Speeding up a loop

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

[Rick Rosson]

Is S a row vector or a column vector?

[Rick Rosson]

What is the goal of this algorithm? What do the variables S and X represent?

[Swisslog]

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]

Is the wanted property of C directly related to the sign of the corresponding element of S?

Answer 1

[Jan]

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);

Answer 2

[Swisslog]

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