(Speed Challenge) Any faster method to calculate distance matrix between rows of two matrices, in terms of Euclidean distance?

First of all, this is NOT the problem of calculating Euclidean distance between two matrices.

Assuming I have two matrices x and y, e.g.,

set.seed(1)
x <- matrix(rnorm(15), ncol=5)
y <- matrix(rnorm(20), ncol=5)

where

> x
           [,1]       [,2]      [,3]       [,4]       [,5]
[1,] -0.6264538  1.5952808 0.4874291 -0.3053884 -0.6212406
[2,]  0.1836433  0.3295078 0.7383247  1.5117812 -2.2146999
[3,] -0.8356286 -0.8204684 0.5757814  0.3898432  1.1249309

> y
            [,1]       [,2]        [,3]       [,4]        [,5]
[1,] -0.04493361 0.59390132 -1.98935170 -1.4707524 -0.10278773
[2,] -0.01619026 0.91897737  0.61982575 -0.4781501  0.38767161
[3,]  0.94383621 0.78213630 -0.05612874  0.4179416 -0.05380504
[4,]  0.82122120 0.07456498 -0.15579551  1.3586796 -1.37705956

Then I want to get distance matrix distmat of dimension 3-by-4, where the element distmat[i,j] is the value from norm(x[1,]-y[2,],"2") or dist(rbind(x[1,],y[2,])).

• My code

distmat <- as.matrix(unname(unstack(within(idx<-expand.grid(seq(nrow(x)),seq(nrow(y))), d <-sqrt(rowSums((x[Var1,]-y[Var2,])**2))), d~Var2)))

which gives

> distmat
         [,1]     [,2]     [,3]     [,4]
[1,] 3.016991 1.376622 2.065831 2.857002
[2,] 4.573625 3.336707 2.698124 1.412811
[3,] 3.764925 2.235186 2.743056 3.358577

but I don't think my code is elegant or efficient enough when with x and y of large number of rows.

• Objective

I am looking forward to a much faster and more elegant code with base R for this goal. Appreciated in advance!

• Benchmark Template (in updating)

For your convenience, you can use the following for benchmark to see if your code is faster:

set.seed(1)
x <- matrix(rnorm(15000), ncol=5)
y <- matrix(rnorm(20000), ncol=5)
# my customized approach
method_ThomasIsCoding_v1 <- function() {
  as.matrix(unname(unstack(within(idx<-expand.grid(seq(nrow(x)),seq(nrow(y))), d <-sqrt(rowSums((x[Var1,]-y[Var2,])**2))), d~Var2)))
}
method_ThomasIsCoding_v2 <- function() {
  `dim<-`(with(idx<-expand.grid(seq(nrow(x)),seq(nrow(y))), sqrt(rowSums((x[Var1,]-y[Var2,])**2))),c(nrow(x),nrow(y)))
}
method_ThomasIsCoding_v3 <- function() {
  `dim<-`(with(idx1<-list(Var1 = rep(1:nrow(x), nrow(y)), Var2 = rep(1:nrow(y), each = nrow(x))), sqrt(rowSums((x[Var1,]-y[Var2,])**2))),c(nrow(x),nrow(y)))
}
# approach by AllanCameron
method_AllanCameron <- function()
{
  `dim<-`(sqrt(rowSums((x[rep(1:nrow(x), nrow(y)),] - y[rep(1:nrow(y), each = nrow(x)),])^2)), c(nrow(x), nrow(y)))
}
# an existing approach by A. Webb from https://stackoverflow.com/a/35107198/12158757
method_A.Webb <- function() {
  euclidean_distance <- function(p,q) sqrt(sum((p - q)**2))
  outer(
    data.frame(t(x)),
    data.frame(t(y)),
    Vectorize(euclidean_distance)
  )
}

# your approach
method_XXX <- function() {
  # fill with your approach
}

bm <- microbenchmark::microbenchmark(
  method_ThomasIsCoding_v1(),
  method_ThomasIsCoding_v2(),
  method_ThomasIsCoding_v3(),
  method_AllanCameron(),
  # method_A.Webb(),
  #method_XXX(),
  unit = "relative",
  check = "equivalent",
  times = 10
)
bm

such that

Unit: relative
                       expr      min       lq     mean   median       uq      max neval
 method_ThomasIsCoding_v1() 1.684104 1.584569 1.611193 1.586154 1.638271 1.622165    10
 method_ThomasIsCoding_v2() 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000    10
 method_ThomasIsCoding_v3() 1.011889 1.017362 1.012064 1.015376 1.026982 1.004885    10
      method_AllanCameron() 1.077066 1.014809 1.029347 1.027979 1.035933 1.029244    10