Using parfor to evaluate integrations

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Probably a stupid question.
I want to integrate a function f(x,y). Can I use parfor() as shown below to compute the sum? I am confused because I consider that each loop over one "ix" value in the following parfor() is run independently from other "ix" values, which means that I must obtain a different "sum_f" for each "ix" value. Right?
clear; clc;
xmin = -2; xmax = 1;
ymin = 1; ymax = 3;
dx = 0.001;
dy = dx;
xs = xmin:dx:xmax; Nx = length(xs);
ys = ymin:dy:ymax; Ny = length(ys);
f = @(x,y) x^2+y^2;
sum_f = 0;
parfor ix = 1:Nx
x = xs(ix);
for iy = 1:Ny
y = ys(iy);
sum_f = sum_f + f(x,y)*dx*dy;
end
end
I = sum(sum_f(:))
  2 个评论
Dyuman Joshi
Dyuman Joshi 2024-2-2
Any particular reason why you are using a double for loop, instead of vectorizing the function handle and the sum?
Luqman Saleem
Luqman Saleem 2024-2-2
@Dyuman Joshi umm.. I don't get you. I am not familiar with "vectorizing" the functions. Do you mean rewriting f(x,y) as
f = @(x,y) arrayfun(@(x,y) f(x,y), x, y)

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采纳的回答

Dyuman Joshi
Dyuman Joshi 2024-2-2
移动:Hans Scharler 2024-2-2
xmin = -2; xmax = 1;
ymin = 1; ymax = 3;
dx = 0.001;
dy = dx;
xs = xmin:dx:xmax; Nx = length(xs);
ys = ymin:dy:ymax; Ny = length(ys);
%vectorizing the function handle
f = @(x,y) x.^2+y.^2;
tic
sum_f = 0;
for ix = 1:Nx
x = xs(ix);
for iy = 1:Ny
y = ys(iy);
sum_f = sum_f + f(x,y)*dx*dy;
end
end
I1 = sum(sum_f(:))
I1 = 32.0317
toc
Elapsed time is 0.364607 seconds.
tic
I2 = sum(f(xs,ys.').*dx.*dy, 'all')
I2 = 32.0317
toc
Elapsed time is 0.033681 seconds.
%Using tolerance to compare floating point numbers
abs(I1-I2)<1e-10
ans = logical
1
  2 个评论
Luqman Saleem
Luqman Saleem 2024-2-2
移动:Hans Scharler 2024-2-2
@Dyuman Joshi :-O areee bhai.. that's much more faster. wow. thank you so much
Mike Croucher
Mike Croucher 2024-2-5
Ahhh loops vs vectors....my old friend. We meet again.
I'm going to let you in on a secret......sometimes loops are faster!
  • Original loop with a function call: 0.218 seconds
  • Optimised loop with function call removed: 0.025 seconds
  • vectorised loop: 0.0275 seconds
So here, my loop version is slightly faster than the vectorised version. What you see might be dependent on machine, problem size and MATLAB version
Observe:
xmin = -2; xmax = 1;
ymin = 1; ymax = 3;
dx = 0.001;
dy = dx;
xs = xmin:dx:xmax; Nx = length(xs);
ys = ymin:dy:ymax; Ny = length(ys);
%vectorizing the function handle
f = @(x,y) x.^2+y.^2;
tic
sum_f = 0;
for ix = 1:Nx
x = xs(ix);
for iy = 1:Ny
y = ys(iy);
sum_f = sum_f + f(x,y)*dx*dy;
end
end
I1 = sum(sum_f(:))
I1 = 32.0317
toc
Elapsed time is 0.218968 seconds.
disp('Inline the function call in the loop')
Inline the function call in the loop
xmin = -2; xmax = 1;
ymin = 1; ymax = 3;
dx = 0.001;
dy = dx;
xs = xmin:dx:xmax; Nx = length(xs);
ys = ymin:dy:ymax; Ny = length(ys);
tic
sum_f = 0;
for ix = 1:Nx
x = xs(ix);
for iy = 1:Ny
y = ys(iy);
sum_f = sum_f + (x.^2+y.^2)*dx*dy;
end
end
I1 = sum(sum_f(:))
I1 = 32.0317
toc
Elapsed time is 0.025205 seconds.
disp('vectorised version')
vectorised version
tic
I2 = sum(f(xs,ys.').*dx.*dy, 'all')
I2 = 32.0317
toc
Elapsed time is 0.027567 seconds.

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