フィルターのクリア

How to reduce number of loops

1 回表示 (過去 30 日間)
David Kusnirak
David Kusnirak 2013 年 4 月 19 日
Hello,
I have code, which assembles coefficients into large sparse matrix. My coding seems to be inefficient as I use 3 loops, so I'm trying to improve the code to get better pefromance. I think it would be nice to reduce the number of loops to single loop and run it parallel (parfor), but I'm struggling to do that..
The main matrix sig is a matrix of size LX,LY,LZ and vectors dx, dy, dz holds the distances between nods along the diretion X, Y an Z.
for i = 2:LX-1
for j = 2:LY-1
for k = 2:LZ-1
Ctop(i,j,k) = (-1/dz(k-1))*((sig(i-1,j ,k-1)*((dx(i-1)*dy(j ))/4))+...
(sig(i ,j ,k-1)*((dx(i )*dy(j ))/4))+...
(sig(i-1,j-1,k-1)*((dx(i-1)*dy(j-1))/4))+...
(sig(i ,j-1,k-1)*((dx(i )*dy(j-1))/4)));
end
end
end
C1v=reshape(Ctop,[],1);
thanks for help!
  4 件のコメント
Cedric
Cedric 2013 年 4 月 19 日
And actually this is correct that Ctop has the size of sig and that Ctop(LX,:,:), Ctop(:,LY,:), Ctop(:,:,LZ), Ctop(1,:,:), Ctop(:,1,:), and Ctop(:,:,1) are 0's ?
David Kusnirak
David Kusnirak 2013 年 4 月 19 日
Values assigned to those matrix positions are evaluated separately by another code as the boundary conditions are applied there.

サインインしてコメントする。

採用された回答

Cedric
Cedric 2013 年 4 月 19 日
Ok, well, you could adapt the following to your case then. I am not checking in-depth that it is working though, and there is room for improvement as I didn't optimize indexing for managing properly boundaries..
n = 100 ;
sig = 20 * rand(n,n,n) ;
LX = size(sig,1) ;
LY = size(sig,2) ;
LZ = size(sig,3) ;
dx = 0.1 + rand(LX, 1) ; % Made irregular for testing..
dy = 0.8 + zeros(LY, 1) ;
dz = 0.3 + zeros(LZ, 1) ;
Ctop = zeros(size(sig)) ;
% - Your code.
tic ;
for i = 2:LX-1
for j = 2:LY-1
for k = 2:LZ-1
Ctop(i,j,k) = (-1/dz(k-1))*((sig(i-1,j ,k-1)*((dx(i-1)*dy(j ))/4))+...
(sig(i ,j ,k-1)*((dx(i )*dy(j ))/4))+...
(sig(i-1,j-1,k-1)*((dx(i-1)*dy(j-1))/4))+...
(sig(i ,j-1,k-1)*((dx(i )*dy(j-1))/4)));
end
end
end
toc
% - Meshgrid based code.
Ctop_mesh = zeros(size(Ctop)) ;
tic ;
[dy_m0, dx_m0] = meshgrid(dy(2:end-1), dx(2:end-1)) ;
[dy_m1, dx_m1] = meshgrid(dy(1:end-2), dx(1:end-2)) ;
sig_m00 = sig(2:end-1,2:end-1,:) ;
sig_m10 = sig(1:end-2,2:end-1,:) ;
sig_m01 = sig(2:end-1,1:end-2,:) ;
sig_m11 = sig(1:end-2,1:end-2,:) ;
for k = 2 : LZ-1
Ctop_mesh(2:end-1,2:end-1,k) = (-1/dz(k-1)) / 4 * ...
(sig_m10(:,:,k-1) .* dx_m1 .* dy_m0 + ...
sig_m00(:,:,k-1) .* dx_m0 .* dy_m0 +...
sig_m11(:,:,k-1) .* dx_m1 .* dy_m1 +...
sig_m01(:,:,k-1) .* dx_m0 .* dy_m1) ;
end
toc
% - Check.
tmp = abs(Ctop_mesh(2:end-1,2:end-1,2:end-1)-Ctop(2:end-1,2:end-1,2:end-1)) ;
all(tmp(:) < 1e-12)
This code outputs:
Elapsed time is 6.281855 seconds. % 3 nested loops
Elapsed time is 0.408983 seconds. % meshgrid
ans = % check
1
Note that it is possible to get rid of the last FOR loop and work with a 3D array for dz built this way:
dz_m1 = permute(repmat(dz(1:end-1),[1,LX-1,LY-1]), [2,3,1]) ;
but I am not sure that it is more efficient (especially if you want to PARFOR the remaining loop).

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeExecution Speed についてさらに検索

Community Treasure Hunt

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

Start Hunting!

Translated by