Code optimization (3 line function)
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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 commentaires
luc
le 28 Jan 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
le 28 Jan 2015
Modifié(e) : Ronald van den Berg
le 28 Jan 2015
Réponse acceptée
Plus de réponses (1)
Titus Edelhofer
le 28 Jan 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 commentaires
Ronald van den Berg
le 28 Jan 2015
Titus Edelhofer
le 28 Jan 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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