How to find 2 closest numbers in a stream of arrays of the same size?

Hi, I want to get the index and the numbers that is closest to 2 points that i select. For example, my input is

My input is:

   12.6600   83.5567
   10.9106   87.1264
    8.5950   91.9967

For each point, i want to find the closest matching point in a larger array and its corresponding index. My larger array is :

   12.0990   83.3643
   12.0963   83.3725
   12.0935   83.3801
   12.0905   83.3883
   10.7051   87.1957
   10.7024   87.2033
   10.6993   87.2115
    8.5992   91.9872
    8.5954   91.9954
    8.5920   92.0029
    8.5882   92.0105
    8.5844   92.0187
    8.5807   92.0263
    8.5772   92.0338
    8.5735   92.0421

The input numbers can also be negative. Can you please suggest if there is any inbuilt function which can do the search?

[yashvin]

These values are actually..latitude and longitude coordinates that i am trying to match

[Image Analyst]

I see the two input arrays, the small 3-row one, and the larger 15-row one. But I don't know which of those 18 points are the "2 points that i select" . Which two are you selecting? And which of the remaining 16 points in the two arrays do you want to check the distance to? For one example, both points are in the small array and you want to compare those to only points in the big array? Or maybe, you select one point in the small array and one point in the big array, and you want to find distances from each of those two points to all the other points in both arrays? Or what??? Please explain.

[yashvin]

For each point in the small array, i want to know to which one of ANY of the large array points , is it the closest. Actually, the left column is latitude and right one is longitude. All the values are in degrees.

I am trying to match each waypoint of the smallest array to see to which point in the largest array does it correspond!

Answer 1

[Walter Roberson]

xdiff = bsxfun(@minus, smallarray(:,1), largearray(:,1).');
ydiff = bsxfun(@minus, smallarray(:,2), largearray(:,2).');
distmat = sqrt(xdiff.^2 + ydiff.^2);
[closestdist, closestidx] = min(distmat,2);

and now closestidx is a vector the same length as the number of rows in smallarray, with the each element giving the index of the row in largearray that is closest (euclidean distance) to the row in the smallarray.

Note that this is a Euclidean distance calculation, and does not take into account crossing +/- 180, and does not take into account the variation in longnitude as latitude increases. If those are important than a different formula would be needed: http://www.mathworks.com/help/map/ref/distance.html

[LAT1, LAT2] = ndgrid(smallarray(:,1), largearray(:,1));
[LONG1, LONG2] = ndgrid(smallarray(:,2), largearray(:,2));
LL1 = [LAT1(:), LONG1(:)];
LL2 = [LAT2(:), LONG2(:)];
distmat = zeros(size(LAT1));
for K = 1 : size(LL1,1)
  distmat(K) = distance(LL1(K,:), LL2(K,:));
end
[closestdist, closestidx] = min(distmat,2);

Untested code; I do not have the mapping toolbox.

[yashvin]

If A is the small array and B is the large array, is this correct?

A=[lat_pl',long_pl']
B=[Lat_act',Long_act']
waypoint_closest_act=[ones(1,length(A))]
for i=1:length(A)
First_pl=A(i,:)
distances = sqrt(sum(bsxfun(@minus, B,First_pl).^2,2));
Index_closest=find(distances==min(distances))
waypoint_closest_act(i)=B(find(distances==min(distances)),:);
end

[yashvin]

Can you please a give a simply differece where the use of euclidean distance and the use of distance from map leads to a different result?

Please find below my code

 A=[-0.8147,0.9058]
 B=[-0.9575,0.7922]
distances_euclidean = sqrt(sum(bsxfun(@minus, B, A).^2,2));
distance_map=distance(A,B)

Both leads to same value? so please give an example where it is different?