r - R 中将向量分成两组的所有组合

Question

我正在尝试找出一种有效的方法来获取向量(向量始终是 1:n 之间的序列)中的元素可以分为两个大小相等的组的所有唯一组合。向量中的每个元素必须表示一次(在两个组之一中)。例如，当 n = 6 时，向量将为 [1,2,3,4,5,6]。这可以分为十种独特的方式来形成两个大小相等的组:

123 | 456
124 | 356
125 | 346
126 | 345
134 | 256
135 | 246
136 | 245
145 | 236
146 | 235
156 | 234

请注意，组内值的顺序并不重要，因此:

156 | 234

等同于:

651 | 342

另请注意，对称解决方案也不重要，因此:

156 | 234

等同于:

234 | 156

当n = 4时，有3种解。当n=6时，有10个解。当n=8时，有35个解。我相信我想出了一种在 R 中获得这些解决方案的方法。但是，一旦 n 变大，它就会有点慢。在大多数情况下，我对我所拥有的感到满意，但想询问是否有人对提高其速度或代码质量等的方法有建议。特别是，我从一个有很多重复的解决方案开始，然后我删除重复。我认为这使得算法相当慢。

library(combinat)
# Number of elements in array
n = 6
# Array
dat <- 1:n

# All possible permutations for array
ans <- permn(dat)
lengthAns <- length(ans)
ansDF <- data.frame()
# Place first permutation in final answer data frame
ansDF <- rbind(ansDF, ans[[1]])

# Look at the rest of the possible permutations. Determine for each one if it is truly unique from all the previously-listed possible permutations. If it is unique from them, then add it to the final answer data frame
for (i in 2:lengthAns){
  j = i
  k = TRUE
  while (k && j > 1){
    j = j-1
    if(setequal(ans[[i]][1:(n/2)], ans[[j]][1:(n/2)]))
      k = FALSE
    if(setequal(ans[[i]][1:(n/2)], ans[[j]][(n/2+1):(n)]))
      k = FALSE
  }
  if (k){
    ansDF <- rbind(ansDF, ans[[i]])
  }
}

# At this point, ansDF contains all unique possible ways to split the array into two-equally sized groups.

Answer 1

N = 6
x = 1:N
x1 = combn(x, N/2) #how many ways can we take half the elements to form the 1st group
NC = NCOL(x1)
x2 = x1[, NC:1] # simplified way to generate the complementary groups that include values not in x1 
grp1 = t(x1[,1:(NC/2)]) # We only need half of the rows, the 2nd half containing the same set in reverse order
grp2 = t(x2[,1:(NC/2)])
all.comb = cbind(grp1, grp2)

#      [,1] [,2] [,3] [,4] [,5] [,6]
# [1,]    1    2    3    4    5    6
# [2,]    1    2    4    3    5    6
# [3,]    1    2    5    3    4    6
# [4,]    1    2    6    3    4    5
# [5,]    1    3    4    2    5    6
# [6,]    1    3    5    2    4    6
# [7,]    1    3    6    2    4    5
# [8,]    1    4    5    2    3    6
# [9,]    1    4    6    2    3    5
#[10,]    1    5    6    2    3    4