Read cell via indexing multiple times, without using for-loop

Hello everyone!

I have a rather simple problem, but can't figure out a 'nice' solution for it, without using a for loop.

I have a cell array, which contains different options. And I have a Matrix, which tells me which of these options to use in different cases.

So I start with something like this:

Options = {'A' 'B' 'C'};
IndexMatrix = logical([
    1 0 0
    1 0 0
    0 0 1
    0 1 0]);

What I want to get is this outcome:

Choice = 

      'A'
      'A'
      'C'
      'B'

This is my (simple) solution for this problem:

Data = cell(size(IndexMatrix,1),1);
for idx = 1:size(IndexMatrix,1)
    Choice(idx) = Options(IndexMatrix(idx,:));
end

It works, but I'm curious if there might be a better solution, especially one without a for loop.

One idea was to use 'arrayfun', but I need to use every row of IndexMatrix on my cell, not every element.

Another idea would be to extend the cell to a second dimension, with the different options 'A' 'B' 'C' in every row, so you could do all the indexing at once. Somehow like this:

Choice = Options_extended(IndexMatrix)

But then again, I did not know how to extend the cell without using a for loop again. I was looking for something similar to what you would do with a vector. E.g.:

ones(4,1) * [1 2 3]

Anyway, I'm very interested, if anyone knows a 'clever' way to do this. Thanks!

Answer 1

[Andrei Bobrov]

C = {'A' 'B' 'C'};
Ind = logical([
    1 0 0
    1 0 0
    0 0 1
    0 1 0]);

ii = bsxfun(@times,Ind,1:numel(C));
out = C(ii(ii~=0))';

or

x = repmat(C,size(Ind,1),1);
out = x(Ind);

[Timo W]

Thanks a lot for the input! Eventually I came up with even another solution:

[idx,~] = find(IndexMatrix');
Choice  = Options(idx)';

Turns out to be even about 5% faster than using bsxfun ;-)

Answer 2

[Jan]

Options = {'A' 'B' 'C'};
Ind = logical([1 0 0;  1 0 0;  0 0 1;  0 1 0]);
C   = Options(Ind * (1:3).');

Here the sum is performed implicitly by the matrix-vector multiplication.

[Timo W]

Nice! This is actually the simple and clean solution I was looking for. When I see it, it's so obvious ;-)

I already accepted Andrei Bobrov's answer, which also helpd a lot, but this one is actually the best one. Thanks.

I have one question though: Why do you use a dot, when transposing the vector (1:3) ?

Answer 3

[Azzi Abdelmalek]

 idx=Options(sum(bsxfun(@times,IndexMatrix,1:3),2))