Vectorizing finding indexes

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!

Answer 1

[Walter Roberson]

[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.

[Sargondjani]

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]

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]

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

Answer 2

[Jamie Rodgers]

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);

[Sargondjani]

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 :-)

Answer 3

[Andrei Bobrov]

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

[Sargondjani]

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

Answer 4

[Sargondjani]

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

[Walter Roberson]

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.