How to pair two set of coordinates together based on their Euclidean distance?

Hi! I got two sets of pixels as my feature points, one is red and another is blue, I want to use Euclidean distance between two features as the metric: if the distance is below a threshold, make the two features as a pair of correspondence. The threshold could be 1 or 2 (pixels). I want to count the number of correspondence pairs. Thank you so much if you can help meT

The image file is the "image.mat", the red pixel coordinates is in "red_pixel_coordinate.mat", and the blue pixel coordinates is in "blue_pixel_coordinate.mat".

Answer 1

[Matt J]

See matchpairs.

The cost matrix can be obtained using pdist2.

[Yuhong]

Thank you very much for the answer! But I'm sorry I don't understand how can this two functions be applied to my problem, could you please give me an example? Thank you!

[Torsten]

pdist2 gives you the distances of each red point from each blue point.

After this, you can sort out the elements of this matrix with values less than 1 (or 2) pixel distance.

[Matt J]

Example:

red=load('red_pixel_coordinate.mat').red_coordinates';
blue=load('blue_pixel_coordinate.mat').blue_coordinates';

Cost=pdist2(red,blue);
Cost(Cost>2)=inf;  %don't match for distance>2

M=matchpairs(Cost,1e5);

[Yuhong]

Hi thank you very much, M returns a 39by2 array, but what does that mean, still little confused;) thank you

[Matt J]

From the the doc:

M — Matches

matrix

Matches, returned as a matrix. M is a p-by-2 matrix, where M(i,1) and M(i,2) are the row and column indices of a matched pair in the cost matrix. The rows of M are sorted with the second column in ascending order.

• Each row and column can be matched a single time only, so each M(i,1) value and each M(i,2) value is unique.
• M contains p matches, and p is less than or equal to the maximum number of matches min(size(Cost)).
• The cost of the matches in M is sum([Cost(M(1,1),M(1,2)), Cost(M(2,1),M(2,2)), ..., Cost(M(p,1),M(p,2))]).

[Yuhong]

Hi Matt, problem solved, thank you so much so much.