Matrix of pairwise distances

Question

I have a set of points coordinates and I want to use it to generate a matrix of distances. More specifically, I have two sets of points, A of size n and B of size m, given as 2d coordinates and I want to have all Euclidean distances between points from A and points from B and no other distances, in a matrix.

Edit: what if the situation is more complicated: what if I have my matrix but now I want to divide each row of it by the sum of Euclidean distances of the first point from A from all the points in set B: that is, normalise each row of distances. Is there an efficient way to do that?

Answer 1

set.seed(101)
n <- 10; m <- 20
A <- data.frame(x=runif(n),y=runif(n))
B <- data.frame(x=runif(m),y=runif(m))

We want

sqrt((x_{1,i}-x_{2,j})^2+(y_{1,i}-y_{2,j})^2)

for every i=1:n and j=1:m .

You can do this via

dists <- sqrt(outer(A$x,B$x,"-")^2 + outer(A$y,B$y,"-")^2)

which in this case is a 10x20 matrix. In words, we're finding the difference ( "-" is a reference to the subtraction operator) between each pair of x values and each pair of y values, squaring, adding, and taking the square root.

If you want to normalize every row by its sum, I would suggest

norm.dists <- sweep(dists,MARGIN=1,STATS=rowSums(dists),FUN="/")

Answer 2

The dist(...) function in base R will not be helpful, because it calculates the auto-distances (distance from every point to every other point in a given dataset). You want cross-distances. There is a dist(...) function in package proxy which is designed for this.

Using the dataset kindly provided by @BenBolker,

library(proxy)   # note that this masks the dist(...) fn in base R...
result <- dist(A,B)
result[1:5,1:5]
#           [,1]      [,2]      [,3]      [,4]      [,5]
# [1,] 0.5529902 0.7303561 0.1985409 0.6184414 0.7344280
# [2,] 0.7109408 0.9506428 0.1778637 0.7216595 0.9333687
# [3,] 0.2971463 0.3809688 0.4971621 0.4019629 0.3995298
# [4,] 0.4985324 0.5737397 0.4760870 0.5986826 0.5993541
# [5,] 0.4513063 0.7071025 0.3077415 0.4289675 0.6761988