How to avoid for nested loops with if condition?

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Mantas Vaitonis 2018-6-25

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评论： Mantas Vaitonis 2018-6-26

采纳的回答： Guillaume

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Hello,

I have big matrixes of same size, and am suing for loop with if statement, which is bottleneck in my code and is very slow. How would it be possible to optimize it?

    for i=1:n1
    for j=1:n2
    if id(i,j)==1
    if RR(i,j)==1
        id(i,x(i,j))=0;
    end
    end
    end
end
 Maybe it is possible to vectorize or use bsxfun?

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Guillaume 2018-6-25

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编辑：Guillaume 2018-6-25

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You certainly don't need the j loop:

for row = 1:size(id)
   id(row, x(id(row, :) == 1 & RR(row, :) == 1)) = 0;
end

You could get rid of the i loop as well at the expense of memory (another array the same size as id)

mask = id == 1 & RR == 1;  %and if id and RR are logical, simply: mask = id & RR;
rows = repmat((1:size(id, 1)', 1, size(id, 2));
id(sub2ind(size(id), rows(mask), x(mask))) = 0;

So it's replicated pairs in a single row that need to be removed, no replicated pairs across the whole matrix. If so, you can indeed use a loop over the rows as you've done but yours is overcomplicated and your transition through a 3d matrix only works by accident (your cat(3,...) is reshaped into a 2d matrix). This would be simpler:

x = [3 1 1 5 3;2 1 1 5 3]; 
a = repmat([1 2 3 4 5], size(x, 1), 1);
for row = 1:size(x, 1)
   [~, tokeep] = unique(sort([x(row, :)', a(row, :)'], 'rows', 'stable');
   a(row, setdiff(1:size(a, 2), tokeep) = 0;
end

Now, there is a way to do it without a loop. It's pretty hairy, having to go through a 4D matrix:

x = [3 1 1 5 3;2 1 1 5 3]; 
a = repmat([1 2 3 4 5], size(x, 1), 1);
sortedpairs = sort(cat(3, x, a), 3);
matchedpairs = all(sortedpairs == permute(sortedpairs, [1 4 3 2]), 3);
matchnotfirst = cumsum(matchedpairs, 2) > 1 & matchedpairs;  %only keep 2nd and subsequent pair
toreset = any(matchnotfirst, 4);
a(toreset) = 0