need help optimizing this code...
Show older comments
how can I do this without a loop:
(consider vectors w and s of length n)...
dTot = 0;
for i=1:n
for j=i+1:n
v = w(i) * w(j) * (s(i)) * (s(j));
dTot = dTot + v;
end
end
dTot = dTot * 2;
Accepted Answer
More Answers (4)
Teja Muppirala
on 28 Mar 2012
You can simplify the loop like this:
for i=1:n
v = w(i)*s(i)*sum( w(i+1:n) .* s(i+1:n) );
dTot = dTot + v;
end
dTot = dTot * 2
[EDITED:]
Aha! I knew there had to be an easier way to do this!
ws = w.*s;
Cws = cumsum(ws(n:-1:2));
dTot = 2 * (sum(ws(n-1:-1:1) .* Cws))
3 Comments
Jan
on 28 Mar 2012
In former days dTot+dTot has been faster than dTot*2. Is this still true on modern processors?
Teja Muppirala
on 28 Mar 2012
Is X + X is faster than 2*X? I'm not really sure... but just trying it out, it seems they are both about the same to me.
Andrei Bobrov
on 28 Mar 2012
+1
Jakob Sørensen
on 27 Mar 2012
Wouldn't it work with:
% Turn vectors to matrix form
W1 = repmat(w, [n 1]);
W2 = repmat(w',[1 n]);
% Multiply these
Wm = W1 .* W2;
% Turn vectors to matrix form
S1 = repmat(s, [n 1]);
S2 = repmat(s',[1 n]);
% Multiply these
Sm = S1 .* S2;
% Multiply Wm and Sm matrices
Vm = Wm .* Sm
% Sum up and multiply by 2
V = 2*sum(Vm(:))
Note that I haven't tested this, but i think it should work...
1 Comment
Jakob Sørensen
on 27 Mar 2012
Kinda missed the j = i+1 part, this changes quite a few things. But you might be able to modify the concept. Sorry about that...
Kevin Holst
on 27 Mar 2012
I don't think that will work actually. Try this, it's a little convoluted, but I think it's correct:
% this assumes w and s are row vectors
wm = triu(repmat(w,[n 1]),1);
sm = triu(repmat(s,[n 1]),1);
wsm = wm .* sm;
wsv = w .* s;
dTot = 2*(sum(wsv*wsm));
If you're wanting a one liner that no one will be able to read or understand then:
dTot = 2*(sum((w.*s)*(triu(repmat(w,[n 1]),1).*triu(repmat(s,[n 1]),1))));
Oleg Komarov
on 27 Mar 2012
I would however use a loop:
ws = w.*s;
wssw = bsxfun(@times,ws,ws.');
sum(wssw(~eye(n)))
Categories
Find more on Loops and Conditional Statements in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!