removing for loop by using 3d matrix

1 vue (au cours des 30 derniers jours)
Parisa Salemi
Parisa Salemi le 15 Fév 2021
Commenté : Jan le 19 Fév 2021
Hi. I wrote a code. I want to calculate pressure(p) of waves which produces by 1d array of elements. p is 2d variable. In the code I used 2d matrix and 2 loop. now I want to make a 3d matrix to remove one of the loops. but the problem that I have is when I creat 3d matrix p also gets a 3d matrix while it must be a 2d matrix.
clc;
close all;
clear all;
a=1;
c=1.5; % mm/us
T=5; dt=0.8; % us
fc=1;BW=3; % MHz
R=10;
xf=20;zf=20;
td=0;
dx=0.3; dz=0.3;pitch=0.3;
% x0=10; z0=0;
f=@(x) exp(-x.^2*BW^2);
g=@(x) sin(2*pi*fc*x);
H=@(x) f(x).*g(x);
Nz=120;
Nx=120;
x=[1:Nx]*dx;
z=[1:Nz]*dz;
xl=[1:Nx]*dx;
[z3,x3,xxl] = ndgrid(z,x,xl);
for t=0:dt:T
d3=sqrt((x3-xxl).^2+(z3).^2);
df=sqrt((xxl-xf).^2+(zf)^2);
td=(R-df)./c;
tt_hat=(d3./c)+td;
p = 1./sqrt(d3).*H(t-tt_hat); %
imagesc(p(:,:,100));
% colormap gray; colorbar;
% set(gca,'clim');
% title(['Time= ',num2str(t)]);
pause(0.05);
end
  3 commentaires
Parisa Salemi
Parisa Salemi le 16 Fév 2021
Modifié(e) : Walter Roberson le 16 Fév 2021
Sorry. Thank you Jan. I want to produce foucsed beam. For this I had 4 loops at first. But by using 2Dmatrix I was able to reduce to two. Now I want to use 4d matrix to write my code without using for loops. I will also sent my previous code. The code I have to write should have the same results as the code below. I hope this time my explaination be clear.
clc;
close all, clear all;
a=1;
c=1.5; % mm/us
T=15; dt=0.8; % us
fc=1;BW=3; % MHz
R=10;
xf=20;zf=20;
td=0;
dx=0.3; dz=0.3;pitch=0.3;
x0=10; z0=0;
f=@(x) exp(-x.^2*BW^2);
g=@(x) sin(2*pi*fc*x);
H=@(x) f(x).*g(x);
Nz=120;
Nx=120;
x=[1:Nx]*dx;
z=[1:Nz]*dz;
%u=zeros(Nz,Nx);
[zz,xx] = ndgrid(z,x);
for t= 0:dt:T
sum1=zeros(Nz,Nx);
for ind= 1:120
x0=ind*pitch;
dd=sqrt((xx-x0).^2+(zz-z0).^2);
df=sqrt((x0-xf)^2+(z0-zf)^2);
td=(R-df)/c;
tt_hat=dd/c+td;
p = 1./sqrt(dd).*H(t-tt_hat); %
%p = 1./sqrt(dd).*(exp(-(t-tt_hat).^2*BW^2).*sin(2*pi*fc*(t-tt_hat)));
sum1=sum1+p;
end
cla;imagesc(sum1.^2);
% colormap gray; colorbar;
% set(gca,'clim');
% title(['Time= ',num2str(t)]);
pause(0.05);
end
Parisa Salemi
Parisa Salemi le 16 Fév 2021
Modifié(e) : Jan le 16 Fév 2021
I could to remove one of the loops. But now the problem that I have is, I do not know how can I remove the time loop in this code.
clc;
close all;
clear all;
a=1;
c=1.5; % mm/us
T=20; dt=0.8; % us
fc=1;BW=3; % MHz
R=10;
xf=20;zf=20;
td=0;
dx=0.3; dz=0.3;pitch=0.3;
% x0=10; z0=0;
f=@(x) exp(-x.^2*BW^2);
g=@(x) sin(2*pi*fc*x);
H=@(x) f(x).*g(x);
Nz=120;
Nx=120;
x=[1:Nx]*dx;
z=[1:Nz]*dz;
xl=[1:Nx]*dx;
[z3,x3,xxl] = ndgrid(z,x,xl);
xf1 = ones(size(xxl))*xf;
zf1 = ones(size(xxl))*zf;
d3=sqrt((x3-xxl).^2+(z3).^2);
df=sqrt((xxl-xf1).^2+(zf1).^2);
td=(R-df)./c;
tt_hat=(d3./c)+td;
for t=0:dt:T
p = 1./sqrt(d3).*H(t-tt_hat);
p1 = sum(p,3).^2;
imagesc(p1);
pause(0.05);
end

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Réponses (1)

Jan
Jan le 16 Fév 2021
Modifié(e) : Jan le 18 Fév 2021
What is your purpose of vectorizing the code? The processing time is dominated by imagesc and pause here. I assume the loop are faster than the vectorizde code, which produces large intermediate arrays. Your original loop takes 0.86 seconds, the code in the comment above 1.33 seconds, if imagesc and pause are removed.
You can add an additional dimension:
dim = [1, size(tt_hat)];
p = H((0:dt:T).' - reshape(tt_hat, dim)) ./ reshape(sqrt(d3), dim);
p1 = sum(p, 4).^2;
But I do not see the advantage compared to clean loops.
If efficiency matters, remember that anonymous functions are expensive. So reduce them to the minimum:
% This:
f=@(x) exp(-x.^2*BW^2);
g=@(x) sin(2*pi*fc*x);
H=@(x) f(x).*g(x);
% slows down the total processing time by 10% compare to this:
H = @(x) exp(-x.^2 * BW^2) .* sin(2 * pi * fc * x);
Note: Please format your code in the forum to improve the readability. I've done this for you today.
The brute clearing header "clc;close all;clear all;" is extremly inefficient: the clear('all') removes all functions from the memory and forces Matlab to reload them from the slow disk. This wastes a lot of time and offers no advantage. Prefer using functions instead to keep your workspace clean.
  4 commentaires
Parisa Salemi
Parisa Salemi le 19 Fév 2021
Modifié(e) : Jan le 19 Fév 2021
Hi Jan. Thanks again. The problem is p1 is 3d matrix. While it must be 2D.
clc;
close all;
clear;
a=1;
c=1.5; % mm/us
T=20; dt=0.8; % us
fc=1;BW=3; % MHz
R=10;
xf=20;zf=20;
td=0;
dx=0.3; dz=0.3;pitch=0.3;
% x0=10; z0=0;
f=@(x) exp(-x.^2*BW^2);
g=@(x) sin(2*pi*fc*x);
H=@(x) f(x).*g(x);
Nz=120;
Nx=120;
x=[1:Nx]*dx;
z=[1:Nz]*dz;
xl=[1:Nx]*dx;
[z3,x3,xxl] = ndgrid(z,x,xl);
d3=sqrt((x3-xxl).^2+(z3).^2);
df=sqrt((xxl-xf).^2+(zf).^2);
td=(R-df)./c;
tt_hat=(d3./c)+td;
dim = [1, size(tt_hat)];
p = H((0:dt:T).' - reshape(tt_hat, dim)) ./ reshape(sqrt(d3), dim);
p1 = sum(p, 4).^2;
imagesc(p1);
Jan
Jan le 19 Fév 2021
Hi. This code produces a 3D array. Of course it does, because the original code produces a set of 2D arrays. Concatenating a bunch of 2D matrices must create a 3D array.
Why do you want to get a 2D matrix? What should happen with the additional information?

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