Speeding up a loop

2013 1 17
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I have made a simple loop that works fine when dealing with small datasets, but takes ages to run when L is 10^6 in size - which unfortunately is the size I need to work with. I'm sure you can tell by the code itself I'm a matlab newbie, so any ideas on how this can be sped up would be very much appreciated. I've read about vectorisation but cannot work out how to vectorise this code.
C=cumsum(S);
L=length(C);
X=zeros(1,L)';
for i=1:L;
if C(i)>min(C(i:L));
X(i);
else
X(i)=1;
end
end

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Rick Rosson
Rick Rosson 2013-1-17
编辑：Rick Rosson 2013-1-17
Is S a row vector or a column vector?
Rick Rosson
Rick Rosson 2013-1-17
编辑：Rick Rosson 2013-1-17
What is the goal of this algorithm? What do the variables S and X represent?
Swisslog
Swisslog 2013-1-17
编辑：Swisslog 2013-1-17
Hi Rick,
I'm dealing with column vectors. I have a column of random numbers (S), which I add up C=(cumsum(S). I want to identify from C intervals where C is increasing (marked as 1 in X), and intervals where it is decreasing (marked as 0 in X). Importantly, if C is decreasing, I also want to set to 0 those values immediately before the decrease which are greater than the minimum value attained during the suceeding decrease - hence: if C(i)>min(C(i:L)). The code works precisely as I want it to, but just takes forever (~1 hour) when Length(S)=10^6.
P.S. As a slight aside, in the parlance of fractals: the aim of this code is esentially to construct a 1D cantor set (X) from a random series (S). In this sense C is akin to a devils staircase
Jan
Jan 2013-1-17
Is the wanted property of C directly related to the sign of the corresponding element of S?

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Jan
Jan 2013-1-17
编辑：Jan 2013-1-17

0 个投票

The main work is done in the repeated determination of the minimum. But this can be avoided easily:
% Cummumaltive minimum in backward direktion:
minC = zeros(size(C));
v = C(end);
for ii = numel(C):-1:1
if C(ii) < v
v = C(ii);
end
minC(ii) = v;
end
X = zeros(L, 1);
X(C <= minC) = 1;
Sorry, I cannot test this currently and I have the feeling that the logic is not correct. But at least the general idea could be helpful.
Btw.: This line wastes time only:
X(i);

更多回答（1 个）

Swisslog
Swisslog 2013-1-17

0 个投票

Thanks Jan, This seems to have done the trick - much faster and I've learned a lot, cheers

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