Assuming the data is read into two variables, d1(1:N1,1:3) and d2(1:N2,1:3) by
d1 = xlsread('Matlab Help.xlsx','Random','B2:D31'); % left Set
d2 = xlsread('Matlab Help.xlsx','Random','H2:J51'); % right set
Then the simplest and fastest solution I can think of is this:
% Invert d1 to 3xN1 and copy along the third dimension
% so that d1b is size 3xN1xN2, where N1/N2 are number of rows
% in d1/d2 respectively.
d1b = repmat(d1',1,1,size(d2,1));
% Reshape d2 to be 3x1xN2
d2b = reshape(d2,[3,1,size(d2,1)]);
% Copy d2 along second dimension N1 times, such that d2b is
% also 3xN1xN2, matching size of d1b
d2b = repmat( d2b, 1,size(d1,1),1);
% Compute the 2nd norm (2, second argument of vecnorm)
% of the difference in coordinates (d1b-d2b, first argument of vecnorm)
% along the first dimension of the 3xN1xN2 matrix (1, third argument of vecnorm)
% such that vecnorm returns 1xN1xN2 matrix, the distance between all combinations of d1
% and d2. Then take the minimum along the third dimension (N2) and
% return the value and index
[min_distance, jmin_index] = min( vecnorm(d1b-d2b,2,1) , [],3);
This might be hard to follow. You could also do something like this, but it will be slower
% Predefine as Not-A-Number
min_distance = NaN(size(d1,1),1);
jmin_index = NaN(size(d1,1),1);
% Loop each value in d1
for i=1:size(d1,1)
% Copy d1 so that it has the same size as d2, but with
% each row being the d1 coordinate
d1b = repmat( d1(i,:), size(d2,1),1 );
% Get the minimum of distance of the vector of distances
[min_distance(i), jmin_index(i)] = min( vecnorm(d1b-d2,2,2) );
end
You could even unroll this even more if this is still confusing:
% Loop each row in d1
for i=1:size(d1,1)
% Get the local coordinate
d1_local = d1(i,:);
% Default min distance to a high value, index to NaN
min_distance(i) = 1e9;
jmin_index(i) = NaN;
% Loop each row in d2
for j=1:size(d2,1)
% Get distance
distance = norm( d1(i,:)-d2(j,:) ,2);
% Check if smaller
if (distance< min_distance(i))
min_distance(i) = distance;
jmin_index(i) = j;
end
end
end
But this is a very inelegant solution compared to the first, and it will be hundreds of times slower for very large datasets.
