Counting frequency of occurrence in matrix

Good day,

I am attempting to count the number of times each number in a matrix occurred in the matrix.

For example, suppose I have a matrix;

x =

22 23 24 23

24 23 24 22

22 23 23 23

I want an output which will tell me 22 occurred 3 times, 23 occurred 6 times, and 24 occurred 2 times. The actual matrix is larger in size.

Is there a specific function which returns such values or are there any other ways I can resolve this challenge?

[Z D]

you can use this command'tabulate'

Syntax

tbl = tabulate(x)
tabulate(x)

[Jerry Olup]

Let x be your vector. Something like this can save you an unnecessary toolbox:

xx = unique(x);       % temp vector of vals
x = sort(x);          % sorted input aligns with temp (lowest to highest)
t = zeros(size(xx)); % vector for freqs
% frequency for each value
for i = 1:length(xx)
    t(i) = sum(x == xx(i));
end

Then the t vector shows freqs of the sorted, unique vector x.

[m_vdv]

Hi, is it also possible to do this without using 'unique' and 'sort'? For example with only using a for loop and/or if loop? Thanks in advance.

Answer 1

[Andrei Bobrov]

x =[
22 23 24 23
24 23 24 22
22 23 23 23];

a = unique(x);
out = [a,histc(x(:),a)];

[Guan Zhao]

This would work. Thanks alot!

[suman mandal]

How to calculate frequency of occurrence for all the element greater than a number e.g what would be the total number of occurrence for all the element greater than equal to 23??

[Réda Abadi]

thx :)

[Darshan Patel]

Awesome

[Valeria Yuraí Ramírez Guatemala Sánchez]

Perfect!

[Sattwama Basu]

can you also show the solution in c?

[Lavanya T]

Thank you

[Brian Derstine]

would be nice if you could just do tabulate(x)

Answer 2

[Aurelien Queffurust]

Using nnz for example:

   nnz(x==22)

will return 3

[Guan Zhao]

Hi,

My actual matrix will have at least 256 elements. I will have to consider zero elements too.

Regards

Guan Zhao

Answer 3

[Thomas]

x=[22 23 24 23

24 23 24 22

22 23 23 23];

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

out=[b' sum((a),2)]

Answer 4

[abdelrahim hashem]

x = [22 23 24 23; 24 23 24 22; 22 23 23 23];

un_x = unique(x);

for i = 1:length(un)

un(i), length(find(x == un_x(i)))

end