Circshift matrix by different amount without for loop

4 Ansichten (letzte 30 Tage)
Hello everyone. I have this 3 dimensional matrix called spectrums and I want to shift the 3rd dimension by different amounts given by vector shifts. Is there a way to do this without using a for loop? something like
shifted_spectrums=arrayfun(@(k) circshift(spectrums(:,:,k),shifts(2:end),2),spectrums);
but a solution that actualy works :P
PS: 3rd dimension of spectrums has 22 elements and shifts has 23 elements, that's why the 2:end.
  1 Kommentar
Jan
Jan am 29 Mär. 2019
Bearbeitet: Jan am 29 Mär. 2019
What is the class and contents of shifts(2:end)? What is not "actually working"?
We cannot guess these details.
A for loop is faster than arrayfun in general. So please mention, why you want to avoid it.
Is this the loop version?
shifted_spectrums = zeros(size(spectrums));
for k = 1:size(spectrums, 3)
shifted_spectrums(:, :, k) = circshift( ...
spectrum(:, :, k), shifts(k+1), 2);
end

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Javier Agustin Romero
Javier Agustin Romero am 29 Mär. 2019
Bearbeitet: Javier Agustin Romero am 29 Mär. 2019
I've just got it working. A reshape is needed so the output has wanted size (sz1=1024, sz2=8192).
shifted_spectrums=reshape(cell2mat(arrayfun(@(k) circshift(spectrum(:,:,k)...
,shifts(k,1),2),1:Nfiles,'uni',0)),sz1/2,sz2,Nfiles);
This version is faster than the for loop
for i=2:Nfiles
shifted_spectrums(:,:,i)=circshift(spectrums(:,:,i),shifts(i-1,1),2);
end
Although not by much (16%, spectrums is a 512x8192x22 matrix)
  1 Kommentar
Jan
Jan am 29 Mär. 2019
Bearbeitet: Jan am 29 Mär. 2019
Did you pre-allocate the output? This is essential for the performance:
shifted_spectrums = zeros(size(spectrums));

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Jan
Jan am 29 Mär. 2019
Bearbeitet: Jan am 29 Mär. 2019
Try:
n = 8192;
spectrums = rand(512, n, 22);
shifts = randi(n, 1, 22);
tic
s1 = cell2mat(arrayfun(@(k) ...
circshift(spectrums(:,:,k), shifts(k), 2),1:22,'uni',0));
toc
tic;
s2 = zeros(size(spectrums));
for k = 1:22
s2(:,:,k)=circshift(spectrums(:,:,k), shifts(k), 2);
end
toc
For n=2000 I get:
arrayfun: 0.22sec
loop: 0.17sec
  3 Kommentare
Jan
Jan am 29 Mär. 2019
Bearbeitet: Jan am 30 Mär. 2019
i7-3770 3.4GHz, 16 GB RAM, n = 8192:
R2009b:
Elapsed time is 2.660240 seconds.
Elapsed time is 2.062384 seconds.
R2016b:
Elapsed time is 2.613608 seconds.
Elapsed time is 2.292270 seconds.
R2018b:
Elapsed time is 2.682782 seconds.
Elapsed time is 1.207014 seconds. ! Nice !
Javier Agustin Romero
Javier Agustin Romero am 30 Mär. 2019
Bearbeitet: Javier Agustin Romero am 30 Mär. 2019
Just ran it again and got
Elapsed time is 1.512578 seconds.
Elapsed time is 1.911443 seconds.
I'm working on a i7-7700HQ, 2.8GHz, 16GB RAM and got Matlab installed in a SSD (good to know that was a good investment). Anyway, what I find weird is that which method works best changes with PC's specs. I mean, shouldn't one method always work best and both times move up or down according to the PC specs? I am talking way over my head here though, have no idea.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange

Tags

Produkte


Version

R2016a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by