r - 如何提取列表列表中所有列表的行名并将它们存储在新的数据框或列表列表中

Question

我有一个列表列表。下面介绍了一个类似的玩具示例。我想从每个列表中提取行名，然后将这些行名存储在新的数据帧或新的列表列表中(与原始结构相同)。

理想情况下，列名或新列表名称应与列表列表中的列表名称相同。

注意。这些列表的长度各不相同，必须予以考虑。我宁愿不用 N/A 填充空白。

---

dput(头(Chars_alive)):

list(FEB_games = list(GAME1 = structure(list(GAME1_Class = structure(c(2L, 
1L, 5L, 4L, 3L), .Label = c("fighter", "paladin", "rouge", "sorcerer", 
"wizard"), class = "factor"), GAME1_Race = structure(c(3L, 1L, 
4L, 3L, 2L), .Label = c("elf", "gnome", "human", "orc"), class = "factor"), 
GAME1_Alignment = structure(c(4L, 2L, 1L, 5L, 3L), .Label = c("CE", 
"CG", "LG", "NE", "NN"), class = "factor"), GAME1_Level = c(6, 
7, 6, 7, 7), GAME1_Alive = structure(c(1L, 1L, 1L, 1L, 1L
), .Label = "y", class = "factor")), row.names = c("Stan", 
"Kenny", "Cartman", "Kyle", "Butters"), class = "data.frame"), 
GAME2 = structure(list(GAME2_Class = structure(c(5L, 2L, 
4L, 1L), .Label = c("bard", "cleric", "fighter", "monk", 
"wizard"), class = "factor"), GAME2_Race = structure(c(3L, 
2L, 4L, 1L), .Label = c("dwarf", "elf", "half-elf", "human"
), class = "factor"), GAME2_Alignment = structure(c(2L, 1L, 
5L, 3L), .Label = c("CE", "CG", "LG", "NE", "NN"), class = "factor"), 
    GAME2_Level = c(5, 5, 5, 5), GAME2_Alive = structure(c(2L, 
    2L, 2L, 2L), .Label = c("n", "y"), class = "factor")), row.names = c("Kenny", 
"Cartman", "Kyle", "Butters"), class = "data.frame")), MAR_games = list(
GAME3 = structure(list(GAME3_Class = structure(c(2L, 1L, 
5L, 3L), .Label = c("barbarian", "cleric", "monk", "ranger", 
"warlock"), class = "factor"), GAME3_Race = structure(c(2L, 
3L, 2L, 1L), .Label = c("dwarf", "elf", "half-elf", "human"
), class = "factor"), GAME3_Alignment = structure(c(2L, 2L, 
1L, 2L), .Label = c("CE", "LG", "LN"), class = "factor"), 
    GAME3_Level = c(1, 1, 1, 1), GAME3_Alive = structure(c(2L, 
    2L, 2L, 2L), .Label = c("n", "y"), class = "factor")), row.names = c("Stan", 
"Kenny", "Cartman", "Butters"), class = "data.frame"), GAME4 = structure(list(
    GAME4_Class = structure(c(1L, 5L, 4L, 3L), .Label = c("fighter", 
    "paladin", "rouge", "sorcerer", "wizard"), class = "factor"), 
    GAME4_Race = structure(c(3L, 2L, 4L, 1L), .Label = c("dwarf", 
    "elf", "half-elf", "human"), class = "factor"), GAME4_Alignment = structure(c(2L, 
    1L, 4L, 3L), .Label = c("CE", "CG", "LG", "LN"), class = "factor"), 
    GAME4_Level = c(5, 5, 5, 5), GAME4_Alive = structure(c(2L, 
    2L, 2L, 2L), .Label = c("n", "y"), class = "factor")), row.names = c("Kenny", 
"Cartman", "Kyle", "Butters"), class = "data.frame")))

---

as.data.frame(rownames(Chars_alive[[1]][[1]])) -> GAME1
as.data.frame(rownames(Chars_alive[[2]][[1]])) -> GAME2

由于 GAME1 和 GAME2 的长度不同，数据帧可能并不理想(我的实际数据在列表列表之间的长度差异很大)。

for (i in Chars_alive) {
  for (j in i)
    rownames(j) -> x
}

for 循环可以工作，但我对循环不熟悉，不知道如何将所有第 j 个元素放入一个新的数据框或列表中。

ls2 <- list(Game1 <- rownames(Chars_alive[[1]][[1]]), Game2 <- rownames(Chars_alive[[1]][[2]]),
                 Game3 <- rownames(Chars_alive[[2]][[1]]), Game4 <- rownames(Chars_alive[[2]][[2]]))

也许直接制作一个新列表会起作用，但如果是这种情况，我想保留原始列表的结构，即 FEB_games > GAME1、GAME2 和 MAR_games > GAME3、GAME4。另外，我希望保持列表名称相同，即 GAME1、GAME2、GAME3 和 GAME4。

---

理想的输出是数据框:

    GAME1    GAME2    GAME3    GAME4
1   Stan     Kenny    Stan     Kenny
2   Kenny    Cartman  Kenny    Cartman
3   Cartman  Kyle     Cartman  Kyle 
4   Kyle     Butters  Butters  Butters
5   Butters   

或者列表:

Listname
    FEB_games
        GAME1
           'Stan', 'Kenny', 'Cartman', 'Kyle', 'Butters'
        GAME2
           'Kenny', 'Cartman', 'Kyle', 'Butters'
    MAR_games
        GAME3
            'Stan', 'Kenny', 'Cartman', 'Butters'
        GAME4
            'Kenny', 'Cartman', 'Kyle', 'Butters'

Answer 1

您好，我会像这样在 lapply 中使用 lapply。我将您的列表称为“list_games”。

lapply(list_games, function(x){lapply(x, row.names)})

这给你

$FEB_games
$FEB_games$GAME1
[1] "Stan"    "Kenny"   "Cartman" "Kyle"    "Butters"

$FEB_games$GAME2
[1] "Kenny"   "Cartman" "Kyle"    "Butters"


$MAR_games
$MAR_games$GAME3
[1] "Stan"    "Kenny"   "Cartman" "Butters"

$MAR_games$GAME4
[1] "Kenny"   "Cartman" "Kyle"    "Butters"

如果行名的长度相同，您可以将其保存为 data.frame

do.call("rbind.data.frame", lapply(list_games, function(x){lapply(x, row.names)}))

这在这里不起作用，因为行名称的长度不同。在这种情况下，您可以执行以下操作:

res <- sapply(list_games, function(x){lapply(x, row.names)})
n.obs <- sapply(res , length)
seq.max <- seq_len(max(n.obs))
df <- data.frame(t(sapply(res, "[", i = seq.max)))
df
     X1      X2      X3      X4      X5
1  Stan   Kenny Cartman    Kyle Butters
2 Kenny Cartman    Kyle Butters    <NA>
3  Stan   Kenny Cartman Butters    <NA>
4 Kenny Cartman    Kyle Butters    <NA>

如果您需要进一步的解释，请告诉我。最后一部分的完成类似于 here