如何动态检查排列而不将结果存储在R中

Question

假设我们对字母“ a”，“ b”和“ c”具有以下排列：

library(combinat)
do.call(rbind, permn(letters[1:3]))
#      [,1] [,2] [,3]
# [1,] "a"  "b"  "c" 
# [2,] "a"  "c"  "b" 
# [3,] "c"  "a"  "b" 
# [4,] "c"  "b"  "a" 
# [5,] "b"  "c"  "a" 
# [6,] "b"  "a"  "c" 

是否可以在给定的“即时”排列（即特定行）上执行某些功能而不存储结果？

即，如果row == "a" "c" "b"或row == "b" "c" "a"，则不存储结果。在这种情况下，期望的结果将是：

#      [,1] [,2] [,3]
# [1,] "a"  "b"  "c" 
# [2,] "c"  "a"  "b" 
# [3,] "c"  "b"  "a" 
# [4,] "b"  "a"  "c" 

我知道我可以使用combinat::permn参数将函数应用于fun中的所有动态排列，例如：

permn(letters[1:3], fun = function(x) {
  res <- paste0(x, collapse = "")
  if (res == "acb" | res == "bca") {
    return(NA)
  } else {
    return(res)
  }
})

但是此剧照存储了一个NA，并且返回的列表包含6个元素，而不是所需的4个元素：

# [[1]]
# [1] "abc"
# 
# [[2]]
# [1] NA
# 
# [[3]]
# [1] "cab"
# 
# [[4]]
# [1] "cba"
# 
# [[5]]
# [1] NA
# 
# [[6]]
# [1] "bac"

注意，我对随后删除NA值不感兴趣；我特别感兴趣的是，对于给定的排列，不要“即时”将结果列表添加到结果中。

Answer 1

我们可以使用magrittr管道，在该管道中，将输入矩阵绑定到要检查的行，并省略重复的行。

library(combinat)
library(magrittr)

Rows <- rbind(c("a", "c", "b"), c("b", "c", "a"))

do.call(rbind, permn(letters[1:3])) %>% 
  subset(tail(!duplicated(rbind(Rows, .)), -nrow(Rows)))

给予：
     [,1] [,2] [,3]
[1,] "a"  "b"  "c" 
[2,] "c"  "a"  "b" 
[3,] "c"  "b"  "a" 
[4,] "b"  "a"  "c" 

Answer 2

对于要忽略的特定条件，您可以return NULL，并且rbind结果将忽略NULL元素并仅绑定所需的组合。

do.call(rbind, combinat::permn(letters[1:3], function(x) 
   if(!all(x == c("a", "c", "b") | x == c("b", "c", "a")))
   return(x)
))

#    [,1] [,2] [,3]
#[1,] "a"  "b"  "c" 
#[2,] "c"  "a"  "b" 
#[3,] "c"  "b"  "a" 
#[4,] "b"  "a"  "c" 

类似地，
do.call(rbind, permn(letters[1:3],function(x) { 
   res <- paste0(x, collapse = "")
  if (!res %in% c("acb","bca")) 
    return(res) 
}))

#      [,1] 
#[1,] "abc"
#[2,] "cab"
#[3,] "cba"
#[4,] "bac"