我如何找到每个单词的前 1-2 个字母的所有可能排列？

Question

我有以下数据：

inputs <- c("Master", "Bachelor", "School")

我想对每个单词的前 1-2 个字母进行所有可能的排列。

first_letter <- sapply(inputs, substr, start = 1, stop = 1)
"M" "B"  "S"

second_letter <- sapply(inputs, substr, start = 1, stop = 2)
"Ma"     "Ba"     "Sc" 

所需的 output：

每个顺序中第一个字母的所有排列，请参见变量“all_order”的列（参见“我尝试过的内容”部分）。 同样在这两种变体中，要么取“first_letter”的第一个值，要么取第一个值“second_letter”，但不能同时取两者。

工商管理硕士、工商管理硕士、工商管理硕士、工商管理硕士

SBM, SBaM, SBaMa, SBaM ScBM, ScBaM, ScBaMa, ScBaM

BSM，BSMa，BScM，BScMa BaSM，BaSMa，BaScM，BaScMa，

......

（让我知道它是否解释得足够好。）

我尝试了什么：

combs <- combn(rep(seq(inputs), 2), 3)
keep <- !colSums(apply(combs, 2, duplicated))
all_order <- combs[, keep]

all_order
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]    1    1    1    1    2    2    3    1
[2,]    2    2    3    2    3    1    1    2
[3,]    3    3    2    3    1    3    2    3

Answer 1

也许我们可以试试下面的代码

d <- do.call(
  rbind,
  combn(
    c(first_letter, second_letter),
    3,
    pracma::perms,
    simplify = FALSE
  )
)

res <- do.call(
  paste0,
  data.frame(d)
)[apply(
  `dim<-`(match(substr(d, 1, 1), first_letter), dim(d)),
  1, 
  function(x) all(!duplicated(x))
)]

这使

> res
 [1] "SBM"    "SMB"    "BSM"    "BMS"    "MBS"    "MSB"    "ScBM"   "ScMB"  
 [9] "BScM"   "BMSc"   "MBSc"   "MScB"   "BaSM"   "BaMS"   "SBaM"   "SMBa"
[17] "MSBa"   "MBaS"   "ScBaM"  "ScMBa"  "BaScM"  "BaMSc"  "MBaSc"  "MScBa"
[25] "MaSB"   "MaBS"   "SMaB"   "SBMa"   "BSMa"   "BMaS"   "ScMaB"  "ScBMa"
[33] "MaScB"  "MaBSc"  "BMaSc"  "BScMa"  "BaMaS"  "BaSMa"  "MaBaS"  "MaSBa"
[41] "SMaBa"  "SBaMa"  "ScBaMa" "ScMaBa" "BaScMa" "BaMaSc" "MaBaSc" "MaScBa"

Answer 2

我们将使用gtools::permutations计算inputs的...排列，然后使用expand.grid显示其中的所有组合。

首先，我们可以通过以下输入的顺序轻松完成：

expand.grid(c("M","Ma"), c("B","Ba"), c("S","Sc"))
#   Var1 Var2 Var3
# 1    M    B    S
# 2   Ma    B    S
# 3    M   Ba    S
# 4   Ma   Ba    S
# 5    M    B   Sc
# 6   Ma    B   Sc
# 7    M   Ba   Sc
# 8   Ma   Ba   Sc
do.call(paste, c(expand.grid(c("M","Ma"), c("B","Ba"), c("S","Sc")), sep = ""))
# [1] "MBS"    "MaBS"   "MBaS"   "MaBaS"  "MBSc"   "MaBSc"  "MBaSc"  "MaBaSc"

