Very slow for loop
조회 수: 4 (최근 30일)
이전 댓글 표시
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
댓글 수: 3
Daniel Shub
2011년 12월 31일
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
2011년 12월 31일
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.
채택된 답변
Daniel Shub
2011년 12월 31일
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).
추가 답변 (3개)
Andrew Newell
2011년 12월 31일
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.
댓글 수: 2
Andrew Newell
2012년 1월 1일
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.
Jan
2012년 1월 1일
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.
참고 항목
카테고리
Help Center 및 File Exchange에서 Waveform Generation에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!