how to find nearest values of all elements of a matrix to another matrix in matlab

hi all, i have 2 matrices of different size(11527*1 and 112813*1) .i want to find the nearest values(distance between the two element <= some condition(say <=2)) of each element of matrix 1 (11527*1) with respect to matrix 2 (112813*1). The result should be of the size of matrix 1. Can anyone help me? one corresponding figure is given.

Answer 1

[Jos (10584)]

A = [1 5 7 3 2 8]
B = [4 12 11 10 9 23 1 15]

TMP = bsxfun(@(x,y) abs(x-y), A(:), reshape(B,1,[]))
[D, idxB] = min(TMP,[],2) 

Result = B(idxB)
TFDiffLessThen3 = D < 3 % different outcome than in your example??

[LUI PAUL]

thanks Jos,it works i want to ask more extension about this.if u r agree ......

[Jos (10584)]

You are free to post questions to this forum :-)

Answer 2

[Jos (10584)]

Take a look at NEARESTPOINT as this function does exactly what you want and is pretty damn fast, if I say so myself :-)

http://www.mathworks.com/matlabcentral/fileexchange/8939-nearestpoint--sep-2012-

Answer 3

[Roger Stafford]

With vectors as large as these it could very well be advisable to sort the second vector. Let A be the first 11527 x 1 vector and B the second 112813 x 1 vector.

   T = sort(B);
   [~,ix] = histc(A,[-inf;(T(1:end-1)+T(2:end))/2;inf]);
   Result = T(ix);
   Diff3 = abs(Result-A)<3;

[LUI PAUL]

Thanks Roger for answer but it is showing Error using ==> vertcat CAT arguments dimensions are not consistent.

[Jos (10584)]

Note that this only works for unique elements of B.

[Roger Stafford]

@Lui: "Thanks Roger for answer but it is showing Error using ==> vertcat CAT arguments dimensions are not consistent."

This probably means that you have used a row vector for your second vector in spite of stating that its dimensions are 112813 x 1. Change the first line of my code to:

   T = sort(B(:);

and the last line to:

   Diff3 = abs(Result-A(:))<3;

@Jos: As Lui has stated this problem it doesn't matter if multiple copies of the same value occur in that second vector. Only the values are listed in 'Result', not their locations. Therefore it should work even if the elements are not unique.

[Jos (10584)]

@Roger. True! Apparently, histc does now accept also non-strictly increasing edges ...

Answer 4

[daniel mitchell]

I think this may help:

A = [1 5 7 3 2 8]';
B = [4 12 11 10 9 23 1 15]';
I = knnsearch(B,A);
R = B(I) % Result

R = 6×1

1 4 9 4 1 9

D = abs(A-R) 

D = 6×1

0 1 2 1 1 1