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How to reduce its execution time?

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Sadiq Akbar
Sadiq Akbar on 26 Dec 2022
Commented: Sadiq Akbar on 29 Dec 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

Accepted Answer

Voss
Voss on 26 Dec 2022
Edited: Voss on 26 Dec 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 Comment
Sadiq Akbar
Sadiq Akbar on 29 Dec 2022
Thanks a lot to both of you dear Karim and Voss.

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More Answers (1)

Karim
Karim on 26 Dec 2022
Edited: Karim on 26 Dec 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 Comments
Voss
Voss on 26 Dec 2022
Don't forget about this:
abc=abc/M;
Karim
Karim on 26 Dec 2022
yes indeed, thank you for pointing it out!

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