Effacer les filtres
Effacer les filtres

speed up the code

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Tiki Tiki
Tiki Tiki le 23 Juil 2018
Commenté : Jan le 3 Août 2018
hi everyone.
can you help me speed up this code?
tic
InputOverlap = magic(64)
SDR_Overlap = InputOverlap;
SDR = (zeros (64,64)) ;
Radius = 2;
InputOverlap = [InputOverlap(:,1:Radius) InputOverlap InputOverlap(:,end+1-Radius:end)];
InputOverlap = [InputOverlap(1:Radius,:) ; InputOverlap ; InputOverlap(end+1-Radius:end,:) ];
for r=1:64
for c=1:64
Neighbour= InputOverlap(r:r+2*Radius,c:c+2*Radius);
Kmax = max(Neighbour(:)) ;
if (SDR_Overlap(r,c)>0)&(SDR_Overlap(r,c)>= Kmax)
SDR(r,c) = 1;
else
SDR(r,c) = 0;
end
end
end
toc
Thanks.
  2 commentaires
Jan
Jan le 23 Juil 2018
Modifié(e) : Jan le 23 Juil 2018
Start with omitting:
else
SDR(r,c) = 0;
SDR is initialized to zero already.
The editor should show a hint that && is more efficient than &. Consider these MLint messages.
The main part of your code happens before the loop. Most of all displaying the magic matrix is slow. I guess, you want to measure the time inside the loop only, don't you?
Tiki Tiki
Tiki Tiki le 26 Juil 2018
Yes. My problem is time in the loop. I remove SDR(r,c) = 0 by setting it is zeros before loop.
But time consumes still high. How can I remove loop in this case?
Please help me. Thank.

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Jan
Jan le 23 Juil 2018
This is slightly faster:
tic
for r = 1:64
for c = 1:64
Neighbour = InputOverlap(r:r+2*Radius, c:c+2*Radius);
Kmax = max(Neighbour(:));
SDR(r,c) = (SDR_Overlap(r,c) >= Kmax);
end
end
toc
Is the SDR_Overlap(r,c)>0 test useful?
  2 commentaires
Tiki Tiki
Tiki Tiki le 26 Juil 2018
Yes. it is a little faster. i also remove SDR_Overlap(r,c)>0. but this code still consume much time.
so i need optimze more. Can you help me how to remove loop in this case?
I have gpu. but dont undertand to use it.
Jan
Jan le 3 Août 2018
Use movmax to replace the loops.
Does the padding of the input matrix belong to the problem? With movmax and 'EndPoints' set to 'shrink' you can omit the padding.
can you post some real input data? Especially the dimensions matter.

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