r - R 中是否有一种有效的方法将矩阵 M2 的每一行 "paste"到矩阵 M1 的每一行以获得所有可能的组合？

Question

假设我有两个矩阵:

M1 <- matrix(letters[1:9], nrow = 3, ncol = 3, byrow = T)
M2 <- matrix(letters[10:18], nrow = 3, ncol = 3, byrow = T)

实际上我有 1000 行的大得多的矩阵。我想要的是在不使用 for 循环的情况下生成以下内容的有效方法，因为这会导致 R 崩溃。

result <- rbind(cbind(M1, matrix(rep(letters[10:12], 3), nrow = 3, ncol = 3, byrow = T)),
                cbind(M1, matrix(rep(letters[13:15], 3), nrow = 3, ncol = 3, byrow = T)),
                cbind(M1, matrix(rep(letters[16:18], 3), nrow = 3, ncol = 3, byrow = T)))

本质上是将 M2 的每一行“粘贴”到 M1 的每一行，这样我就有了所有可能的组合。

Answer 1

您可以使用 expand.grid 在 M1 和 M2 中创建所有可能的行索引组合。

mat <- expand.grid(1:nrow(M1), 1:nrow(M2))
cbind(M1[mat[[1]], ], M2[mat[[2]], ])

#     [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] "a"  "b"  "c"  "j"  "k"  "l" 
# [2,] "d"  "e"  "f"  "j"  "k"  "l" 
# [3,] "g"  "h"  "i"  "j"  "k"  "l" 
# [4,] "a"  "b"  "c"  "m"  "n"  "o" 
# [5,] "d"  "e"  "f"  "m"  "n"  "o" 
# [6,] "g"  "h"  "i"  "m"  "n"  "o" 
# [7,] "a"  "b"  "c"  "p"  "q"  "r" 
# [8,] "d"  "e"  "f"  "p"  "q"  "r" 
# [9,] "g"  "h"  "i"  "p"  "q"  "r"