如何计算R中两个维度不等的矩阵之间的欧氏距离

Question

如何在 R 中计算矩阵 A 和矩阵 B 之间的欧氏距离，如下所示：

我有两个矩阵，矩阵 A 和矩阵 B

矩阵A：

     [,1][,2]
[1,]   1   1   
[2,]   1   2   
[3,]   2   1   
[4,]   2   2   
[5,]   10  1   
[6,]   10  2   
[7,]   11  1   
[8,]   11  2   
[9,]   5   5   
[10,]  5   6

矩阵 B:

     [,1][,2][,3][,4][,5][,6]
[1,]   2   1   5   5  10   1
[2,]   1   1   2   1  10   1
[3,]   5   5   5   6  11   2
[4,]   2   2   5   5  10   1
[5,]   2   1   5   6  5    5
[6,]   2   2   5   5  11   1
[7,]   2   1   5   5  10   1
[8,]   1   1   5   6  11   1
[9,]   2   1   5   5  10   1
[10,]  5   6   11  1  10   2


I want the Result matrix (euclidean distance) to be as per below:

        [1,]  [,2]  [,3]

    [1,] 1.00  5.66  9.00
    [2,] 1.00  1.41
    [3,]
    [4,]
    [5,]
    [7,]
    [8,]
    [9,]
    [10]

对于矩阵 A 中的每一行，计算矩阵 B 中每一行中每两列的欧式距离。

例如，要在结果矩阵中获取以下内容的答案：

        [,1]
    [1,]

计算是：

    A(1,1) - From Matrix A
    B(2,1) - From Matrix B

    = sqrt((xA -xB)^2 + (yA -yB)^2)
    = sqrt((1-2)^2 + (1-1)^2)
    = 1.00

    xA and yA from Matrix A
    xB and yB from Matrix B

要在结果矩阵中得到以下答案：

        [,2]
    [1,] 5.66

计算是：

    A(1,1) - From Matrix A
    B(5,5) - From Matrix B

    = sqrt((xA -xB)^2 + (yA -yB)^2)
    = sqrt((1-5)^2 + (1-5)^2)
    = 5.66

要在结果矩阵中得到以下答案：

        [,3]
    [1,] 9.00

计算是：

    A(1,1) - From Matrix A
    B(10,1) - From Matrix B

    = sqrt((xA -xB)^2 + (yA -yB)^2)
    = sqrt((1-10)^2 + (1-1)^2)
    = 9.00

目前，我的以下代码仅在矩阵 A 和 B 的维数相等时才有效：

    distance <- function(MatrixA, MatrixB) {
      resultMatrix <- matrix(NA, nrow=dim(MatrixA)[1], ncol=dim(MatrixB)[1])
      for(i in 1:nrow(MatrixB)) {
         resultMatrix[,i] <- sqrt(rowSums(t(t(MatrixA)-MatrixB[i,])^2))
      }
         resultMatrix
      }

Answer 1

您只需要更改您的 for 循环，它会为每一行计算结果矩阵的所有三列：

for(i in 1:nrow(matA)) 
{
  resultMatrix[i,1] <- sqrt(rowSums((t(MatrixA[i,])-MatrixB[i,1:2])^2))
  resultMatrix[i,2] <- sqrt(rowSums((t(MatrixA[i,])-MatrixB[i,3:4])^2))
  resultMatrix[i,3] <- sqrt(rowSums((t(MatrixA[i,])-MatrixB[i,5:6])^2))

}

泛化为任意数量的列：

for(i in 1:nrow(MatrixA)) 
{
  for(j in 1:((dim(MatrixB)[2])/2)) 
  {  
    k = (j * 2) - 1
    resultMatrix[i,j] <- sqrt(rowSums((t(MatrixA[i,])-MatrixB[i,k:(k+1)])^2))
  }
}

如何计算R中两个维度不等的矩阵之间的欧氏距离

How to calculate the euclidean distance in R between two matrices each with unequal dimensions

r

euclidean-distance