Question

x <- letters[1:4]
x
# [1] "a" "b" "c" "d"

t(combn(x, 2))
#   [,1] [,2]
# [1,] "a"  "b" 
# [2,] "a"  "c" 
# [3,] "a"  "d" 
# [4,] "b"  "c" 
# [5,] "b"  "d" 
# [6,] "c"  "d"

如果我还有与b-a，c-a ... d-c的逆组合，我应该如何编写代码。总共12种组合。

Answer 1

您可以使用基础R中的expand.grid来获取向量的所有可能组合（即16种组合），然后使用subset（或[.data.frame）以便使用两列永远不会在一行中相等（产生预期的12种组合）：

x <- letters[1:4]
subset(expand.grid(rep(list(x),2)), Var1 != Var2)
#   Var1 Var2
#2     b    a
#3     c    a
#4     d    a
#5     a    b
#7     c    b
#8     d    b
#9     a    c
#10    b    c
#12    d    c
#13    a    d
#14    b    d
#15    c    d

使用data.table交叉加入（CJ）功能的替代方案：

libraray(data.table)
CJ(x, x)[V1 != V2]
#    V1 V2
# 1:  a  b
# 2:  a  c
# 3:  a  d
# 4:  b  a
# 5:  b  c
# 6:  b  d
# 7:  c  a
# 8:  c  b
# 9:  c  d
#10:  d  a
#11:  d  b
#12:  d  c

Answer 2

包gtools具有permutations功能。

x <- letters[1:4]
library(gtools)
permutations(length(x), 2, x, repeats = FALSE)
#       [,1] [,2]
#  [1,] "a"  "b" 
#  [2,] "a"  "c" 
#  [3,] "a"  "d" 
#  [4,] "b"  "a" 
#  [5,] "b"  "c" 
#  [6,] "b"  "d" 
#  [7,] "c"  "a" 
#  [8,] "c"  "b" 
#  [9,] "c"  "d" 
# [10,] "d"  "a" 
# [11,] "d"  "b" 
# [12,] "d"  "c"

Answer 3

dplyr/tidyr

的另一个选项

library(dplyr)
library(tidyr)
data_frame(x1=x,x2=x) %>% 
        expand(x1, x2) %>%
        filter(x1!=x2)
#      x1    x2
#    (chr) (chr)
#1      a     b
#2      a     c
#3      a     d
#4      b     a
#5      b     c
#6      b     d
#7      c     a
#8      c     b
#9      c     d
#10     d     a
#11     d     b
#12     d     c

矢量与顺序的可能组合

3 个答案: