(Speed Challenge) Any faster way to compute distance matrix in terms of generic Hamming distance?

Question

I am looking for a more efficient way to get the distance matrix in terms of Hamming distance.

Backgrounds

I know there is a function hamming.distance() from package e1071 to compute the distance matrix, but I suspect it might be very slow when involving a large matrix with many rows, since it applied nested for loops for computation.

So far I have a faster way (see methodB) in the code below. However, it is only suitable for in the binary domain, i.e., {0,1}^n. However, it is unavailable when encountering domains consisting of more than 2 elements, i.e., {0,1,2,...,K-1}^n. In this sense, methodB is not for generic hamming distance.

Objective

My objective is to find a approach having the following features:

• composed by functions only from base R (not using Rcpp to rewrite function for speeding up)
• faster than my approach methodB() for the special case k=2
• can be generalized for any positive integer k
• outperform the speed of hamming.distance() from package e1071

My code

library(e1071)
# vector length, i.e., number of matrix
n <- 7
# number of elements to consist of domain {0,1,...,k-1}^n
k <- 2
# matrix for computing hamming distances by rows
m <- as.matrix(do.call(expand.grid,replicate(n,list(0:k-1))))

# applying `hamming.distance()` from package "e1071", which is generic so it is available for any positive integer `k`
methodA <- function(M) hamming.distance(M)
# my customized method from base R function `dist()`, which is not available for cases `k >= 2`
methodB <- function(M) as.matrix(round(dist(M,upper = T,diag = T)**2))

and the benchmark gives

microbenchmark::microbenchmark(
  methodA(m),
  methodB(m),
  unit = "relative",
  check = "equivalent",
  times = 50
)

Unit: relative
       expr      min       lq   mean   median       uq      max neval
 methodA(m) 33.45844 33.81716 33.963 34.30313 34.92493 14.92111    50
 methodB(m)  1.00000  1.00000  1.000  1.00000  1.00000  1.00000    50

Appreciated in advance!

Answer 1

methodM <- function(x) {
  xt <- t(x)
  sapply(1:nrow(x), function(y) colSums(xt != xt[, y]))
}
microbenchmark::microbenchmark(
  methodB(m), methodM(m),
  unit = "relative", check = "equivalent", times = 50
)
# Unit: relative
#       expr  min       lq     mean   median       uq      max neval cld
# methodB(m) 1.00 1.000000 1.000000 1.000000 1.000000 1.000000    50  a 
# methodM(m) 1.25 1.224827 1.359573 1.219507 1.292463 4.550159    50   b