R中的特定排序

我正在为一个特定的输入制作所有可能的组合，但它也必须根据输入的顺序进行排序。由于这些组合的大小不同，我很难找到之前发布的答案

我想知道这是否可行

输入：

D N A 3

这意味着我需要以最多3个字符串的所有组合输出它：

D
DD
DDD
DDN
DDA
DND
DNA
. 
.

---

基本上是升序，如果我们考虑<代码> d> p>一个解决方案，使用我刚才发现的一个随机库（我可能使用错了）被称为<代码> ITEPC。p> 生成所有的组合，对元素进行因子分析，排序，然后生成字符串

ordered_combn = function(elems) {
  require(data.table)
  require(iterpc)

  I = lapply(seq_along(elems), function(i) iterpc::iterpc(table(elems), i, replace=TRUE, ordered=TRUE))
  I = lapply(I, iterpc::getall)
  I = lapply(I, as.data.table)

  dt = rbindlist(I, fill = TRUE)
  dt[is.na(dt)] = ""

  cols = paste0("V", 1:length(elems))
  dt[, (cols) := lapply(.SD, factor, levels = c("", elems)), .SDcols = cols]

  setkey(dt)
  dt[, ID := 1:.N]
  dt[, (cols) := lapply(.SD, as.character), .SDcols = cols]
  dt[, ord := paste0(.SD, collapse = ""), ID, .SDcols = cols]

  # return dt[, ord] as an ordered factor for neatness
  dt
}

elems = c("D", "N", "A")
combs = ordered_combn(elems)
combs

输出

    V1 V2 V3 ID ord
 1:  D        1   D
 2:  D  D     2  DD
 3:  D  D  D  3 DDD
 4:  D  D  N  4 DDN
 5:  D  D  A  5 DDA
 6:  D  N     6  DN
 7:  D  N  D  7 DND
 8:  D  N  N  8 DNN
...

---

这里有一个可能的解决方案：

generateCombs <- function(x, n){
  if (n == 1) return(x[1]) # Base case
  # Create a grid with all possible permutations of 0:n. 0 == "", and 1:n correspond to elements of x
  permutations = expand.grid(replicate(n, 0:n, simplify = F)) 
  # Order permutations
  orderedPermutations = permutations[do.call(order, as.list(permutations)),] 
  # Map permutations now such that 0 == "", and 1:n correspond to elements of x
    mappedPermutations = sapply(orderedPermutations, function(y) c("", x)[y + 1])
  # Collapse each row into a single string
  collapsedPermutations = apply(mappedPermutations, 1, function(x) paste0(x, collapse = ""))
  # Due to the 0's, there will be duplicates. We remove the duplicates in reverse order
  collapsedPermutations = rev(unique(rev(collapsedPermutations)))[-1] # -1 removes blank
  # Return as data frame
  return (as.data.frame(collapsedPermutations))
}

x = c("D", "N", "A")
n = 3
generateCombs(x, n)

---

订购的原因是什么？我不懂DND，这不是DDN吗？你可以对整数进行排序，然后对字符进行细分。

chartr（'123'，'DNA'，c（'1'，'11'，'123'，'122'，'133'））

@rawr这是升序，DDN在DDDYeah之后。这给了我一个想法，我可能需要做一个矩阵，其中每个字符都由一个特征表示（转换为数字），然后按特征排序，因为你没有对每个字符串中的字母进行排序。。

generateCombs <- function(x, n){
  if (n == 1) return(x[1]) # Base case
  # Create a grid with all possible permutations of 0:n. 0 == "", and 1:n correspond to elements of x
  permutations = expand.grid(replicate(n, 0:n, simplify = F)) 
  # Order permutations
  orderedPermutations = permutations[do.call(order, as.list(permutations)),] 
  # Map permutations now such that 0 == "", and 1:n correspond to elements of x
    mappedPermutations = sapply(orderedPermutations, function(y) c("", x)[y + 1])
  # Collapse each row into a single string
  collapsedPermutations = apply(mappedPermutations, 1, function(x) paste0(x, collapse = ""))
  # Due to the 0's, there will be duplicates. We remove the duplicates in reverse order
  collapsedPermutations = rev(unique(rev(collapsedPermutations)))[-1] # -1 removes blank
  # Return as data frame
  return (as.data.frame(collapsedPermutations))
}

x = c("D", "N", "A")
n = 3
generateCombs(x, n)

   collapsedPermutations
1                      D
2                     DD
3                    DDD
4                    DDN
5                    DDA
6                     DN
7                    DND
8                    DNN
9                    DNA
10                    DA
11                   DAD
...