How to vectorize a for-loop?

Hello!

I would appreaciate any help vectorizing the for loop below. The problem i cannot solve is how to take out the constants from wdrive and multiply it with the rest, according to the loop below. I want to become better at vectorizing my MATLAB code so I would really appreciate any help! Thanks alot!

%S and R are square matrices (404*404).
%fc is a column vector

wdrive=0:1:800;
a2=zeros(2*dofs,(length(wdrive)));
b2=zeros(2*dofs,(length(wdrive)));
Amp=zeros(1,length(wdrive));

 SRS=S*(R\S);
SRfc=S*(R\fc);

for i=1:length(wdrive)
  a2(:,i)=(wdrive(i)^2*SRS+R)\(fs-wdrive(i)*SRfc);
  b2(:,i)=(R)\(fc+wdrive(i)*S*a2(:,i));
  Amp(i)=sqrt(a2(dofs-1,i)^2+b2(dofs-1,i)^2);
end

[Matt J]

What is "fs"?

[Samuel Hammarberg]

Ops! Forgot to specify that. Fc is a column vector with the same size as Fs.

Answer 1

[Matt J]

I don't think you can vectorize everything, but you can cut down on the operations done inside the loop,

     Rfc=R\fc;
     SRS=S*(R\S);
     SRfc=S*(Rfc);

     rhs=bsxfun(@minus,fs,SRfc*wdrive);

    for i=1:length(wdrive)
      a2(:,i)=(wdrive(i)^2*SRS+R)\rhs(:,i);
    end

     b2=Rfc+R\S*bsxfun(@times, wdrive, a2);
     Amp=abs(a2(dofs-1,:)+i*b2(dofs-1,:));

[Samuel Hammarberg]

Tank you for answering! I will make sure to try it your way once I get home!

[Samuel Hammarberg]

I have now tried and had to change just one little thing. This is what i ended up with, it's not too pretty but atleast it's faster now.

    Rfc=R\fc;

    SRS=S*(R\S);

    SRfc=S*(Rfc);

    rhs=bsxfun(@minus,fs,SRfc*wdrive);

        for i=1:length(wdrive)
            a2(:,i)=(wdrive(i)^2*SRS+R)\(rhs(:,i));
        end

    b2=R\(bsxfun(@plus,fc,(bsxfun(@times,wdrive,S*a2))));

    Amp=abs(a2(dofs-1,:)+i*b2(dofs-1,:));

[Matt J]

Samuel Hammarberg commented:

It would still be sweet if you, somehow, could get rid of the for loop altogether. So if some one could crack that, that would be great!

Thank you very much!

[Matt J]

The for-loop isn't slowing you down significantly. You just have a lot of matrix inversions

 (wdrive(i)^2*SRS+R)

to do. There's no simplification to this that I can see, unless there is some special structure in S and R that you haven't mentioned.

[Matt J]

You could also try parfor, if you have the Parallel Computing Toolbox.

Answer 2

[Samuel Hammarberg]

Relocated to Comment by Matt J