a simpler way of answering this now using groupcounts(R2019a) or grouptransform(R2018b):
>> 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

Show 4 older comments
Razvan Carbunescu on 6 Jun 2019
For the example you gave above how does the solution look and what does 'similar number' for the first column mean?
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
This seems like a very different type of problem so unlikely the functions in this topic will help you directly. I'd post this question as a separate thread with the example input/output

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)

#### 1 Comment

Johannes Korsawe on 1 Jun 2017
the second solution exhibits pathological tendencies...

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

#### 1 Comment

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))

#### 1 Comment

Walter Roberson on 16 May 2015
When using hist() pay attention to Dan's comment above pointing out a flaw in the approach. This flaw is not shared by Andrei's histc approach above.

lamghari on 30 Nov 2015
Hi, I want to count the number of followed occurrences of each element in a vector.
So if my input is
x = [1 1 1 2 2 1 1 2 5 5]
I need an output
y = [1 2 1 2 5;3 2 2 1 2] How do I do this?

Andrei Bobrov on 30 Nov 2015
y = [x(t); diff([find(t),numel(x)+1])]
Jan on 29 Jan 2017
[B, N] = RunLength(X);
Y = [B; N]

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'.