How to avoid for nested loops with if condition?
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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
编辑:Guillaume
2018-6-25
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;
10 个评论
Guillaume
2018-6-26
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
更多回答(1 个)
Nithin Banka
2018-6-25
I don't think you need to use 'for' loop. The 2 'if' statements can be easily handled in MATLAB.
index_of_1_in_id_and_RR = (id==1&RR==1); %valid as you told they have same size
You can use this variable in further steps.
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