來自兩個向量的不同數量元素的組合

我有兩個向量：

group1 <- c("a", "b")
group2 <- c("c", "d", "e")

我想從“group1”中獲取一個元素的所有組合，從“group2”中獲取一個或兩個元素。

期望的結果是：

"a" "c" # one element from group1, one element from group 2
"b" "c"
"a" "d"
"b" "d"
"a" "e"
"b" "e"
"a" "c" "d" # one element from group1, two elements from group 2
"b" "c" "d"
"a" "c" "e"
"b" "c" "e"
"a" "d" "e"
"b" "d" "e"

uj5u.com熱心網友回復：

您可以像這樣獲得第二組的組合：

g = c(as.list(group2), combn(group2,2,simplify=F))

然后使用lapply()這些，每次c()與以下元素一起使用group1：

unlist(lapply(g, \(i) lapply(group1,c,i)), recursive=F)

輸出：

[[1]]
[1] "a" "c"

[[2]]
[1] "b" "c"

[[3]]
[1] "a" "d"

[[4]]
[1] "b" "d"

[[5]]
[1] "a" "e"

[[6]]
[1] "b" "e"

[[7]]
[1] "a" "c" "d"

[[8]]
[1] "b" "c" "d"

[[9]]
[1] "a" "c" "e"

[[10]]
[1] "b" "c" "e"

[[11]]
[1] "a" "d" "e"

[[12]]
[1] "b" "d" "e"

uj5u.com熱心網友回復：

允許任何輸入向量以及從每個向量中提取的最大/最小字符數的完全通用解決方案是：

comb <- function(a, b, min_a = 1, max_a = 1, min_b = 1, max_b = 2) {
  as <- do.call(c, lapply(min_a:max_a, \(i) combn(a, i, simplify = FALSE)))
  bs <- do.call(c, lapply(min_b:max_b, \(i) combn(b, i, simplify = FALSE)))
  apply(expand.grid(as, bs), 1, unlist, use.names = FALSE)
}

默認情況下，這將給出：

comb(group1, group2)
#> [[1]]
#> [1] "a" "c"
#> 
#> [[2]]
#> [1] "b" "c"
#> 
#> [[3]]
#> [1] "a" "d"
#> 
#> [[4]]
#> [1] "b" "d"
#> 
#> [[5]]
#> [1] "a" "e"
#> 
#> [[6]]
#> [1] "b" "e"
#> 
#> [[7]]
#> [1] "a" "c" "d"
#> 
#> [[8]]
#> [1] "b" "c" "d"
#> 
#> [[9]]
#> [1] "a" "c" "e"
#> 
#> [[10]]
#> [1] "b" "c" "e"
#> 
#> [[11]]
#> [1] "a" "d" "e"
#> 
#> [[12]]
#> [1] "b" "d" "e"

^{使用reprex v2.0.2創建于 2022-09-03}

uj5u.com熱心網友回復：

兩次呼叫expand.grid將執行此操作：

eg1 <- expand.grid(group1, group2)
eg1
#   Var1 Var2
# 1    a    c
# 2    b    c
# 3    a    d
# 4    b    d
# 5    a    e
# 6    b    e

和

eg2 <- expand.grid(group1, seq_len(ncol(combn(group2, 2))))
eg2 <- cbind(eg2, t(combn(group2, 2)[, eg2$Var2]))[,-2]
eg2
#   Var1 1 2
# 1    a c d
# 2    b c d
# 3    a c e
# 4    b c e
# 5    a d e
# 6    b d e

