Azzera filtri
Azzera filtri

Very slow for loop

4 visualizzazioni (ultimi 30 giorni)
Rafael
Rafael il 31 Dic 2011
Hi, I am new to Matlab, i have more experience with C/C++
I am trying run this simple loop:
x=zeros([10001,10001,2]);
for t=1:2
for z =1:10000
for i =1:10000
x(z,i,t)=i+z+t+2;
end
end
end
x(5,5,1)
But is taking a while..imagine many of then inside a single algorithm. in C++ this loop runs in a few seconds...
Is there any other way to run this loop in Matlab faster???
Sorry if my question is silly, first time running Matlab. I tried some vectorization but for nested loops things get really complicated, especially with different loop sizes.
Thanks
  3 Commenti
Daniel Shub
Daniel Shub il 31 Dic 2011
Why should that be any faster/slower than 1:10000? The variable x is already intialized to a size that handles either case.
Andrew Newell
Andrew Newell il 31 Dic 2011
The point of my question is that the values in each row and column increase until they suddenly drop to zero at the end. I'm guessing that is not what Rafael intended.

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Risposta accettata

Daniel Shub
Daniel Shub il 31 Dic 2011
You might be able to get a slight performance boost by changing how the memory is acessed ...
Basically x(z,i,t) might not be the same as x(t,z,i) or x(i,z,t).
  1 Commento
Rafael
Rafael il 31 Dic 2011
Hi Daniel, Thanks!!! Indeed improved the permornce by 2-4 seconds but is still very slow, almost 25 seconds, on my machine (not a powerful one though). Running it using C++ on GCC 4.6 (with architecture and performance flags optimized) it runs in 3 seconds. The difference is huge. Thanks anyway and happy new year (not yet in Brazil - almost 5 hours to go)!!!!

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Più risposte (3)

the cyclist
the cyclist il 1 Gen 2012
x = bsxfun(@plus,bsxfun(@plus,1:10000,(1:10000)'),permute(1:2,[1 3 2]))+2;
  1 Commento
Rafael
Rafael il 1 Gen 2012
Thanks!, indeed much faster!

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Andrew Newell
Andrew Newell il 31 Dic 2011
For a problem where the numbers count upward monotonically in each direction, here is a compact and fast solution:
y = repmat(2:10002,10001,1);
x = cat(3,y+y',y+y'+1);
On my computer, your code takes about 25 seconds and mine takes 6 seconds.
  2 Commenti
Rafael
Rafael il 31 Dic 2011
Hi, thanks.
As I said before, according to your feedback I have to avoid for loops in my code. Will be hard to translate C++ codes for Matlab if this is true.
Andrew Newell
Andrew Newell il 1 Gen 2012
I think that you mostly can leave loops in because MATLAB uses just-in-time compiling on loops. However, as far as I know this compiling isn't done on arrays of dimension greater than 2.

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Jan
Jan il 1 Gen 2012
If the shown code is your original problem, and not a simplification, I'm in doubt, that it is an efficient approach: You occupy 1.6GB memory with values, which are very easy to calculate dynamically.
Note: The reference "x(z,i,t)" requires two multiplications, while "i+z+t+2" uses 3 additions only.
  1 Commento
Rafael
Rafael il 1 Gen 2012
You are right. The intention is to make a n-by-n-2 array.

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