How to return a true/false logical array from a string array of repeating numbers?

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Liv
Liv 2022 年 7 月 7 日
編集済み: Jon 2022 年 7 月 7 日
ans = 3 2 1 5 1 4 0
I want this to return a 7x1 logical array 0 0 1 0 1 0 0. Corresponding to the repeating “1” in the ans variable. Or 1 1 0 1 0 1 1. Whichever is easier to program. How do I do this?

採用された回答

Rohit Kulkarni
Rohit Kulkarni 2022 年 7 月 7 日
I think this may work:
A = [3 3 2 1 5 1 0 4];
[uniqueA i j] = unique(A,'first');
idRep = find(not(ismember(1:numel(A),i)))
idRep = 1×2
2 6
rep_var = A(idRep)
rep_var = 1×2
3 1
ll = ismember(A,rep_var)
ll = 1×8 logical array
1 1 0 1 0 1 0 0
  1 件のコメント
Liv
Liv 2022 年 7 月 7 日
Yes, this does exactly what I’m looking for! Thank you!!

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その他の回答 (1 件)

Jon
Jon 2022 年 7 月 7 日
x = [3 2 1 5 1 4 0]
L = x == 1
  5 件のコメント
Liv
Liv 2022 年 7 月 7 日
No, I seem to be getting an array bounds error. I’m sure if I went through the troubleshooting process that it would work, but for now I will be using the other user’s answer. If I ever go back to it again then I will update this thread.
Jon
Jon 2022 年 7 月 7 日
編集済み: Jon 2022 年 7 月 7 日
That's fine as long as you have a solution, but I'm puzzled, as to why you would have array bounds errors, when as you can see it ran without issues in the small example I show above. As you had an array bounds error, are you sure you used :
[N,edges,bin]= histcounts(x,[u,u(end)+1])
and not:
[N,edges,bin]= histcounts(x,[u,u(end+1)])

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