How to reduce its execution time?

2 visualizaciones (últimos 30 días)
Sadiq Akbar
Sadiq Akbar el 26 de Dic. de 2022
Comentada: Sadiq Akbar el 29 de Dic. de 2022
I have the following piece of code. It works but takes time. How can we reduce its time to a very minimum value?
u=[1 3 5 7 20 30 40 50];
b=u;
Noise=5;
[R,C]=size(b);
P=C/2;
M=2*C;
%%%%%%%%%%%%%%%%%%%%
% Swapping vector b
%%%%%%%%%%%%%%%%%%%%
[~, ix] = sort(u); % u is my desired vector
[~, ix1(ix)] = sort(b);
b = b(ix1);
%%%%%%%%%%%%%%%%%%%%%%%%%%
% calculate xo
%%%%%%%%%%%%%%%%%%%%%%%%%%
xo=zeros(1,M);
for k=1:M
for i=1:P
xo(1,k)=xo(1,k)+u(i)*exp(-1i*(k-1)*pi*cosd(u(P+i)));
end
end
xo=awgn(xo,Noise);% add Noise
%%%%%%%%%%%%%%%%%%%%%
% Calculate xe
%%%%%%%%%%%%%%%%%%%%%
xe=zeros(1,M);
for k=1:M
for i=1:P
xe(1,k)=xe(1,k)+b(i)*exp(-1i*(k-1)*pi*cosd(b(P+i)));
end
end
%%%%%%%%%%%%%%%%%
% MSE
%%%%%%%%%%%%%%%%%
abc=0.0;
for m1=1:M
abc=abc+(abs(xo(1,m1)-xe(1,m1))).^2;
end
abc=abc/M;
e=abc

Respuesta aceptada

Voss
Voss el 26 de Dic. de 2022
Editada: Voss el 26 de Dic. de 2022
u=[1 3 5 7 20 30 40 50];
b=u;
Noise=5;
[R,C]=size(b);
P=C/2;
M=2*C;
%%%%%%%%%%%%%%%%%%%%
% Swapping vector b
%%%%%%%%%%%%%%%%%%%%
[~, ix] = sort(u); % u is my desired vector
[~, ix1(ix)] = sort(b);
b = b(ix1);
% not sure why you sort b and u, since you have b=u above. seems like you
% can just sort one of them. anyway, I don't know what the "swapping vector
% b" code is supposed to do, so I left it alone. see below for the rest of
% it:
%%%%%%%%%%%%%%%%%%%%%%%%%%
% calculate xo
%%%%%%%%%%%%%%%%%%%%%%%%%%
% xo=zeros(1,M);
% for k=1:M
% for i=1:P
% xo(1,k)=xo(1,k)+u(i)*exp(-1i*(k-1)*pi*cosd(u(P+i)));
% end
% end
xo = sum(u(1:P).*exp(-1i.*(0:M-1).'.*pi.*cosd(u(P+1:C))),2).';
xo=awgn(xo,Noise);% add Noise
%%%%%%%%%%%%%%%%%%%%%
% Calculate xe
%%%%%%%%%%%%%%%%%%%%%
% xe=zeros(1,M);
% for k=1:M
% for i=1:P
% xe(1,k)=xe(1,k)+b(i)*exp(-1i*(k-1)*pi*cosd(b(P+i)));
% end
% end
xe = sum(b(1:P).*exp(-1i.*(0:M-1).'.*pi.*cosd(b(P+1:C))),2).';
%%%%%%%%%%%%%%%%%
% MSE
%%%%%%%%%%%%%%%%%
% abc=0.0;
% for m1=1:M
% abc=abc+(abs(xo(1,m1)-xe(1,m1))).^2;
% end
% abc=abc/M
e = mean(abs(xo-xe).^2,2);
  1 comentario
Sadiq Akbar
Sadiq Akbar el 29 de Dic. de 2022
Thanks a lot to both of you dear Karim and Voss.

Iniciar sesión para comentar.

Más respuestas (1)

Karim
Karim el 26 de Dic. de 2022
Editada: Karim el 26 de Dic. de 2022
Hi, I vectorized your code. Using tic/toc procedure the run time is about 0.04 seconds.
Hope it helps.
tic
% transpose u and b
% we want these to be column vectors for easy vectorization
u = [1 3 5 7 20 30 40 50]';
b = u;
Noise = 5;
[R,~] = size(b);
P = R/2;
M = 2*R;
I commented the lines below, since they don't do anything. You start with u=b and you end up with u=b. Hence these steps do nothing, you will need to provide details on what you mean with "swapping b"
% % Swapping vector b
% [~, ix] = sort(u); % u is my desired vector
% [~, ix1(ix)] = sort(b);
% b = b(ix1);
% calculate xo
u1 = u( 1:P );
u2 = u( P+(1:P));
k = 1:M;
xo = sum( u1.*exp( -1i.*pi*cosd(u2).*(k-1) ) , 1);
% add Noise
xo = awgn(xo,Noise);
% Calculate xe
b1 = b( 1:P );
b2 = b( P+(1:P));
xe = sum( b1.*exp( -1i.*pi*cosd(b2).*(k-1) ) , 1);
% MSE
e = mean( abs( xo(1,:) - xe(1,:) ) .^2 )
e = 0.2862
toc
Elapsed time is 0.040369 seconds.
  2 comentarios
Voss
Voss el 26 de Dic. de 2022
Don't forget about this:
abc=abc/M;
Karim
Karim el 26 de Dic. de 2022
yes indeed, thank you for pointing it out!

Iniciar sesión para comentar.

Categorías

Más información sobre Creating and Concatenating Matrices en Help Center y File Exchange.

Community Treasure Hunt

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

Start Hunting!

Translated by