Row sums by vector indices without for loop?
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Best illustrated by example
M=rand(5,10);
s=[1 3; 5 8; 9 10];
for i=1:size(s,1)
sm(:,i)=sum(M(:,s(i,1):s(i,2)),2);
end
Is there a one-liner or short-cut for the for loop? I do not know how large M or s is ahead of time, or their values. Above code shown only for illustration of problem trying to solve.
Thanks!
edit: Corrected typo of 'm' in for-loop to 'M'
采纳的回答
Walter Roberson
2011-2-7
0 个投票
If your accuracy requirements are not high and your values are not extreme, you can use cumsum(), after which it becomes the difference of the cumsum indexed at two positions -- a transformation that gets around the problem that your indices do not always designate the same number of elements to be summed per row.
If I recall collectly, Bruno, and Matt Fig, both have posted non-cumsum() versions of this task in the past (it has come up before in cssm discussions.)
2 个评论
Oleg Komarov
2011-2-7
There's a submission by Bruno, mcolon, which will allow you to create the indexes and use accumarray avoiding the multiple call to sum.
Mike Sheppard
2011-2-8
更多回答（1 个）
Bruno Luong
2011-2-7
Yes, to be more specific my mcolon on FEX
can be used like this:
% Data
M = ceil(10*rand(5,10))
s = [1 3; 5 8; 9 10];
% Engine
[a i] = mcolon(s(:,1),s(:,2));
[r c] = ndgrid(1:size(M,1),i);
Ma = M(:,a);
sm = accumarray([r(:) c(:)], Ma(:))
Bruno
2 个评论
Jan
2011-2-7
Did you compare the speed with the OP's FOR loop method with adding a pre-allocation?
Bruno Luong
2011-2-8
Following Jan's question: I make the following test and I get the 25% acceleration by using MCOLON
Time relative for-loop/mcol = 1.26013/1
% Code
ntest = 10;
t = zeros(2,ntest);
for n=1:ntest
s = ceil(10*rand(1,2*10000));
s = cumsum(s);
s = reshape(s, 2, [])';
M=rand(5,s(end));
% for loop Engine
tic
sm = zeros(size(M,1),size(s,1));
for i=1:size(s,1)
sm(:,i)=sum(M(:,s(i,1):s(i,2)),2);
end
t(1,n) = toc;
% mcolon Engine
tic
[a c] = mcolon(s(:,1),s(:,2));
[r c] = ndgrid(1:size(M,1),c);
Ma = M(:,a);
sm = accumarray([r(:) c(:)], Ma(:));
t(2,n) = toc;
end
t = min(t,[],2);
t = t/min(t);
fprintf('Time relative for-loop/mcol = %g/%g\n', t);
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2011-2-7
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