使用分組變數確定所有可能的專案組合

這個問題與這個問題非常相似，但不是將每個原始組中的一個專案放入每個輸出組中，而是希望每個輸出組中每個原始組中的多個專案。

我有以下資料。

list1 <- list(Group_1 = c("1", "2", "3", "4", "5", "6"), Group_2 = c("13", "14", "15", "16", "17", "18"), Group_3 = c("19", "20", "21", "22", "23", "24", "25"))
Number_of_Items_From_Each_Original_Group_to_End_up_in_Each_Output_Group <- 2

Group_1包含 6 項，Group_2包含 6 項，Group_3包含 7 項。基于這些初始組，我想將專案放入 3 個新組中，確保每個新組包含Number_of_Items_From_Each_Original_Group_to_End_up_in_Each_Output_Group每個原始組中相同數量（2 或 ）的專案。此外，這 3 個新組中的任何一個專案都不能重復 - 每個專案只能在每個新組中使用一次。例如，一個可能的輸出可能是下面的這個輸出。

(list(New_Group_1 = c("1", "2", "13", "14", "19", "20"), New_Group_2 = c("3", "4", "15", "16", "21", "22"), New_Group_3 = c("5", "6", "17", "18", "23", "24")))

我實際上想生成一個list顯示每個可能輸出（每個可能的組合）的輸出。對于每個串列元素，應該有三個新組，每個組中應該有不同的專案。例如，除了上面的解決方案之外，另一個可能是以下解決方案。

(list(New_Group_1 = c("3", "4", "13", "14", "19", "20"), New_Group_2 = c("5", "6", "15", "16", "21", "22"), New_Group_3 = c("1", "2", "17", "18", "23", "24")))

我在這里問的問題非常相似，只是在那個問題中，我只將每個原始組中的一項放入每個新組中。在這里，我想選擇將每個原始組中的多個專案放入每個新組中。

uj5u.com熱心網友回復：

這是相當棘手的，而且相當緩慢，因為指定的三個組有 23,625 種可能的組合

n <- 2

all_pairs <- lapply(list1, function(group) {
  all_combs <- apply(combn(ncol(combn(length(group), n)), length(list1)), 
                     2,
        function(x) c(combn(length(group), n)[,x]), simplify = FALSE)
  
  all_combs[sapply(all_combs, function(x) !any(duplicated(x)))]
})

index_df <- do.call(expand.grid, lapply(all_pairs, function(x) seq(length(x))))

result <- apply(index_df, 1, function(x) {
  m <- t(sapply(seq_along(x), function(y) list1[[y]][all_pairs[[y]][[x[y]]]]))
  setNames(lapply(split(m, (seq_along(m)-1) %/% (length(list1) * n)),
         function(x) c(t(matrix(x, ncol = n)))), names(list1))
}, simplify = FALSE)

結果串列如下所示：

head(result)
#> [[1]]
#> [[1]]$Group_1
#> [1] "1"  "2"  "13" "14" "19" "20"
#> 
#> [[1]]$Group_2
#> [1] "3"  "4"  "15" "16" "21" "22"
#> 
#> [[1]]$Group_3
#> [1] "5"  "6"  "17" "18" "23" "24"
#> 
#> 
#> [[2]]
#> [[2]]$Group_1
#> [1] "1"  "2"  "13" "14" "19" "20"
#> 
#> [[2]]$Group_2
#> [1] "3"  "5"  "15" "16" "21" "22"
#> 
#> [[2]]$Group_3
#> [1] "4"  "6"  "17" "18" "23" "24"
#> 
#> 
#> [[3]]
#> [[3]]$Group_1
#> [1] "1"  "2"  "13" "14" "19" "20"
#> 
#> [[3]]$Group_2
#> [1] "3"  "6"  "15" "16" "21" "22"
#> 
#> [[3]]$Group_3
#> [1] "4"  "5"  "17" "18" "23" "24"
#> 
#> 
#> [[4]]
#> [[4]]$Group_1
#> [1] "1"  "3"  "13" "14" "19" "20"
#> 
#> [[4]]$Group_2
#> [1] "2"  "4"  "15" "16" "21" "22"
#> 
#> [[4]]$Group_3
#> [1] "5"  "6"  "17" "18" "23" "24"
#> 
#> 
#> [[5]]
#> [[5]]$Group_1
#> [1] "1"  "3"  "13" "14" "19" "20"
#> 
#> [[5]]$Group_2
#> [1] "2"  "5"  "15" "16" "21" "22"
#> 
#> [[5]]$Group_3
#> [1] "4"  "6"  "17" "18" "23" "24"
#> 
#> 
#> [[6]]
#> [[6]]$Group_1
#> [1] "1"  "3"  "13" "14" "19" "20"
#> 
#> [[6]]$Group_2
#> [1] "2"  "6"  "15" "16" "21" "22"
#> 
#> [[6]]$Group_3
#> [1] "4"  "5"  "17" "18" "23" "24"

^{由reprex 包(v2.0.1)創建于 2022-06-09}

uj5u.com熱心網友回復：

可能你可以定義一個像下面這樣的函式（它生成按元素f分組的所有排列）nv

f <- function(v, n = 2) {
    if (length(v) < n) {
        return(list(NULL))
    }
    if (length(v) == n) {
        return(list(v))
    }
    x <- combn(v, n, simplify = FALSE)
    unlist(lapply(x, function(p) Map(rbind, list(p), f(v[!v %in% p], n))), recursive = FALSE)
}

res <- apply(expand.grid(lapply(list1, f)), 1, function(x) asplit(do.call(cbind, x), 1))

結果的一個子集看起來像

> apply(head(expand.grid(lapply(list1, f))), 1, fun .... [TRUNCATED]
$`1`
$`1`[[1]]
[1] "1"  "2"  "13" "14" "19" "20"

$`1`[[2]]
[1] "3"  "4"  "15" "16" "21" "22"

$`1`[[3]]
[1] "5"  "6"  "17" "18" "23" "24"


$`2`
$`2`[[1]]
[1] "1"  "2"  "13" "14" "19" "20"

$`2`[[2]]
[1] "3"  "5"  "15" "16" "21" "22"

$`2`[[3]]
[1] "4"  "6"  "17" "18" "23" "24"


$`3`
$`3`[[1]]
[1] "1"  "2"  "13" "14" "19" "20"

$`3`[[2]]
[1] "3"  "6"  "15" "16" "21" "22"

$`3`[[3]]
[1] "4"  "5"  "17" "18" "23" "24"


$`4`
$`4`[[1]]
[1] "1"  "2"  "13" "14" "19" "20"

$`4`[[2]]
[1] "4"  "5"  "15" "16" "21" "22"

$`4`[[3]]
[1] "3"  "6"  "17" "18" "23" "24"


$`5`
$`5`[[1]]
[1] "1"  "2"  "13" "14" "19" "20"

$`5`[[2]]
[1] "4"  "6"  "15" "16" "21" "22"

$`5`[[3]]
[1] "3"  "5"  "17" "18" "23" "24"


$`6`
$`6`[[1]]
[1] "1"  "2"  "13" "14" "19" "20"

$`6`[[2]]
[1] "5"  "6"  "15" "16" "21" "22"

$`6`[[3]]
[1] "3"  "4"  "17" "18" "23" "24"

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