finding repetition numbers in array.

A=[1;1;1;2;2;2;2;3;3;3]; %double
%I wanna known how many times 1,2,3 are exist in A matrix with orderly like that;

rep=[3;4;3]; %w.r.t A matrix (3 times 1, 4 times 2 and 3 times 3)

[Yao Li]

I'm not sure I've understood your question. My current understanding is you have a matrix A, and wanna calculate the array rep. Am I right?

[sermet]

I wanna determine the repetition times of each number in A matrix for example 1 repeated 3 times, 2 repeated 4 times and so on.

[Jos (10584)]

Your question title (finding repetition numbers) and your question text ("how many times exist") are open for ambiguity.

For A = [1 1 4 1 1 1] should the algorithm return [5 1], [5 0 0 1] or [2 1 3]?

[Yao Li]

He accepted Neuroscientist's answer below. Seems [5,1] is the correct answer.

[Jos (10584)]

Sure Yao, but the ambiguity remains ...

Answer 1

[Neuroscientist]

you can have something like this:

A=[1;1;1;2;2;2;2;3;3;3];

B = unique(A); % which will give you the unique elements of A in array B

Ncount = histc(A, B); % this willgive the number of occurences of each unique element

best NS

[shubham gupta]

simple and clear explaination. thank you sir, now i am able to solve my problem.

Answer 2

[Jos (10584)]

Your question title (finding repetition numbers) and your question text ("how many times exist") are open for ambiguity.

For A = [1 1 4 1 1 1] should the algorithm return [5 1], [5 0 0 1] or [2 1 3]?

A = [1 1 4 1 1 1]

% [5 1] case
R = histc(A,unique(A))
% [5 0 0 1] case
R = histc(A,1:max(A))
% [2 1 3] case
N = diff([0 find(diff(A)) numel(A)])

[omran alshalabi]

hi, thank you for your detailed answer,

I have another question, can I get some case like,

% [1 4 1] case

[MarKf]

removing duplicates then

A = [1 1 4 1 1 1];
b = A([true, diff(A)~=0])

b = 1×3

1 4 1