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

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?

Answer 1

[Rohit Kulkarni]

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

[Liv]

Yes, this does exactly what I’m looking for! Thank you!!

Answer 2

[Jon]

x =  [3 2 1 5 1 4 0]
L = x == 1

[Liv]

I should have specified that these numbers change depending on the output of other variables in my program. For example the ans variable could be “3 3 2 1 5 1 0 4” instead. Then I would want it to return “1 1 0 1 0 1 0 0”. I need a logical array that shows if there are repeats.

[Jon]

Sorry I misunderstood what you were looking for. I think this does what you want:

x = [3 3 2 1 5 1 0 4]

x = 1×8

3 3 2 1 5 1 0 4

u = unique(x)

u = 1×6

0 1 2 3 4 5

[N,edges,bin]= histcounts(x,[u,u(end)+1])

N = 1×6

1 2 1 2 1 1

edges = 1×7

0 1 2 3 4 5 6

bin = 1×8

4 4 3 2 6 2 1 5

N(bin)>1

ans = 1×8 logical array
   1   1   0   1   0   1   0   0

[Jon]

Does this also do what you want?

[Liv]

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]

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)])