Vectorized alterlative to identify duplicate elements in an array

Hello everyone,

I have a 1xN array, A, of random integers. Repetitions are permitted. Let's consider only the integers 1:5 to make things simple:

A = [1, 5, 3, 4, 4, 2, 3, 1, 4, 5, 3, 2, 3, 4, 2, 3];

I generate a second 1xN array, B, which represents an arbitrary subset of unique(A):

B = [1, 3, 4];

For each element in B, I would like to find all of the corresponding indexes in A. So far I am using a simple FOR loop to achieve this:

indexes = [];
for i = 1:length(B)
    indexes = [indexes, find(A == B(i))];
end

I would like to write an alternative which does not require a FOR loop. My problem is that for each element in B, I want all of the corresponding indexes in A. Functions like ISMEMBER and INTERSECT do not include repetitions.

I don't need to distinguish between the elements in B, so the output can be a single 1xN vector containing all of the indexes in A. For the above example, the output would be:

indexes = [1, 8, 3, 7, 11, 13, 16, 4, 5, 9, 14];

The final order of INDEXES is arbitrary. It can be sorted or random.

Thanks!

EDIT: I found the solution:

[LIA, ~] = ismember(A, B);
indexes = find(A(LIA));

Answer 1

[fsgeek]

[LIA, ~] = ismember(A, B);

indexes = find(LIA);

[Stephen23]

Simpler on one line:

>> indices = find(ismember(A,B))
indices =
    1    3    4    5    7    8    9   11   13   14   16

[fsgeek]

Looks good, thanks for the tip!

Answer 2

[Jakob B. Nielsen]

Logical indexing for every instance, then convert the logical output to doubles and sum it (since you cant sum logicals), and find the indexes where there is a 1, at a glance. It wont give the same order of indexes as in your example but I guess that doesnt matter too much if you dont need to distinguish the elements in B.

logic=(A==B(:));
logic=sum(double(logic));
find(logic==1)