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

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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.

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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]);

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