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improve speed --- image matrix replace

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Bob
Bob 2013-5-20
i have a code which has three "for". it's so slow.
if nx=2000,ny=2000. it takes me two minutes to finish replacement.
xfig(nx,ny,3) is a image matrix.
please help me!!
w0=xfig;
nx=length(xfig(:,1,1));%coordinate x
ny=length(xfig(1,:,1));%coordinate y
w1=w0;
for k=1:4
for x=1:nx
for y=1:ny
w1(x,y,:)=w0(mod1(x+y,nx),mod1(x+2*y,ny),:);
end;
end;
%t=1:1:nx;v=1:1:ny;
%w1(t,v,:)=w0(mod1(t+v,nx),mod1(t+2*v,ny),:);
w0=w1;
end;
% if mod(8,4)=0 then it isn't 0, instead of 4
function T=mod1(x,y)
if mod(x,y)==0
T=y;
else
T=mod(x,y);
end
end
  1 个评论
Sean de Wolski
Sean de Wolski 2013-5-20
Have you used the profiler to run your code and identify bottlenecks?

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回答（1 个）

David Sanchez
David Sanchez 2013-5-20
1st: initialize w1
w1 = zeros(nx,ny,3);
The two inner loops do not make use of k, the outer loop index. Why do you have them there? Take them out.
for x=1:nx
for y=1:ny
w1(x,y,:)=w0(mod1(x+y,nx),mod1(x+2*y,ny),:);
end;
end;
what's this for?
for k=1:4
%t=1:1:nx;v=1:1:ny;
%w1(t,v,:)=w0(mod1(t+v,nx),mod1(t+2*v,ny),:);
w0=w1;
end;
  1 个评论
Bob
Bob 2013-5-20
the outer loops k=1:4 is also important,it means iteration of the w1. I must iterate w1 for more than 50 times. so the speed must improve. now if we run the code
for x=1:nx
for y=1:ny
w1(x,y,:)=w0(mod1(x+y,nx),mod1(x+2*y,ny),:);
end;
end;
20 seconds is necessary. it's so bad. please help me!! i have already had no idea!!

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