Matlab matrix operations without loops

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Florin
Florin 2013-1-28
Hello. I have an issue with a code performing some array operations. It is getting to slow, because I am using loops. I am trying for some time to optimize this code and to re-write it with less or without loops. Until now unsuccessful. Can you please help me solve this:
YVal = 1:1:100000;
M_MAX = 1000;
N_MAX = 2000;
clear YTemp
tic
for M=1:1:M_MAX
for N = 1:1:N_MAX
YTemp(M,N) = sum(YVal (N+1:N+M) ) - sum(YVal (1:M) );
end
end
For large N_MAX and M_MAX the execution time of these two loops is very high. How can I optimize this?
Thank you,
Florin

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采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2013-1-28
编辑：Azzi Abdelmalek 2013-1-28
Try his code, 200 faster
YVal = 1:1:100000;
M_MAX = 1000;
N_MAX = 2000;
tic
som1=zeros(1,M_MAX+N_MAX);
YTemp=zeros(M_MAX,N_MAX);
for k=1:M_MAX+N_MAX
som1(k)=sum(YVal(1:k));
end
for M=1:1:M_MAX % Number of accumulated periods
som2=som1(M+1)-YVal(1);
for N = 1:1:N_MAX % statistic
YTemp(M,N) = som2- som1(M);
som2=som2+YVal(N+M+1)-YVal(N+1);
end
end
toc

更多回答（5 个）

Teja Muppirala
Teja Muppirala 2013-1-28
Your entire script is equivalent to this:
M_MAX = 1000;
N_MAX = 2000;
YTemp = (1:M_MAX)'*(1:N_MAX);
  3 个评论
显示 1更早的评论
Teja Muppirala
Teja Muppirala 2013-1-28
Ah, I see, I should have read that. This works quickly, but it's not very readable.
YVal = rand(1,1e6);
M_MAX = 1000;
N_MAX = 2000;
YS = cumsum(YVal);
YTemp = bsxfun(@minus, YS( bsxfun(@plus,(1:N_MAX),(1:M_MAX)') ) , YS(1:N_MAX));
YTemp = bsxfun(@minus, YTemp, YS(1:M_MAX)');
Florin
Florin 2013-1-29
This is what I was looking for (regarding code optimization without using loops)! Thank you.

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Thorsten
Thorsten 2013-1-28
Ytemp = [1:M_MAX]'*[1:N_MAX];
  1 个评论
Florin
Florin 2013-1-28
Hello!
Thank you for the fast answer. This works perfectly in case if
YVal = 1:1:100000;
Though, if I change this to
YVal = 1:2:100000 or rand(1, 100000)
this will not work any more
I Tried:
Ytemp = [YVal(1:M_MAX)]'*[YVal(1:N_MAX)];

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Azzi Abdelmalek
Azzi Abdelmalek 2013-1-28
编辑：Azzi Abdelmalek 2013-1-28
You can begin by pre-allocating
YTemp=zeros(M_MAX,N_MAX);
YVal = 1:1:100000;
M_MAX = 1000;
N_MAX = 2000;
YTemp=zeros(M_MAX,N_MAX);
for k=1:M_MAX
som1(k)=sum(YVal(1:k));
end
tic
for M=1:1:M_MAX % Number of accumulated periods
for N = 1:1:N_MAX % statistic
YTemp(M,N) = sum(YVal(N+1:N+M)) - som1(M);
end
end
toc
This code is three times faster
Jan
Jan 2013-1-28
编辑：Jan 2013-1-28
No, the code is not slow due to the loops, but due to a missing pre-allocation and repeated work. Please measure the speed of this:
YVal = 1:100000;
M_MAX = 1000;
N_MAX = 2000;
YTemp = zeros(M_MAX, N_MAX);
tic
a = 0;
for M = 1:M_MAX % Number of accumulated periods
a = a + YVal(M); % sum(YVal(1:M))
b = a;
for N = 1:N_MAX % statistic
b = b - YVal(N) + YVal(N+M); % sum(YVal(N+1:N+M))
YTemp(M, N) = b - a;
end
end
toc
[EDITED] In fact, the code can be simplified:
YVal = 1:100000;
M_MAX = 1000;
N_MAX = 2000;
YTemp = zeros(M_MAX, N_MAX);
for M = 1:M_MAX % Number of accumulated periods
b = 0;
for N = 1:N_MAX % statistic
b = b - YVal(N) + YVal(N+M); % sum(YVal(N+1:N+M))
YTemp(M, N) = b;
end
end
  2 个评论
Florin
Florin 2013-1-28
编辑：Florin 2013-1-28
Hmmm... At some point I tried to to some pre-allocation, but not so deep.
Your code, on my machine, is ~18 times faster. Thank you! Next time I will remember this lesson.
Jan
Jan 2013-1-28
Here only the ZEROS call pre-allocates. Avoiding the repeated summation helps also according to the simple rule, that all repeated work wastes time - as in the real life also.

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Florin
Florin 2013-1-28
编辑：Florin 2013-1-28
Both answers from Azzi Abdelmaleka and Jan Simon are very helpful, regarding the matter of runtime optimization. Though, I am still curious if you can do this without loops.
Thank you guys and have a nice day!
  1 个评论
Jan
Jan 2013-1-28
Does my answer reply the correct result? I do not have access to Matlab currently, but if it is correct, there is no dependency to "a" in the inner loop: At first "b" is initialized to "a", than "a" is subtracted in each iteration. Therefore I could imagine, that 2 CUMSUMs could be sufficient, but I cannot try this at the moment.

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