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

1 vue (au cours des 30 derniers jours)
Luqman Saleem
Luqman Saleem le 29 Sep 2024
Modifié(e) : Matt J le 29 Sep 2024
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 commentaire
Matt J
Matt J le 29 Sep 2024
Modifié(e) : Matt J le 29 Sep 2024
It is not clear why you expect fun(x,y) to be an 8x8 matrix. As posted, fun() returns a 1x12 vector.

Connectez-vous pour commenter.

Réponse acceptée

Matt J
Matt J le 29 Sep 2024
Modifié(e) : Matt J le 29 Sep 2024
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]);

Plus de réponses (0)

Catégories

En savoir plus sur Loops and Conditional Statements dans Help Center et File Exchange

Tags

Produits


Version

R2024b

Community Treasure Hunt

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

Start Hunting!

Translated by