距离矩阵计算

Question

例如，我们有两个矩阵：

A=
     [,1] [,2] [,3]
[1,]   -2    1    0
[2,]    1    0   -1
[3,]    0   -1    4

和

B=
     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    1    2    3
[3,]    3    1    2
[4,]    2    3    1
[5,]    3    1    2

我要计算A行和B行之间的距离矩阵，换句话说，用我们的例子：

      [,1]                   [,2]                    [,3]
[1,]    dist(B[1,]-A[1,])      dist(B[1,]-A[2,])       dist(B[1,]-A[3,])
[2,]    dist(B[2,]-A[1,])      dist(B[2,]-A[2,])       dist(B[2,]-A[3,])
[3,]    dist(B[3,]-A[1,])      dist(B[3,]-A[2,])       dist(B[3,]-A[3,])
[4,]    dist(B[4,]-A[1,])      dist(B[4,]-A[2,])       dist(B[4,]-A[3,])
[5,]    dist(B[5,]-A[1,])      dist(B[5,]-A[2,])       dist(B[5,]-A[3,])

我正在考虑使用"dist“函数，该函数计算并返回数据矩阵的行间距离矩阵。但它只适用于一个矩阵。我尝试使用命令"rbind(A，B)“并将"dist”应用于结果矩阵，但在这种情况下，我还获得了同一矩阵的行之间的距离。

Answer 1

也许这条路？

A = matrix(c(-2, 1, 0, 1, 0, -1, 0, -1, 4), ncol = 3, byrow = TRUE)
B = matrix(c(1, 2, 3, 1, 2, 3, 3, 1, 2, 2, 3, 1, 3, 1, 2), ncol = 3, byrow = TRUE)

t(sapply(1:nrow(B), function(x) {
  sqrt(rowSums(t(t(A)-B[x,])^2))
}))

         [,1]     [,2]     [,3]
[1,] 4.358899 4.472136 3.316625
[2,] 4.358899 4.472136 3.316625
[3,] 5.385165 3.741657 4.123106
[4,] 4.582576 3.741657 5.385165
[5,] 5.385165 3.741657 4.123106

Answer 2

# First build a matrix
  Dist.M <- matrix(0, nrow = nrow(A), ncol = ncol(B))

# Complete the matrix
  for (i in 1:nrow(A)){
    for (j in 1:nrow(B)){
         Dist.M[i,j] <- dist(rbind(A[i,],B[j,])
    }
  }