Find row in Matrix A which is closest two Matrix B

Question

Lets say I have two matrices A and B given by

set.seed(123)
m1 = matrix(runif(10*5), nrow = 10, ncol = 5)
m2 = matrix(runif(10*5), nrow = 10, ncol = 5)

I want to find for each row in matrix A the row in matrix B which is closest to the row in matrix A. I know I can do this by looping over each row in A and comparing it to each row in B like this:

for(i in 1:nrow(m1)){
  dist = 9999
  index = -1
  for(j in 1:nrow(m2)){
    test = sqrt(sum(abs(m1[i,]-m2[j,])))
    
    if (test < dist) {
      dist = test
      index = j
    }
  }
  print(index)
}

I have however a million rows and it takes forever. I'm struggeling to find a efficient way. Any ideas?

Answer 1

Here is one base R solution using apply :

apply(m1, 1, \(x) which.min(sqrt(colSums(abs(x - t(m2)))))) 
#[1]  8  3  2  3  3  1  2  3  6 10

Comparing it with your current solution it fares well :

set.seed(123)
m1 = matrix(runif(10 * 5), nrow = 10, ncol = 5)
m2 = matrix(runif(10 * 5), nrow = 10, ncol = 5)

baseR_sol <- function(m1, m2) {
  apply(m1, 1, \(x) which.min(sqrt(colSums(abs(x - t(m2))))))  
}

for_loop_sol <- function(m1, m2) {
  for(i in 1:nrow(m1)){
    dist = 9999
    index = -1
    for(j in 1:nrow(m2)){
      test = sqrt(sum(abs(m1[i,]-m2[j,])))
      
      if (test < dist) {
        dist = test
        index = j
      }
    }
    print(index)
  }
}


microbenchmark::microbenchmark(
  baseR_sol = baseR_sol(m1, m2), 
  for_loop_sol = for_loop_sol(m1, m2), times = 10L
)

#         expr   min    lq    mean median     uq    max neval
#    baseR_sol 158.0 185.2  865.81 195.35  224.8 6902.8    10
# for_loop_sol 764.6 830.2 1051.29 973.45 1312.0 1348.9    10