speed up the code

1 次查看（过去 30 天）
Tiki Tiki
Tiki Tiki 2018-7-23
评论： Jan 2018-8-3
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 个评论
Jan
Jan 2018-7-23
编辑：Jan 2018-7-23
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 2018-7-26
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.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Jan
Jan 2018-7-23
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 个评论
Tiki Tiki
Tiki Tiki 2018-7-26
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 2018-8-3
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.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by