# Vectorization of a for loop to increase speed

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Howard Wilton on 25 Nov 2022
Edited: Bruno Luong on 25 Nov 2022
I am looking to embed an logical expression in an equation. The code example below is to show the problem and is not a reflection of the underlying math I am trying to solve!
t_ = -1:0.1:1;
for t = t_
flag = (t > a); % a is a vector, defined seperately
res = t - a(flag(:));
end
For each value of t, I would get a vector that has a length that corresponds to the number of logical 1s in .
What I would very much like to do is eliminate the for loop and do something kind of like this,
t = -1:0.1:1;
res = t - a(t > a)(:) % i presume this looks AWFUL
And as a result i end up with a matrix with one dimension the size of length t and the other dimension the size of the number of logical 1s in .
Would welcome any ideas or comments.
Stephen23 on 25 Nov 2022
Edited: Stephen23 on 25 Nov 2022
"And as a result i end up with a matrix with one dimension the size of length t and the other dimension the size of the number of logical 1s in flag."
Your concept does not have a general solution because the "number of logical 1s in flag" could be different on each loop iteration, so that res vectors have different lengths, and therefore there is no such matrix (because each "other dimension" could be different).
Would you allow for padding of a matrix, or a 1xN vector, or use of a container array?
"The code example below is to show the problem and is not a reflection of the underlying math I am trying to solve!"
It would probably be much faster to explain the actual problem.

Bruno Luong on 25 Nov 2022
Edited: Bruno Luong on 25 Nov 2022
This returns a matrix, whose elements which do not satisfy the condition and was not computed in your orginal code are marked by NaN. Each row corresponds to one iteration of the original code
resmat = t_(:) - a(:).';
resmat(resmat<=0) = NaN;
% Move NaN towards the last columns
[~,js] = sort(isnan(resmat),2);
m = size(resmat,1);
resmat = resmat((js-1)*m+(1:m)')

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