Comparing matrices of different length

I have two matrices of different length. For both matrices the first column is x-coordinate, second column is y-coordinate and third column is height. I would like to compare all points for both matrices and for those points where the distance is less than 100 it should calculate the height difference. Is there a smart way of doing this?

[KL]

Do you want to apply this 'less than 100' condition along both axes?

[KSSV]

Question need to be more clear. Let A and B be the set of points; pick one point from A..say A(i,:), you want to pick all the points in B which are less then 100 units of distance from Ai..and like for each and very point for A?

[JVM]

I will try to make it more clear. Imagine both matrices being x(first column) and y(second column) coordinates with a given height(third column). I want to compare the two matrices for all points and for those points were the distance is less than 100 i want to calculate the height difference. Notice that the coordinates are not given in lat/lon coordinates but in an metric system.

I have already made a simpel script that works based on two for loops but my two matrices are very large so it takes a lot of time to run. My script is as follows

counter = 0;
A1 = % first matrix 450000x3 double
A2 = % second matrix 650000x3 double
for i = 1:length(A1)
  for j = 1:length(A2)
      if sqrt((A2(j,1)-A1(i,1))^2+(A2(j,2)-A1(i,2))^2)<100
          counter = counter + 1;
          height(counter,1) = A1(i,1);
          height(counter,2) = A1(i,2);
          height(counter,3) = abs(A1(i,3)-A2(j,3));
      end
  end
end

I haven't written the actual matrices but I hope you get the idea.

Answer 1

[Stephen23]

If the matrices are not large then you can easily use permute and bsxfun. Here I put the A points down the first dimension, and the B points along the second dimension of the output matrices.

>> A = [0,0,1;10,10,0]; % [X,Y,H]
>> B = [1,1,0;11,11,2;111,111,3]; % [X,Y,H]
>> A3 = permute(A,[1,3,2]);
>> B3 = permute(B,[3,1,2]);
>> H = bsxfun(@minus,A3(:,:,3),B3(:,:,3)); % all height differences

Actually all height differences are in H. If you want to identify the distances between the points, then do this:

>> M = bsxfun(@minus,A3(:,:,1:2),B3(:,:,1:2)); % all X & Y differences
>> D = sqrt(sum(M.^2,3)) % euclidean distance from X & Y differences
D =
   1.4142    15.5563   156.9777
  12.7279     1.4142   142.8356
>> H(D>=100) = NaN % optional
H =
   1    -1   NaN
   0    -2   NaN

[JVM]

Thanks, but my matrices are unfortunately very large.

Answer 2

[Stephen23]

For large matrices you could (possibly) speed up your code by only using one loop, and vectorizing the operations in the inner loop:

A = randi(10000,450000,3);
B = randi(10000,650000,3);
C = {};
for k = 1:size(A,1)
    D = sqrt(...
        (B(:,1)-A(k,1)).^2 + ...
        (B(:,2)-A(k,2)).^2);
    idx = D<100;
    C{end+1} = B(idx,3)-A(k,3);
end

[JVM]

Of course, I haven't even though about that. I will try to run it now, but isn't there a way to vectorizing both operators?

[Stephen23]

If memory was not limited, then the fully vectorized approach would be exactly as I showed in my other answer, using bsxfun. The reason to use one loop is simply because (most likely) you are using a PC and have some GB of memory. Using one loop means that the intermediate arrays are of a size that can actually be stored in memory.