A=[ 1 1 2 2 3 3 3];
unique(A)=[1 2 3]; but I want to find the duplicates that are not the first occurrence. i.e x=[2 4 6 7]; I typed help unique but I couldn't figure out if I and J reported by this function helps with my purpose.I know that I can program it but i want to be as efficient as possible in my codes to reduce the running time.

 採用された回答

the cyclist
the cyclist 2011 年 8 月 5 日

4 投票

Here is one way:
[uniqueA i j] = unique(A,'first');
indexToDupes = find(not(ismember(1:numel(A),i)))

その他の回答 (1 件)

Jan
Jan 2011 年 8 月 5 日

12 投票

Another solution:
A = [1 1 2 2 3 3 3];
[U, I] = unique(A, 'first');
x = 1:length(A);
x(I) = [];

2 件のコメント

Oleg Komarov
Oleg Komarov 2011 年 8 月 5 日
Clever and simple.
Jan
Jan 2011 年 8 月 5 日
I'm inspired by Marsaglia's KISS random number generator: "Keep It Simple Stupid".

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