我在嵌套列表中有一组数据,这样顶层列表的每个成员都有描述符(但不一定是相同的描述符),然后是子列表,这些子节点可以有子节点,依此类推.孩子们的深度是任意的.一个小小的例子:
family <- list(
list(name = "Alice", age = 40, eyes = "blue", children = list(
list(name = "Bob", age = 20, eyes = "blue"),
list(name = "Charlie", age = 18, eyes = "brown")
)),
list(name = "Dan", age = 12, eyes = "green"),
list(name = "Erin", age = 69, eyes = "green", children = list(
list(name = "Frank", age = 45, eyes = "blue", children = list(
list(name = "George", age = 24, eyes = "blue", children = list(
list(name = "Harry", age = 2, eyes = "green")
)),
list(name = "Ingrid", age = 22, eyes = "brown", hair = "brown"),
list(name = "Jack", age = 29, eyes = "brown")
)),
list(name = "Karen", age = 43),
list(name = "Larry", age = 21, eyes = "blue")
))
)
> str(family, max.level = 2)
List of 3
$ :List of 4
..$ name : chr "Alice"
..$ age : num 40
..$ eyes : chr "blue"
..$ children:List of 2
$ :List of 3
..$ name: chr "Dan"
..$ age : num 12
..$ eyes: chr "green"
$ :List of 4
..$ name : chr "Erin"
..$ age : num 69
..$ eyes : chr "green"
..$ children:List of 3
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理想情况下,我想创建一个数据框,使每行都是一个系列成员,每列都是列表中的一个属性:
name age eyes
1 Alice 40 blue
2 Bob 20 blue
3 Charlie 18 brown
(etc)
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但目前还不清楚如何递归到任意深度.我设法让顶级成员使用map(family, ~ .$name),但我不明白如何更深入.一些成员没有任何孩子这一事实使事情复杂化.
我已经阅读了purrr文档,但我找不到任何看起来会有所帮助的内容.也许我没有使用正确的术语.
建议表示赞赏.谢谢!
您始终可以递归地遍历列表并收集信息.除了map_dfr从purrr我的解决方案采用defaults从plyr处理缺失眼睛或头发的颜色的情况下.
get_people <- function(x) {
if (is.null(x)) return(NULL)
if (!is.null(x$name)) { # if there's a name, we're at the level of a person
children <- x$children
x$children <- NULL
row <- data.frame(plyr::defaults(x, list(age = NA, eyes = NA, hair = NA)),
stringsAsFactors = FALSE)
rbind(row, get_people(children))
}
else {
purrr::map_dfr(x, get_people)
}
}
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当您将其应用于您的列表时:
> get_people(family)
name age eyes hair
1 Alice 40 blue <NA>
2 Bob 20 blue <NA>
3 Charlie 18 brown <NA>
4 Dan 12 green <NA>
5 Erin 69 green <NA>
6 Frank 45 blue <NA>
7 George 24 blue <NA>
8 Harry 2 green <NA>
9 Ingrid 22 brown brown
10 Jack 29 brown <NA>
11 Karen 43 <NA> <NA>
12 Larry 21 blue <NA>
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