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

5 次查看(过去 30 天)
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.

采纳的回答

Jos (10584)
Jos (10584) 2015-3-29
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??

更多回答(3 个)

Jos (10584)
Jos (10584) 2015-3-30
Take a look at NEARESTPOINT as this function does exactly what you want and is pretty damn fast, if I say so myself :-)

Roger Stafford
Roger Stafford 2015-3-29
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;
  4 个评论
Roger Stafford
Roger Stafford 2015-3-30
@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.

请先登录,再进行评论。


daniel mitchell
daniel mitchell 2022-1-2
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

类别

Help CenterFile Exchange 中查找有关 Matrix Indexing 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by