r - R中没有重复的组合

Question

我正在尝试获取变量元素的长度为 3 的所有可能组合。虽然它部分与 combn() 一起工作，但我并没有完全得到我正在寻找的输出。这是我的例子

x <- c("a","b","c","d","e")
t(combn(c(x,x), 3)) 

我得到的输出看起来像这样

       [,1] [,2] [,3]
  [1,] "a"  "b"  "c" 
  [2,] "a"  "b"  "d" 
  [3,] "a"  "b"  "e" 

出于两个原因，我对这个命令并不满意。我想得到一个输出，上面写着“a+b+c”“a+b+b”......，不幸的是我无法使用 paste() 或其他东西来编辑输出。

我也期待每组字母的一个组合，即我要么得到“a+b+c”或“b+a+c”，但不能同时得到两者。

Answer 1

尝试类似:

x <- c("a","b","c","d","e")
d1 <- combn(x,3) # All combinations

d1 

#     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,] "a"  "a"  "a"  "a"  "a"  "a"  "b"  "b"  "b"  "c"  
# [2,] "b"  "b"  "b"  "c"  "c"  "d"  "c"  "c"  "d"  "d"  
# [3,] "c"  "d"  "e"  "d"  "e"  "e"  "d"  "e"  "e"  "e"

nrow(unique(t(d1))) == nrow(t(d1))
# [1] TRUE

d2 <- expand.grid(x,x,x) # All permutations 

d2

#     Var1 Var2 Var3
# 1      a    a    a
# 2      b    a    a
# 3      c    a    a
# 4      d    a    a
# 5      e    a    a
# 6      a    b    a
# 7      b    b    a
# 8      c    b    a
# 9      d    b    a
# ...

nrow(unique(d2)) == nrow(d2)
# [1] TRUE