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

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Timo W
Timo W 2013-5-31
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!

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2013-5-31
编辑：Andrei Bobrov 2013-5-31
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);
  1 个评论
Timo W
Timo W 2013-5-31
编辑：Timo W 2013-5-31
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 ;-)

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更多回答（2 个）

Jan
Jan 2013-5-31
编辑：Jan 2013-5-31
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.
  1 个评论
Timo W
Timo W 2013-6-1
编辑：Timo W 2013-6-1
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) ?

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Azzi Abdelmalek
Azzi Abdelmalek 2013-5-31
idx=Options(sum(bsxfun(@times,IndexMatrix,1:3),2))

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