Найдите строку в матрице A, которая является ближайшей к матрице B

Вот одно из базовых решений R с использованием apply:

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

Сравнивая его с вашим текущим решением, он работает хорошо:

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

Последующий тест, основанный на решении @Ronak Shah

# Ronak Shah's original solution
f1 <- \() {
    apply(m1, 1, \(x) which.min(sqrt(colSums(abs(x - t(m2))))))
}

# a variant to Ronak Shah's solution, by removing `sqrt` and moving `t(m2)` out of `apply`
f2 <- \() {
    tm2 <- t(m2)
    apply(m1, 1, \(x) which.min(colSums(abs(x - tm2))))
}

# Friede's solution
f3 <- \() {
    tm2 <- t(m2)
    collapse::dapply(X = m1, FUN = \(x) which.min(Rfast::colsums((abs(x - tm2)))), MARGIN = 1L)
}

# G. Grothendieck's solution
f4 <- \(){
    apply(Rfast::dista(m1, m2, type = "manhattan"), 1, which.min)
}

# a variant to G. Grothendieck's solution, by using `dapply`
f5 <- \(){
    collapse::dapply(X = Rfast::dista(m1, m2, type = "manhattan"), which.min, MARGIN = 1)
}

# Onyambu's solution
Rcpp::cppFunction("IntegerVector closestRcpp(NumericMatrix A, NumericMatrix B) {
  int n = A.nrow();
  IntegerVector closest(n);
  for (int i = 0; i < n; ++i) {
    double prev_dist = 1e308, dist;
    for (int j = 0; j < B.nrow(); ++j){
      double dist = sum(abs(A(i, _) - B(j, _)));
      if (dist < prev_dist) {
        prev_dist = dist;
        closest[i] = j+1;
      }
    }
  }
  return closest;
}")
f6 <- \() {
    closestRcpp(m1, m2)
}


microbenchmark(
    f1(),
    f2(),
    f3(),
    f4(),
    f5(),
    f6(),
    unit = "relative",
    check = "equivalent"
)

шоу

Unit: relative
 expr       min       lq     mean    median        uq       max neval
 f1() 28.222593 28.05963 7.522702 28.221577 29.144915 2.4924392   100
 f2() 18.704444 19.37280 5.715675 20.383970 21.840241 2.1410204   100
 f3() 11.630370 11.87556 4.233875 12.320916 12.728101 2.2702689   100
 f4() 12.555926 13.06498 3.480127 13.401387 14.256614 1.0021835   100
 f5()  9.074074  9.30748 2.714870  9.563296  9.612287 0.9920494   100
 f6()  1.000000  1.00000 1.000000  1.000000  1.000000 1.0000000   100

Мы можем использовать dista из Rfast

library(Rfast)

apply(dista(m1, m2, type = "manhattan"), 1, which.min)
## [1]  8  3  2  3  3  1  2  3  6 10

Рассмотрите возможность использования Rcpp/C:

Rcpp::cppFunction("IntegerVector closestRcpp(NumericMatrix A, NumericMatrix B) {
  int n = A.nrow();
  IntegerVector closest(n);
  for (int i = 0; i < n; ++i) {
    double prev_dist = 1e308, dist;
    for (int j = 0; j < B.nrow(); ++j){
      double dist = sum(abs(A(i, _) - B(j, _)));
      if (dist < prev_dist) {
        prev_dist = dist;
        closest[i] = j+1;
      }
    } 
  }
  return closest;
}")

closestRcpp(m1, m2)
[1]  8  3  2  3  3  1  2  3  6 10