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Subset of an array with a moving window

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Alex Nickerson
Alex Nickerson on 16 Apr 2021
Edited: Alex Nickerson on 16 Apr 2021
I have the following code for computing structure functions. I have provided examples of "ULs" and "UTs" merely for simplicity.
ULs = rand(1,145); % called in from function
UTs = rand(1,145); % called in from function
nrmax = 145; % called in from function, I just happen to be using 145
for nr = 1:nrmax
Nts = numel(ULs); % number of elements
Umat_L = nan(Nts-nr,nr+1); % must be NaN, so zeros() won't work
Umat_T = nan(Nts-nr,nr+1); % must be NaN, so zeros() won't work
for i = 1:(Nts-nr) % this is
Umat_L(i,:)=ULs(i:i+nr); % the for loop
Umat_T(i,:)=UTs(i:i+nr); % I want to remove
end
end
The nested loop thus takes subsets ULs(1:1+nr), then ULs(2:2+nr), ULs(17:17+nr), etc. Is it possible to replicate this subsetting process without using a loop?
It would improve my code run time if I could remove the nested "for i = 1:(Nts-nr)". However, as I see it, there is no way around this because of the nature of the moving window. I tried the following substitution, but it didn't work.
i = (1:Nts-nr);
inr = i + nr;
Umat_L(i,:) = ULs(i:inr);
Umat_T(i,:) = UTs(i:inr);
  2 Comments
Alex Nickerson
Alex Nickerson on 16 Apr 2021
Well, as can be seen in the code, there is no moving mean. I don't have access to the Image Processing Toolbox. I just need to be able to get these variable subsets of the code. I will add some details to the original question to further clarify.

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Accepted Answer

Rik
Rik on 16 Apr 2021
The code below relies on implicit expansion to create an index array, which replicates the results from the inner for-loop. Since your code overwrites the results every iteration, you can also have a substantial speed increase by only doing the last one.
nrmax = 145; % called in from function, I just happen to be using 145
ULs = rand(1,nrmax); % called in from function
UTs = rand(1,nrmax); % called in from function
%tic,toc is only an approximate, use timeit for real performance testing
tic,[Umat_L1,Umat_T1]=fun_original(nrmax,ULs,UTs);toc
Elapsed time is 1.689398 seconds.
tic,[Umat_L2,Umat_T2]=fun_unlooped(nrmax,ULs,UTs);toc
Elapsed time is 0.994556 seconds.
if ~isequal(Umat_L1,Umat_L2) || ~isequal(Umat_T1,Umat_T2)
error('results don''t match')
else
disp('everything OK')
end
everything OK
function [Umat_L,Umat_T]=fun_original(nrmax,ULs,UTs)
for slowdown=1:100
for nr = 1:nrmax
Nts = numel(ULs); % number of elements
Umat_L = nan(Nts-nr,nr+1); % must be NaN, so zeros() won't work
Umat_T = nan(Nts-nr,nr+1); % must be NaN, so zeros() won't work
for i = 1:(Nts-nr) % this is
Umat_L(i,:)=ULs(i:i+nr); % the for loop
Umat_T(i,:)=UTs(i:i+nr); % I want to remove
end
end
end
end
function [Umat_L,Umat_T]=fun_unlooped(nrmax,ULs,UTs)
for slowdown=1:100
for nr = 1:nrmax
Nts = numel(ULs); % number of elements
ind=( 0:nr ) + ( 1:(Nts-nr) ).';
Umat_L=ULs(ind);
Umat_T=UTs(ind);
end
end
end
  1 Comment
Alex Nickerson
Alex Nickerson on 16 Apr 2021
I have more lines of code that go far beyond the nested loop... I just didn't include them here to save space. This code did exactly what I needed, @Rik, and it cut my code run time (the sample on which I tested it) in half. Thank you for your help.

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