Question

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，那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

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)

但目前还不清楚如何递归到任意深度。我已经设法使用map(family, ~ .$name)让顶级成员，但我不明白如何更深入。一些成员没有孩子，这很复杂。

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！

Answer 1

您始终可以递归地遍历列表并收集信息。除了来自map_dfr的{{1}}我的解决方案使用purrr中的defaults来处理失踪的眼睛或头发颜色的情况。

plyr

将此内容应用到列表中时：

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)
  }
}

Answer 2

这是一个迭代children元素的选项，将人们收集到一个扁平列表中：

flatten_family <- function(x){
    ppl <- list()

    rflatten <- function(y){
        lapply(y, function(z){
            if(exists('children', z)) {
                ppl <<- c(ppl, list(z[-which(names(z) == 'children')]))
                rflatten(z$children)
            } else {
                ppl <<- c(ppl, list(z))
            }
        })
        ppl
    }

    rflatten(x)
}

dplyr::bind_rows(flatten_family(family))
#> # A tibble: 12 x 4
#>    name      age eyes  hair 
#>    <chr>   <dbl> <chr> <chr>
#>  1 Alice   40.0  blue  <NA> 
#>  2 Bob     20.0  blue  <NA> 
#>  3 Charlie 18.0  brown <NA> 
#>  4 Dan     12.0  green <NA> 
#>  5 Erin    69.0  green <NA> 
#>  6 Frank   45.0  blue  <NA> 
#>  7 George  24.0  blue  <NA> 
#>  8 Harry    2.00 green <NA> 
#>  9 Ingrid  22.0  brown brown
#> 10 Jack    29.0  brown <NA> 
#> 11 Karen   43.0  <NA>  <NA> 
#> 12 Larry   21.0  blue  <NA>

从多级嵌套列表中提取信息

2 个答案: