如何计算成对的组合？

Question

从许多项(n)中，如何计算R中所有可能的对组合？我可以用(n-1)*(n-3)*(n-5)*…计算出可能的组合数，但是如何获得组合本身？

例如，我有

x <- c("A", "B", "C", "D") 

这将使4-1=3组合成对，所以我希望有这样的东西

("AB","CD")
("AC","BD")
("AD","BC")

非常感谢!

Answer 1

首先，让我们得到所有对对：

dat = t(combn(combn(x, 2, paste, collapse=""), 2))

[,1] [,2] [1,] "AB" "AC" [2,] "AB" "AD" [3,] "AB" "BC" [4,] "AB" "BD" [5,] "AB" "CD" [6,] "AC" "AD" [7,] "AC" "BC" [8,] "AC" "BD" [9,] "AC" "CD" [10,] "AD" "BC" [11,] "AD" "BD" [12,] "AD" "CD" [13,] "BC" "BD" [14,] "BC" "CD" [15,] "BD" "CD"

现在我们有了所有的对。但看起来你只想要一对，其中每个字母只出现一次。下面的代码标识了符合这个条件的行(尽管看起来代码比它所需要的要复杂)。

dat[sapply(strsplit(apply(dat, 1, paste, collapse=""), ""), function(i) length(unique(i)) == 4), ]

[,1] [,2] [1,] "AB" "CD" [2,] "AC" "BD" [3,] "AD" "BC"

Answer 2

为此，可以使用combn或expand.grid。

x <- c("A","B","C","D") 

expand.grid(x, x) # generates both way combinations

   Var1 Var2
1     A    A
2     B    A
3     C    A
4     D    A
5     A    B
6     B    B
7     C    B
8     D    B
9     A    C
10    B    C
11    C    C
12    D    C
13    A    D
14    B    D
15    C    D
16    D    D

t(combn(x, 2)) # order does not matter

     [,1] [,2]
[1,] "A"  "B" 
[2,] "A"  "C" 
[3,] "A"  "D" 
[4,] "B"  "C" 
[5,] "B"  "D" 
[6,] "C"  "D"