How can calcuate the lines of code efficiently?

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Amir Torabi
Amir Torabi 2019-12-4
评论： Ahmad Nadeem 2021-7-14
Hello. This code calculates the Laplacian. I am looking for reducing the time running of it. My main aim is improving the performance of loop running time.
Thanks
%%%code
nx=512;
ny=nx;
dx=1;
dy=dx;
format long;
nxny=nx*ny;
r=zeros(1,nx);
r(1:2)=[2,-1];
T=toeplitz(r);
E=speye(nx);
grad=-(kron(T,E)+kron(E,T));%Tensor
%-- for periodic boundaries
for i=1:nx
ii=(i-1)*nx+1;
jj=ii+nx-1;
grad(ii,jj)=1.0;
grad(jj,ii)=1.0;
kk=nxny-nx+i;
grad(i,kk)=1.0;
grad(kk,i)=1.0;
end%for
grad = grad /(dx*dy);
  1 个评论
Ahmad Nadeem
Ahmad Nadeem 2021-7-14
hello brother!
I want to contact you. Can you please share your email or any other ID?
Thank You
Best Regards
Ahmad Nadeem
Hongik University, Seoul, South Korea

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采纳的回答

Stephan
Stephan 2019-12-4
Yes, you can if you avoid using a loop and vectorize your code - see here:
tic
%%%code
nx=512;
ny=nx;
dx=1;
dy=dx;
format long;
nxny=nx*ny;
r=zeros(1,nx);
r(1:2)=[2,-1];
T=toeplitz(r);
E=speye(nx);
grad=-(kron(T,E)+kron(E,T));%Tensor
%-- for periodic boundaries
ii=(0:nx-1).*nx+1;
jj = ii+nx-1;
grad(ii,jj)=1.0;
grad(jj,ii)=1.0;
kk=(nxny-nx+1):nxny;
grad(1:nx,kk)=1.0;
grad(kk,1:nx)=1.0;
grad = grad /(dx*dy);
toc
Elapsed time is 0.145899 seconds.
Your code:
tic
%%%code
nx=512;
ny=nx;
dx=1;
dy=dx;
format long;
nxny=nx*ny;
r=zeros(1,nx);
r(1:2)=[2,-1];
T=toeplitz(r);
E=speye(nx);
grad=-(kron(T,E)+kron(E,T));%Tensor
%-- for periodic boundaries
for i=1:nx
ii=(0:nx-1).*nx+1;
jj=ii+nx-1;
grad(ii,jj)=1.0;
grad(jj,ii)=1.0;
kk=nxny-nx+i;
grad(i,kk)=1.0;
grad(kk,i)=1.0;
end
toc
Elapsed time is 11.333450 seconds.
Are the results correct? - to check this we call one of the results grad1 and subtract grad1 from grad:
nx=512;
ny=nx;
dx=1;
dy=dx;
format long;
nxny=nx*ny;
r=zeros(1,nx);
r(1:2)=[2,-1];
T=toeplitz(r);
E=speye(nx);
grad=-(kron(T,E)+kron(E,T));%Tensor
%-- for periodic boundaries
for i=1:nx
ii=(0:nx-1).*nx+1;
jj=ii+nx-1;
grad(ii,jj)=1.0;
grad(jj,ii)=1.0;
kk=nxny-nx+i;
grad(i,kk)=1.0;
grad(kk,i)=1.0;
end
ii=(0:nx-1).*nx+1;
jj = ii+nx-1;
grad(ii,jj)=1.0;
grad(jj,ii)=1.0;
kk=(nxny-nx+1):nxny;
grad(1:nx,kk)=1.0;
grad(kk,1:nx)=1.0;
grad1 = grad /(dx*dy);
expect_all_zeros = grad1-grad
expect_all_zeros =
All zero sparse: 262144×262144
Appears to work fine...
  4 个评论
显示 2更早的评论
Amir Torabi
Amir Torabi 2019-12-5
i checked it again. yes your answer is correct. Thanks.
Ahmad Nadeem
Ahmad Nadeem 2021-7-14
I didnt understand this code can you please help me to understand this code?

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