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

1.400 visualizaciones (últimos 30 días)
MathWorks Support Team el 11 de En. de 2012
Comentada: Walter Roberson el 26 de Abr. de 2022
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 el 27 de Feb. de 2020
Editada: MathWorks Support Team el 27 de Feb. de 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.
##### 3 comentariosMostrar 1 comentario más antiguoOcultar 1 comentario más antiguo
Walter Roberson el 19 de Mayo de 2020
S = 'aaabb'
nnz(triu((S'=='a') & S=='b'))
Walter Roberson el 26 de Abr. de 2022
https://www.mathworks.com/help/matlab/ref/double.groupcounts.html#mw_92fbcf5a-2ab5-45d0-ac09-68c1986c269f
when you use groupcounts then the groups are in the order returned by unique('sorted')

Iniciar sesión para comentar.

### Más respuestas (3)

Andrei Bobrov el 14 de Ag. de 2014
[a,b] = histc(x,unique(x));
y = a(b);
##### 1 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
Andrei Bobrov el 3 de Ag. de 2015
y = accumarray(x(:),1)

Iniciar sesión para comentar.

Razvan Carbunescu el 9 de Mayo de 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
##### 4 comentariosMostrar 2 comentarios más antiguosOcultar 2 comentarios más antiguos
Razvan Carbunescu el 6 de Jun. de 2019
For the example you gave above how does the solution look and what does 'similar number' for the first column mean?
Razvan Carbunescu el 7 de Jun. de 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

Iniciar sesión para comentar.

Julian Hapke el 1 de Jun. de 2017
Editada: Julian Hapke el 1 de Jun. de 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 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
Johannes Korsawe el 1 de Jun. de 2017
the second solution exhibits pathological tendencies...

Iniciar sesión para comentar.

### Categorías

Más información sobre Performance and Memory en Help Center y File Exchange.

R14SP1

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by