finding repetition numbers in array.

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sermet
sermet 2014-5-15
评论： MarKf 2022-10-22
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)
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Yao Li
Yao Li 2014-5-15
He accepted Neuroscientist's answer below. Seems [5,1] is the correct answer.
Jos (10584)
Jos (10584) 2014-5-15
Sure Yao, but the ambiguity remains ...

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采纳的回答

Neuroscientist
Neuroscientist 2014-5-15
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
  1 个评论
shubham gupta
shubham gupta 2019-2-26
simple and clear explaination. thank you sir, now i am able to solve my problem.

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更多回答（1 个）

Jos (10584)
Jos (10584) 2014-5-15
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)])
  2 个评论
omran alshalabi
omran alshalabi 2022-8-28
hi, thank you for your detailed answer,
I have another question, can I get some case like,
% [1 4 1] case
MarKf
MarKf 2022-10-22
Ran in:
removing duplicates then
A = [1 1 4 1 1 1];
b = A([true, diff(A)~=0])
b = 1×3
1 4 1

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