然后它們可以與c. 出于演示目的，我將在這里取消它們的名稱，盡管呼叫unname純粹是裝飾性的。

out <- c(asplit(eg1, 1), asplit(eg2, 1))
out <- lapply(out, function(z) unname(c(z)))
str(out)
# List of 12
#  $ : chr [1:2] "a" "c"
#  $ : chr [1:2] "b" "c"
#  $ : chr [1:2] "a" "d"
#  $ : chr [1:2] "b" "d"
#  $ : chr [1:2] "a" "e"
#  $ : chr [1:2] "b" "e"
#  $ : chr [1:3] "a" "c" "d"
#  $ : chr [1:3] "b" "c" "d"
#  $ : chr [1:3] "a" "c" "e"
#  $ : chr [1:3] "b" "c" "e"
#  $ : chr [1:3] "a" "d" "e"
#  $ : chr [1:3] "b" "d" "e"

uj5u.com熱心網友回復：

1)此解決方案僅使用基礎 R 并且僅使用一次combn. 它的作業原理是采用 3 個元素的所有組合，c(group1, group2, "")并只保留 group1 中的一個元素和 group2 中的 1 或 2 個元素。"" 元素被洗掉，留下所需的向量串列。我們還展示了如何選擇性地將其表示為矩陣。

ok如果出現inix元素，則將函式定義為 TRUE，否則定義為FALSE。如果 group1 中有一個元素并且 group2 中有 1 或 2 個元素，則定義回傳其引數而不包含任何 "" 組件；否則，它回傳 NULL。適用于 的 3 個元素的所有組合。然后洗掉任何長度為零的行并回傳串列。如果您想要一個字符向量串列作為結果，請使用它。ghggc(group1, group2, "")L

如果需要矩陣結果而不是串列結果，則也使用最后一行。

ok <- function(g, h, ix) sum(g %in% h) %in% ix
g <- function(x) if (ok(x, group1, 1) && ok(x, group2, 1:2)) x[nchar(x) > 0]

L <- Filter(length, combn(c(group1, group2, ""), 3, g, simplify = FALSE))

# next line is only if you want a matrix result
do.call("rbind", lapply(L, `[`, 1:3))

給出這個矩陣：

      [,1] [,2] [,3]
 [1,] "a"  "c"  "d" 
 [2,] "a"  "c"  "e" 
 [3,] "a"  "c"  NA  
 [4,] "a"  "d"  "e" 
 [5,] "a"  "d"  NA  
 [6,] "a"  "e"  NA  
 [7,] "b"  "c"  "d" 
 [8,] "b"  "c"  "e" 
 [9,] "b"  "c"  NA  
[10,] "b"  "d"  "e" 
[11,] "b"  "d"  NA  
[12,] "b"  "e"  NA  

1a)這也可以用這樣的管道來表示。 g是從上面。

c(group1, group2, "z") |>
  combn(3, g, simplify = FALSE) |>
  Filter(f = length) |>
  lapply(`[`, 1:3) |> # this & next line if matrix wanted
  do.call(what = "rbind")

2）使用字串操作的方法是將組合輸出的每個組件粘貼在一起，然后使用 str_count 對其進行過濾以計算出現次數。這給出了一個字符向量結果，或者我們可以使用將其拆分為一個串列strsplit(res, "")。

library(stringr)

cc <- combn(c(group1, group2, ""), 3, paste, collapse = "")
res <- cc[str_count(cc, "[ab]") == 1 & str_count(cc, "[cde]") %in% 1:2]
res
## [1] "acd" "ace" "ac"  "ade" "ad"  "ae"  "bcd" "bce" "bc"  "bde" "bd"  "be" 

uj5u.com熱心網友回復：

我們可以用

a <- expand.grid(group1 , group2 , stringsAsFactors = F)
a$Var3 -> ""

b <- expand.grid(group1 , group2[-3] , group2[-1] ,
       stringsAsFactors = F) |>subset(Var2 != Var3)

rbind(a,b) |> unname() |> apply(1,list) |> unlist(F)

• 輸出

[[1]]
[1] "a" "c"

[[2]]
[1] "b" "c"

[[3]]
[1] "a" "d"

[[4]]
[1] "b" "d"

[[5]]
[1] "a" "e"

[[6]]
[1] "b" "e"

$`1`
[1] "a" "c" "d"

$`2`
[1] "b" "c" "d"

$`5`
[1] "a" "c" "e"

$`6`
[1] "b" "c" "e"

$`7`
[1] "a" "d" "e"

$`8`
[1] "b" "d" "e"

標籤：r 列表 组合

上一篇：對整數字串進行排序并保存原始索引

下一篇：串列中的多個搜索詞