如何在R中对两个维数不相等的矩阵进行成对计算

How to do pairwise calculation for two matrices with unequal dimensions in R

当矩阵A和矩阵B的尺寸不相等时，如何计算R在矩阵R中的欧式距离：

我有两个矩阵，分别是矩阵A和矩阵B

矩阵A：

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[,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：

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[,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 for List 1 to store result of
the euclidean distance between row 1 to row 10 in matrix A and every two
columns of row 1 in Matrix B as per below:

List [[1]]

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

[1,] 1.00 5.66 9.00
[2,] 0.00 1.00 9.00
[3,] 5.66 6.40 10.05
[4,]
[5,]
[7,]
[8,]
[9,]
[10]

For List 2, I want the Result matrix to store the result of the euclidean
distance between row 1 to row 10 in matrix A and every two columns of row 2
in Matrix B as per below:

List [[2]]

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

[1,] 1.41 5.00 9.06
[2,] 1.00 1.41 8.00
[3,]
[4,]
[5,]
[7,]
[8,]
[9,]
[10]

接下来，列表3用于矩阵B中的第3行

这应该一直进行到列表10

例如，要在结果矩阵列表1中获得以下答案：

1 2	[,1] [1,] 1.00

计算公式为：

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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

获取以下内容的答案：

1 2	[,2] [1,] 5.66

计算公式为：

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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

获取以下内容的答案：

1 2	[,3] [1,] 9.00

计算公式为：

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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

这是我目前拥有的，但是它在矩阵A的第一行之间进行计算
矩阵B中的第1行，依此类推。我想要的是将矩阵A中的每一行都移到列表1中矩阵B的第一行中，将矩阵A中的每一行都移到矩阵B中的第二行，依此类推，直到矩阵B中的第10行；

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ObjCentDist <- function(matrixA, matrixB) {
resultMatrix <- matrix(NA, nrow=dim(matrixA)[1],ncol=dim(matrixB[2]/2)
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))
}
}
resultMatrix
}

matrixA <- matrix(c(1,1,1,2,2,1,2,2,10,1,10,2,11,1,11,2,5,5,5,6), ncol = 2, byrow = TRUE)
matrixB <- matrix(c(2,1,5,5,10,1,1,1,2,1,10,1,5,5,5,6,11,2,2,2,5,5,10,1,2,1,5,6,5,5,2,2,5,5,11,1,2,1,5,5,10,1,1,1,5,6,11,1,2,1,5,5,10,1,5,6,11,1,10,2), nrow=10, ncol=6, byrow=TRUE)

我注意到了pdist()，但是我不确定如何在循环中使用它并在mycase中获得所需的输出，因为我对R还是很陌生，并且仍在学习