现在这是一个订单（ M < B < S ），现在我们需要重新排列它们。 我们可以手动调用所有排序，或者我们可以使用gtools::permutations来提供帮助。

inputlist <- lapply(inputs, substring, 1, 1:2)
str(inputlist)
# List of 3
#  $ : chr [1:2] "M" "Ma"
#  $ : chr [1:2] "B" "Ba"
#  $ : chr [1:2] "S" "Sc"

perms <- gtools::permutations(3, 3)
perms
#      [,1] [,2] [,3]
# [1,]    1    2    3
# [2,]    1    3    2
# [3,]    2    1    3
# [4,]    2    3    1
# [5,]    3    1    2
# [6,]    3    2    1
inputlist[perms[2,]]
# [[1]]
# [1] "M"  "Ma"
# [[2]]
# [1] "S"  "Sc"
# [[3]]
# [1] "B"  "Ba"
inputlist[perms[3,]]
# [[1]]
# [1] "B"  "Ba"
# [[2]]
# [1] "M"  "Ma"
# [[3]]
# [1] "S"  "Sc"

导致

allperms <- do.call(rbind,
  apply(gtools::permutations(3, 3), 1, 
        function(ind) do.call(expand.grid, inputlist[ind]))
)
head(allperms); tail(allperms)
#   Var1 Var2 Var3
# 1    M    B    S
# 2   Ma    B    S
# 3    M   Ba    S
# 4   Ma   Ba    S
# 5    M    B   Sc
# 6   Ma    B   Sc
#    Var1 Var2 Var3
# 43    S   Ba    M
# 44   Sc   Ba    M
# 45    S    B   Ma
# 46   Sc    B   Ma
# 47    S   Ba   Ma
# 48   Sc   Ba   Ma
do.call(paste, c(allperms, list(sep = "")))
#  [1] "MBS"    "MaBS"   "MBaS"   "MaBaS"  "MBSc"   "MaBSc"  "MBaSc"  "MaBaSc" "MSB"    "MaSB"   "MScB"  
# [12] "MaScB"  "MSBa"   "MaSBa"  "MScBa"  "MaScBa" "BMS"    "BaMS"   "BMaS"   "BaMaS"  "BMSc"   "BaMSc" 
# [23] "BMaSc"  "BaMaSc" "BSM"    "BaSM"   "BScM"   "BaScM"  "BSMa"   "BaSMa"  "BScMa"  "BaScMa" "SMB"   
# [34] "ScMB"   "SMaB"   "ScMaB"  "SMBa"   "ScMBa"  "SMaBa"  "ScMaBa" "SBM"    "ScBM"   "SBaM"   "ScBaM" 
# [45] "SBMa"   "ScBMa"  "SBaMa"  "ScBaMa"

Answer 3

您可以使用 package e1071 （统计系的其他功能，概率论组（以前：E1071））中的 function permutations 。

library(e1071)

res <- c(substr(inputs,1,1), substr(inputs,1,2))
res
[1] "M"  "B"  "S"  "Ma" "Ba" "Sc"

perm <- unique(matrix(e1071::permutations(6), ncol=3))

# to exclude repetitions find 1,4 2,5 and 3,6
apply(matrix(res[perm], ncol=3), 1, paste, collapse="")[
  rowSums(cbind(rowSums(perm == 1 | perm == 4)==2,
                rowSums(perm == 2 | perm == 5)==2,
                rowSums(perm == 3 | perm == 6)==2))==0]
 [1] "MSBa"   "SMBa"   "SMaBa"  "MaSBa"  "SMaB"   "BMaS"   "SMB"    "SBM"   
 [9] "MBS"    "BMS"    "BSM"    "MSB"    "MaBS"   "MaSB"   "BSMa"   "SBMa"  
[17] "SBaM"   "MBaS"   "MaBaS"  "SBaMa"  "BaSMa"  "BaSM"   "BaMS"   "BaMaS" 
[25] "BMaSc"  "MBSc"   "BMSc"   "MaBSc"  "MBaSc"  "MaBaSc" "BaMSc"  "BaMaSc"
[33] "MaScB"  "BScM"   "MScB"   "BScMa"  "BaScM"  "BaScMa" "MScBa"  "MaScBa"
[41] "ScBMa"  "ScMB"   "ScBM"   "ScMaB"  "ScMBa"  "ScMaBa" "ScBaM"  "ScBaMa"