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Loop that I am trying to get rid off

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To follow up on my previous thread regarding a very time consuming routine, the main source of slowness I believe is the following: Rating agencies such as Moody and Standard and Poors report ratings for bonds outstanding. This rating changes from time to time. So I have in a matrix X0: column 1 is a bond ID Column 2 is the rating beginning date Column 3 is the rating Column 4 is the rating ending date for example for Bond ID 1 I have the following inputs: Ratings=[1 726834 3 728000;1 728000 2 729500;1 729500 3 731000]; then I have a vector date of 20 year daily data so I am running a loop that looks like this for a single bond (I have 15000 bonds):
for i=1:length(date)
x1=find(X0(:,2)<=date(i) & X0(:,4)>date(i));
if isempty(x1)
R(i)=nan;%Rating
else R(i)=X0(x1,3);
end
end
Is there a way to get rid of this loop ?

Accepted Answer

Oleg Komarov
Oleg Komarov on 6 Aug 2011
% Preallocate R already with NaNs
R = NaN(size(date));
% Now we want to test how many dates belong to each rating period, to do this construct the edges of each period
edges = unique(X0(:,[2,4]));
% Total length or rating history (excluded last day of last period)
idx = date >= edges(1) & date < edges(end);
% Count ho many dates fall per period (last bucket is last day of last rating period, which is excluded)
len = histc(date, edges);
% Use run-length decoding (vectirized)
R(idx) = rude(len(1:end-1), X0(:,3));
Rude on FEX: here
  3 Comments
joseph Frank
joseph Frank on 6 Aug 2011
thanks. I eliminated the duplication

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More Answers (2)

Fangjun Jiang
Fangjun Jiang on 6 Aug 2011
I wonder if you can use interp1(). Basically, it is a look up operation, right? It will fill NaNs for dates outside of Ratings. But there should be no gap of dates in Ratings.
Use your own discretion. BTW, don't use 'date' as the variable name. It's a function.
Ratings=[1 726834 3 728000;1 728000 2 729500;1 729500 3 731000];
X=[Ratings(:,2),Ratings(:,4)-1]';
X=X(:);
Y=[Ratings(:,3),Ratings(:,3)]';
Y=Y(:);
Dates=726830:731005;
R=interp1(X,Y,Dates,'nearest');
  1 Comment
Oleg Komarov
Oleg Komarov on 6 Aug 2011
I like your approach, more comprehensible.

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Daniel Shub
Daniel Shub on 6 Aug 2011
Loops in MATLAB are much faster nowadays then they used to be. Without timing solutions, it is not always obvious what way is fastest. Sometimes you just want to get to fast enough.
It looks to me like you could loop over the rows of X0 instead of the dates.

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