How to convert two nested for-loops to one parfor loop.

5 次查看（过去 30 天）
Luqman Saleem
Luqman Saleem 2024-9-29
编辑： Matt J 2024-9-29
I have the following code. I want to get it in one parfor loop.
clear; clc;
% number of points:
Nx = 80;
Ny = 90;
xs = linspace(-2*pi/(3),4*pi/(3),Nx);
ys = linspace(-2*pi/(sqrt(3)),2*pi/(sqrt(3)),Ny);
% Allocate memory
ZZ = zeros(Nx,Ny,8);
XX = zeros(Nx,Ny);
YY = zeros(Nx,Ny);
for ix = 1:Nx
x = xs(ix);
for iy = 1:Ny
y = ys(iy);
FUN = fun(x,y);
% sort eigenvalues:
[~,D] = eig(FUN);
[D,I] = sort(diag(real(D)),'descend');
evals = diag(D);
% store data:
ZZ(ix,iy,:) = diag(evals);
XX(ix,iy) = x;
YY(ix,iy) = y;
end
end
%% Plot figure
figure;
tiled = tiledlayout(2,2,"TileSpacing","tight","Padding","compact");
for q = 1:2:8
nexttile
surf(XX,YY,ZZ(:,:,q),'LineStyle','none','FaceColor','interp');
view(2)
axis([-2*pi/(3) 4*pi/(3) -2*pi/(sqrt(3)) +2*pi/(sqrt(3))])
box on
grid off
axis square
colorbar
end
%%
function out = fun(x,y)
out = [ 3, 0, - 2*cos(x/4 + (3^(1/2)*y)/4) - (3*cos((3*x)/4 - (3^(1/2)*y)/4))/5, 0, - 2*cos(x/4 - (3^(1/2)*y)/4) - (3*cos((3*x)/4 + (3^(1/2)*y)/4))/5, 0, - 2*cos(x/2) - (3*cos((3^(1/2)*y)/2))/5, 0
0, -3, 0, - 2*cos(x/4 + (3^(1/2)*y)/4) - (3*cos((3*x)/4 - (3^(1/2)*y)/4))/5, 0, - 2*cos(x/4 - (3^(1/2)*y)/4) - (3*cos((3*x)/4 + (3^(1/2)*y)/4))/5, 0, - 2*cos(x/2) - (3*cos((3^(1/2)*y)/2))/5
- 2*cos(x/4 + (3^(1/2)*y)/4) - (3*cos((3*x)/4 - (3^(1/2)*y)/4))/5, 0, -1, -2*2^(1/2), - 2*cos(x/2) - (3*cos((3^(1/2)*y)/2))/5, 0, - 2*cos(x/4 - (3^(1/2)*y)/4) - (3*cos((3*x)/4 + (3^(1/2)*y)/4))/5, 0
0, - 2*cos(x/4 + (3^(1/2)*y)/4) - (3*cos((3*x)/4 - (3^(1/2)*y)/4))/5, -2*2^(1/2), 1, 0, - 2*cos(x/2) - (3*cos((3^(1/2)*y)/2))/5, 0, - 2*cos(x/4 - (3^(1/2)*y)/4) - (3*cos((3*x)/4 + (3^(1/2)*y)/4))/5
- 2*cos(x/4 - (3^(1/2)*y)/4) - (3*cos((3*x)/4 + (3^(1/2)*y)/4))/5, 0, - 2*cos(x/2) - (3*cos((3^(1/2)*y)/2))/5, 0, -1, 2^(1/2) - 6^(1/2)*1i, - 2*cos(x/4 + (3^(1/2)*y)/4) - (3*cos((3*x)/4 - (3^(1/2)*y)/4))/5, 0
0, - 2*cos(x/4 - (3^(1/2)*y)/4) - (3*cos((3*x)/4 + (3^(1/2)*y)/4))/5, 0, - 2*cos(x/2) - (3*cos((3^(1/2)*y)/2))/5, 2^(1/2) + 6^(1/2)*1i, 1, 0, - 2*cos(x/4 + (3^(1/2)*y)/4) - (3*cos((3*x)/4 - (3^(1/2)*y)/4))/5
- 2*cos(x/2) - (3*cos((3^(1/2)*y)/2))/5, 0, - 2*cos(x/4 - (3^(1/2)*y)/4) - (3*cos((3*x)/4 + (3^(1/2)*y)/4))/5, 0, - 2*cos(x/4 + (3^(1/2)*y)/4) - (3*cos((3*x)/4 - (3^(1/2)*y)/4))/5, 0, -1, 2^(1/2) + 6^(1/2)*1i
0, - 2*cos(x/2) - (3*cos((3^(1/2)*y)/2))/5, 0, - 2*cos(x/4 - (3^(1/2)*y)/4) - (3*cos((3*x)/4 + (3^(1/2)*y)/4))/5, 0, - 2*cos(x/4 + (3^(1/2)*y)/4) - (3*cos((3*x)/4 - (3^(1/2)*y)/4))/5, 2^(1/2) - 6^(1/2)*1i, 1
];
end
  1 个评论
Matt J
Matt J 2024-9-29
编辑：Matt J 2024-9-29
It is not clear why you expect fun(x,y) to be an 8x8 matrix. As posted, fun() returns a 1x12 vector.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Matt J
Matt J 2024-9-29
编辑：Matt J 2024-9-29
It does not seem advisable to use parfor. Everything in your code is vectorizable. However, here is what a parfor approach could look like:
[XX,YY,ZZ]=meshgrid(xs,ys);
[M,N]=size(XX);
ZZ=zeros(M*N,8);
parfor i=1:M*N
ZZ(i,:)=sort( eig( fun(XX(i),YY(i)) ),'descend');
end
ZZ=reshape(ZZ,[M,N,8]);

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

标签

产品


版本

R2024b

Community Treasure Hunt

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

Start Hunting!

Translated by