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Vectorizing finding indexes

Asked by Sargondjani on 24 Jun 2012
hi, i have a vector X and a vector Y (both in ascending order). for every element in X i want the indexes of the values in Y between which this element lies. Actually, 1 index suffices (preferably the higher bound).
For example if have
Y=[0, 1, 2, 3, 4, 5]
X=[0.1, 2.5, 2.8, 4.1];
then I want to get as a result:
IND = [2 4 4 6]; %the higher bounds of the interval in Y in which the elements of X fall
I can do this with a for loop:
for ix=1:length(X);
IND(1,ix)=min(find(Y>X(ix)))
end
My question is whether it is possible to vectorize this, and how... Many thanks in advance!

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4 Answers

Answer by Walter Roberson
on 24 Jun 2012
 Accepted Answer

[counts, tIND] = histc( X, Y );
IND = tIND + 1; %to get the higher index
Note that this has questionable results in the case where an X is exactly equal to a Y: it will return the bin number of the next Y. You could, though, correct for this with
t = (X == Y(tIND));
IND(t) = IND(t) - 1;
It is also possible to correct for it at the time of histc(), but the code becomes notably more difficult to read.

  3 Comments

Sargondjani on 24 Jun 2012
thanks Walter!! :-)
i havent worked out the rest of the computations, so i dont know what to do when X is equal to some value of Y, but im sure i can work something out...
by the way, it is weird that i can never find this type of functions when i search for it....
Walter Roberson
on 24 Jun 2012
While you are thinking about that, also think about what to do if an X is before or after all of the Y.
See the comments in http://www.mathworks.com/matlabcentral/answers/41814-interpolation-with-high-frequency-financial-data
Sargondjani on 24 Jun 2012
that's not going to happen. i am doing value function iteration using Y as the grid (for the state variable), and i force my policy function, X, to be in/on the grid

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Answer by Jamie Rodgers on 24 Jun 2012

Try This: Your vector
Y=[0:1:1000];
X=[0.1, 2.5, 2.8, 4.1];
Vectorised code
Z1=arrayfun(@(a)intersect(Y,ceil(a)),X);
idx=arrayfun(@(x)find(Y==x),Z1);

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Sargondjani on 24 Jun 2012
this probably works best, because i would actually like to have index of lower bound and index of higher bound... this way i can get them both directly with an understandable method :-)

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Andrei Bobrov
Answer by Andrei Bobrov
on 24 Jun 2012

[idx,~] = find(bsxfun(@eq,ceil(X(:)'),Y(:)))

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Sargondjani on 24 Jun 2012
i really like this but if i change small thing i get results that i dont get!
if i remove the transpose (i would prefer the solution to have the same dimension as X and Y) then i get a weird result
and i get only ones if i remove the (:)
...so unfortunately this solution is probably out of my league

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Answer by Sargondjani on 24 Jun 2012

Hmmmmm, andrei and jamie: do your solutions only work when Y consists of integers??? In my example Y was just integers, but in fact Y is not integers...
when i tried your solutions with Y not being integers i got results that were not correct

  1 Comment

Walter Roberson
on 24 Jun 2012
You are correct: the expressions given by Jamie and Andrei as of the time of my writing this comment, only work for integer Y. My histc() based code does not depend on Y being integer.

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