Code optimization (3 line function)

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Ronald van den Berg
Ronald van den Berg el 28 de En. de 2015
Comentada: Titus Edelhofer el 28 de En. de 2015
I need a function that efficiently applies a reflective bound to a signal: when the signal reaches the bound, it stays there until the signal increases again (example below).
I wrote a short function that applies such a bound to all signals in a matrix (code below).
My question is: does someone know a way to speed up this code? I'm applying it to matrices of size 5000x5000, which takes about a second per matrix. Since i have to do this on thousands of matrices, it would be nice to speed up the function. Thanks!
function Y = reflect(Y,b)
% apply reflecting bound to signals in matrix Y; columns are signals, rows time steps
mhist = zeros(1,size(Y,2)); % history of shifts applied in prev time steps
for ii=1:size(Y,1) % loop over time steps
m = max(b-Y(ii,:)-mhist,0); % compute required shift for current time step
Y(ii,:) = Y(ii,:) + mhist + m; % apply shift
mhist = mhist + m; % add shift to history
end
Here is an example of its output (red=original, black=after applying reflective bound at Y=-50):
The plot was produced using the following code:
rng(1);
X=0:.1:100;
Y=cumsum(normrnd(0,5,size(X)))-X;
plot(X,Y,'r');
hold on;
plot(X,reflect(Y',-50),'k-');
plot([X(1) X(end)],[-50 -50],'b');
  2 comentarios
luc
luc el 28 de En. de 2015
Pre-allocating matrices with nan vallues.
m, Y and mhist change within the loop, you know their final size so before entering the loop try to create them and fill them with nan vallues.
function:nan(3,3)=[nan nan nan;nan nan nan;nan nan nan]
Ronald van den Berg
Ronald van den Berg el 28 de En. de 2015
Editada: Ronald van den Berg el 28 de En. de 2015
Luc, thanks for your response, but i'm afraid this won't help. Preallocation helps when variables change size within a loop, but m, nor Y, nor mhist does so (m and mhist are vectors, Y is the input matrix which is changed row by row in the loop).

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Sean de Wolski
Sean de Wolski el 28 de En. de 2015
Editada: Sean de Wolski el 28 de En. de 2015
If you have MATLAB Coder, this could be a potentially good candidate for C code generation and MEXing. It involves a loop with a persistent state that's updated on each iteration, something compiled languages are often faster with.
If you don't, post a zip file with the matrix and I'll benchmark it for you.
  1 comentario
Ronald van den Berg
Ronald van den Berg el 28 de En. de 2015
Thanks, this is exactly what we're trying now. And it seems to be pretty fast (about to do a comparison now) -- i don't think any clever .m function would be able to beat this good old C solution :)

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Titus Edelhofer
Titus Edelhofer el 28 de En. de 2015
Hi,
one thing that comes to my mind: can you swap rows and columns, i.e., transpose your input matrix?
function Y = reflect(Y,b)
% apply reflecting bound to signals in matrix Y; columns are signals, rows time steps
mhist = zeros(size(Y,1),1); % history of shifts applied in prev time steps
for ii=1:size(Y,2) % loop over time steps
m = max(b-Y(:,ii)-mhist,0); % compute required shift for current time step
Y(:,ii) = Y(:,ii) + mhist + m; % apply shift
mhist = mhist + m; % add shift to history
end
That should be faster, since MATLAB likes columns better than rows ...
Titus
  2 comentarios
Ronald van den Berg
Ronald van den Berg el 28 de En. de 2015
Oh, i didn't know that column operations are faster than row (this may have to do with the order in which matrix elements are stored in memory?). I'll try this. Even if it doesn't help here (since we have a mex solution now), this could be very useful to keep in mind for future functions. Thanks!
Titus Edelhofer
Titus Edelhofer el 28 de En. de 2015
I'm curious to see the speed up you gain with the MEX/MATLAB Coder approach. And yes, matrices are stored columnwise in MATLAB (you see this e.g. with
A = magic(4);
A(:)
Therefore both from programming standpoint but also from memory access (cache!) working on columns is faster.
Titus

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