2 回答

• 2

  像这样的东西？

  n <- 3
  g <- 2 # g must be < n 
  m <- combn(n, g)
  mm <- as.numeric(m)
  mat <- matrix(0, nrow = g * ncol(m), ncol = n)
  mat[ cbind(1:nrow(mat), mm)] <- 1
  
  mat
  #       [,1] [,2] [,3]
  #[1,]    1    0    0
  #[2,]    0    1    0
  
  #[3,]    1    0    0
  #[4,]    0    0    1
  
  #[5,]    0    1    0
  #[6,]    0    0    1
  
  # mat is half the answer :)
  # the other half is
  mat[nrow(mat):1, ]
  
  #      [,1] [,2] [,3]
  #[1,]    0    0    1
  #[2,]    0    1    0
  
  #[3,]    0    0    1
  #[4,]    1    0    0
  
  #[5,]    0    1    0
  #[6,]    1    0    0
  
  soln <- rbind(mat, mat[nrow(mat):1, ])
  
  # as suggested by the OP to split the soln 
  d <- split(data.frame(soln), rep(1:(nrow(soln)/g), each=g))
  

  回复于 2024-05-04T07:06:52+08:00
• 1

  精明的读者会注意到这个问题可以简化为：“如何生成 permutations 的所有成对 permutations ？”通过这种方式查看，我们可以避免最初处理二进制向量并将其保存到最后一步 .

  使用基本R函数 intToBits ，this answer到问题How to convert integer numbers into binary vector?，以及任何可以生成特定长度排列的函数（有很多软件包： gtools::permutations ， RcppAlgos::permuteGeneral ， iterpc 和 arrangements::permutations ），我们可以在一个中获得所需的结果线 .

  library(gtools)
  t(sapply(t(gtools::permutations(3, 2, 2^(0:2))),  
           function(x) {as.integer(intToBits(x))})[1:3, ])
  
        [,1] [,2] [,3]
   [1,]    1    0    0
   [2,]    0    1    0
  
   [3,]    1    0    0
   [4,]    0    0    1
  
   [5,]    0    1    0
   [6,]    1    0    0
  
   [7,]    0    1    0
   [8,]    0    0    1
  
   [9,]    0    0    1
  [10,]    1    0    0
  
  [11,]    0    0    1
  [12,]    0    1    0
  

  推广很容易 .

  bitPairwise <- function(numBits, groupSize) {
      t(sapply(t(gtools::permutations(numBits, groupSize, 2^(0:(numBits-1)))), 
                   function(x) {as.integer(intToBits(x))})[1:numBits, ])
  }
  
   bitPairwise(numBits = 6, groupSize = 3)[1:12, ]
        [,1] [,2] [,3] [,4] [,5] [,6]
   [1,]    1    0    0    0    0    0
   [2,]    0    1    0    0    0    0
   [3,]    0    0    1    0    0    0
  
   [4,]    1    0    0    0    0    0
   [5,]    0    1    0    0    0    0
   [6,]    0    0    0    1    0    0
  
   [7,]    1    0    0    0    0    0
   [8,]    0    1    0    0    0    0
   [9,]    0    0    0    0    1    0
  
  [10,]    1    0    0    0    0    0
  [11,]    0    1    0    0    0    0
  [12,]    0    0    0    0    0    1
  

  更新

  我只是张贴这个来指出@Suren的答案是如何正确的 .

  OP正在寻找排列而不是组合

  从评论中的对话中，您会看到@Suren的解决方案在组数增加时没有给出正确的结果（"I am also trying to get groupings of three instead of 2 (or any number)"和"This is cutting off some solutions"） .

  似乎@ Suren的回答用 g = 2 给出了正确的结果 . 这是如此，因为 1:n choose 2 的排列等于 1:n choose 2 的组合与 n:1 choose 2 的组合（注意 1:n 被颠倒） . 这正是@ Suren的答案正在做的事情（即生成组合选择2，以相反的顺序编写它们并组合） .

  ## original version
  surenFun <- function(n, g) {
      m <- combn(n, g)
      mm <- as.numeric(m)
      mat <- matrix(0, nrow = g * ncol(m), ncol = n)
      mat[ cbind(1:nrow(mat), mm)] <- 1
      soln <- rbind(mat, mat[nrow(mat):1, ])
      split(data.frame(soln), rep(1:(nrow(soln)/g), each=g))
  }
  
  ## Here is the corrected version
  surenFunCorrected <- function(n, g) {
      ## changed combn to gtools::permutations or any other
      ## similar function that can generate permutations
      m <- gtools::permutations(n, g)
      ## you must transpose m
      mm <- as.numeric(t(m))
      ## change ncol(m) to nrow(m)
      mat <- matrix(0, nrow = g * nrow(m), ncol = n)
      mat[ cbind(1:nrow(mat), mm)] <- 1
      ## removed soln
      split(data.frame(mat), rep(1:(nrow(mat)/g), each=g))
  }
  

  使用OP中的给定示例，它以不同的顺序给出相同的结果：

  ## The order is slightly different
  match(surenFunCorrected(3, 2), surenFun(3, 2))
  [1] 1 2 6 3 5 4
  
  all(surenFunCorrected(3, 2) %in% surenFun(3, 2))
  [1] TRUE
  
  all(surenFun(3, 2) %in% surenFunCorrected(3, 2))
  [1] TRUE
  

  让我们用 g = 3 和 n = 4 测试一下 .

  ## N.B. all of the original output is
  ## contained in the corrected output
  all(surenFun(4, 3) %in% surenFunCorrected(4, 3))
  [1] TRUE
  
  ## However, there are 16 results
  ## not returned in the original
  leftOut <- which(!(surenFunCorrected(4, 3) %in% surenFun(4, 3)))
  leftOut
  [1]  3  5  6  7  8  9 11 12 13 14 16 17 18 19 20 22
  
  ## E.g. 3 examples that were left out
  surenFunCorrected(4, 3)[leftOut[c(1,8,16)]]
  $`3`
    X1 X2 X3 X4
  7  1  0  0  0
  8  0  0  1  0
  9  0  1  0  0
  
  $`12`
     X1 X2 X3 X4
  34  0  1  0  0
  35  0  0  0  1
  36  0  0  1  0
  
  $`22`
     X1 X2 X3 X4
  64  0  0  0  1
  65  0  1  0  0
  66  0  0  1  0
  

  回复于 2024-05-04T07:06:52+08:00