Effacer les filtres
Effacer les filtres

What is the fastest, circshift or indices ?

9 vues (au cours des 30 derniers jours)
Stéphane
Stéphane le 5 Nov 2014
Modifié(e) : Matt J le 5 Nov 2014
Hi
I'm trying to improve the efficiency of my algorithm. I need to compute something like :
Y(2:end-1,:) = X(2:end-1,:) + X(1:end-2,:) - X(3:end,:)
For Y and X n by m matrices. I was wondering if it was faster to do it as it is above, or using circshift :
Y = X + circshift(X,[-1,0]) - circshift(X,[+1,0]).
I've done a few tests but I can't get a clear idea... The aim is to do this on pretty big matrices on a cluser with multiple cores...
Any idea ? Thanks.
Stéphane
  1 commentaire
Adam
Adam le 5 Nov 2014
I regularly create small test scripts with as many alternative implementations as I can think of, each wrapped in a
timeit(f)
instruction for function handle, f. It is extremely useful for improving understanding as well as getting the (hopefully conclusive) final answer as to which method is fastest. Sounds like you did kind of do that though...

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Matt J
Matt J le 5 Nov 2014
Modifié(e) : Matt J le 5 Nov 2014
Just use conv2
Y=conv2(X,[-1;1;1],'valid');
As for circshift, it is not a builtin function and, if you look inside it
>>type circshift
you will see that it does pretty much the same indexing as you are doing. So, I wouldn't expect it to be a competitive approach.
  4 commentaires
Stéphane
Stéphane le 5 Nov 2014
Second thought...
I've tried in 1D to understand what conv (conv2) does and it seems conv(X,[-1;1]) and the same with circshift doesn't produce the same result. Is it normal or am I doing something wrong ?
(What I want is X(2:end) - X(1:end-1) )
Matt J
Matt J le 5 Nov 2014
Modifié(e) : Matt J le 5 Nov 2014
(What I want is X(2:end) - X(1:end-1) )
That would correspond to
conv(X,[1;-1],'valid')
You could also do
diff(X)

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