using colon instead of loops

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Baz
Baz on 18 Mar 2013
Dear Matlab users,
I have recently realised that using loops instead of built-in functions makes my code really slow. And I tried to solve some problem without using the loops.
here we create a fluid dynamics 3d vector with a number of spatial points M
w = zeros(3, M);
u = zeros(3, M);
f = zeros(3, M);
w = calc_w_from_init(ro, U, E, M);
u = calc_u_from_w (w(:,1), gamma, M);
f = calc_f_from_w (w(:,1), gamma, M);
dt = calc_dt(w(:,1),dx,cfl,gamma,M);
% here comes the main calculation
wtilde = 0.5*(w(:,1:M-1)+w(:,2:M)) - 0.5*dt*(1/dx)*(f(:,2:M)-f(:,1:M-1));
ftilde = calc_f_from_w (wtilde(:,1:M-1), gamma,M);
w(:,2:M) = w(:,1:M-1) - dt*(1/dx)*(ftilde(:,2:M) - ftilde(:,1:M-1));
the point is how can I use the three previous statements and run them they were in one for 1:M loop.???**
  4 Comments
Baz
Baz on 18 Mar 2013
Thank you for your quick reply.
My task was to calculate the vector w(the solution of an advection equation Lax-Wendroff scheme)
here w(: <- this is the 3 element array for ro - density, ro*U <- velocity, E- energy
here is my initial code
for m=1:M
wtilde = 0.5*(w(:,m)+w(:,m+1)) - 0.5*dt*(1/dx)*(f(:,m+1)-f(:,m));
ftilde = calc_f_from_w (wtilde(:,m), gamma,M);
w(:,m+1) = w(:,m) - dt*(1/dx)*(ftilde(:,m+1) - ftilde(:,m));
end
*I want to write that without using the for loop *
Jan
Jan on 19 Mar 2013
But why? Do you assume that a vectorized code is faster here? If so, why?
At first I'd avoid unnecessary calculations:
% Pre-allocate w!!!
c2 = dt / dx;
c1 = 0.5 * c2;
for m = 1:M
wtilde = 0.5*(w(:,m)+w(:,m+1)) - c1 * (f(:,m+1)-f(:,m));
ftilde = calc_f_from_w (wtilde(:,m), gamma, M);
w(:,m+1) = w(:,m) - c2 * (ftilde(:,m+1) - ftilde(:,m));
end
But wait: wtilde is a column vector, such that wtilde(:,m) must crash. So I stop to try optimizing a crashing code.

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Answers (1)

Image Analyst
Image Analyst on 18 Mar 2013
I'm not sure what you're looking for, but they're already vectorized so I don't think you'll get much more speed up by making them into a single statement.
How big is M anyway? Is it like hundreds of millions of voxels or something? Do you have a complete solid volume you need to deal with, or just a short list of a few thousand locations in that volume? How long does the calculation take?

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