The fastest method to perform sum of function
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Suppose that I have a function fun(x,y,z) whose output is a 2 by 2 matrix (one example is given below). I want to calculate the sum of this function over a 3D grid of x, y, and z. Of course I can use for() loops, but that will be very slow. What is the fastest way to perform this task? Vectorizing, but how?
clear; clc;
dx = 0.1;
dy = 0.15;
dz = 0.05;
xmin = -2;
xmax = 4;
ymin = -2*pi;
ymax = 2*pi;
zmin = -1;
zmax = 2;
xs = xmin:dx:xmax;
ys = ymin:dy:ymax;
zs = zmin:dz:zmax;
answer = sum(fun(xs,ys,zs),'All')
% answer should be a 2-by-2 matrix
function out = fun(x,y,z)
J = 1;
S = 1;
D = 0.75;
Sigma0_R = -1i*0.1*eye(2);
h0 = eye(2)*(3*J*S);
hx = [0,1; 1,0]*(-J*S*(cos(y/2 - (3^(1/2)*x)/2) + cos(y/2 + (3^(1/2)*x)/2) + cos(y)));
hy = [0, -1i; 1i, 0]*(-J*S*(sin(y/2 - (3^(1/2)*x)/2) + sin(y/2 + (3^(1/2)*x)/2) - sin(y)));
hz = [1, 0; 0, -1]*(-2*D*S*(sin(3^(1/2)*x) + sin((3*y)/2 - (3^(1/2)*x)/2) - sin((3*y)/2 + (3^(1/2)*x)/2)));
out = inv( z*eye(2) - (h0 + hx + hy + hz) - Sigma0_R );
end
5 commentaires
Matt J
le 3 Fév 2024
Suppose that I have a function fun(x,y,z) whose output is a 2 by 2 matrix... I want to calculate the sum of this function over a 3D grid of x, y, and z.
Does that mean the final result will also be a 2x2 matrix, or will it be a scalar representing the sum over all 2*2*Nx*Ny*Nz values?
Réponses (3)
Walter Roberson
le 2 Fév 2024
The fastest way of adding the values depends upon the fine details of your hardware, and on the details of how the data is laid out in memory, and depends upon what else is happening on your system to determine how calculations are scheduled by the CPU.
These are not suitable topics for discussion here.
5 commentaires
Aaron
le 2 Fév 2024
In that case, you'd do sum(A, [3,4,5]) to just sum over the dimensions you're storing the dimensional dependence in.
Torsten
le 2 Fév 2024
Modifié(e) : Torsten
le 2 Fév 2024
I tried precomputing the output of your function symbolically and then evaluate it for numerical input. But it seems there is no advantage over a simple loop with the numerical function called.
I don't know of a fast way to rewrite your function so that it can deal with array inputs for x,y and z.
dx = 0.1;
dy = 0.15;
dz = 0.05;
xmin = -2;
xmax = 4;
ymin = -2*pi;
ymax = 2*pi;
zmin = -1;
zmax = 2;
xs = xmin:dx:xmax;
ys = ymin:dy:ymax;
zs = zmin:dz:zmax;
tic
% Compute result in a simple loop
s=0;
for i=1:numel(xs)
for j=1:numel(ys)
for k=1:numel(zs)
s = s+fun(xs(i),ys(j),zs(k));
end
end
end
toc
s
syms x y z
J = 1;
S = 1;
D = 0.75;
Sigma0_R = -1i*0.1*eye(2);
h0 = eye(2)*(3*J*S);
hx = [0,1; 1,0]*(-J*S*(cos(y/2 - (3^(1/2)*x)/2) + cos(y/2 + (3^(1/2)*x)/2) + cos(y)));
hy = [0, -1i; 1i, 0]*(-J*S*(sin(y/2 - (3^(1/2)*x)/2) + sin(y/2 + (3^(1/2)*x)/2) - sin(y)));
hz = [1, 0; 0, -1]*(-2*D*S*(sin(3^(1/2)*x) + sin((3*y)/2 - (3^(1/2)*x)/2) - sin((3*y)/2 + (3^(1/2)*x)/2)));
out = inv( z*eye(2) - (h0 + hx + hy + hz) - Sigma0_R )
out = matlabFunction(out)
[Xs,Ys,Zs]=meshgrid(xs,ys,zs);
% Compute result from the precomputed function
tic
answer = arrayfun(@(x,y,z)out(x,y,z),Xs(:),Ys(:),Zs(:),'UniformOutput',0);
s=sum(cat(3,answer{:}),3)
toc
function out = fun(x,y,z)
J = 1;
S = 1;
D = 0.75;
Sigma0_R = -1i*0.1*eye(2);
h0 = eye(2)*(3*J*S);
hx = [0,1; 1,0]*(-J*S*(cos(y/2 - (3^(1/2)*x)/2) + cos(y/2 + (3^(1/2)*x)/2) + cos(y)));
hy = [0, -1i; 1i, 0]*(-J*S*(sin(y/2 - (3^(1/2)*x)/2) + sin(y/2 + (3^(1/2)*x)/2) - sin(y)));
hz = [1, 0; 0, -1]*(-2*D*S*(sin(3^(1/2)*x) + sin((3*y)/2 - (3^(1/2)*x)/2) - sin((3*y)/2 + (3^(1/2)*x)/2)));
out = inv( z*eye(2) - (h0 + hx + hy + hz) - Sigma0_R );
end
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Matt J
le 2 Fév 2024
Modifié(e) : Matt J
le 3 Fév 2024
dx = 0.1;
dy = 0.15;
dz = 0.05;
xmin = -2;
xmax = 4;
ymin = -2*pi;
ymax = 2*pi;
zmin = -1;
zmax = 2;
xs = xmin:dx:xmax;
ys = ymin:dy:ymax;
zs = zmin:dz:zmax;
tic;
answer = sum(fun(xs,ys,zs),3);
toc
function out = fun(x,y,z)
J = 1;
S = 1;
D = 0.75;
fnc=@(q) reshape(q,1,1,[]);
fn=@(q) shiftdim(q,-3);
x=x(:); y=y(:).'; z=fnc(z);
Sigma0_R = -1i*0.1*eye(2);
h0 = eye(2)*(3*J*S);
hx = [0,1; 1,0].*fn((-J.*S.*(cos(y./2 - (3.^(1./2).*x)./2) + cos(y./2 + (3.^(1./2).*x)./2) + cos(y))));
hy = [0, -1i; 1i, 0].*fn((-J.*S.*(sin(y./2 - (3.^(1./2).*x)./2) + sin(y./2 + (3.^(1./2).*x)./2) - sin(y))));
hz = [1, 0; 0, -1].*fn((-2.*D.*S.*(sin(3.^(1./2).*x) + sin((3.*y)./2 - (3.^(1./2).*x)./2) - sin((3.*y)./2 + (3.^(1./2).*x)./2))));
tmp=fn(z);
out = inv2( reshape( tmp.*eye(2) - (h0 + hx + hy + hz) - Sigma0_R ,2,2,[]));
end
function X=inv2(A)
%Inverts 2x2xN matrix stack A.
%Based on FEX contribution
%https://www.mathworks.com/matlabcentral/fileexchange/27762-small-size-linear-solver?s_tid=srchtitle
%by Bruno Luong
A = reshape(A, 4, []).';
X = [A(:,4) -A(:,2) -A(:,3) A(:,1)];
% Determinant
D = A(:,1).*A(:,4) - A(:,2).*A(:,3);
X = X./D;
X = reshape(X.', 2, 2, []);
end
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