r - 数据框到嵌套列表

Question

我有一个从 .csv 文件中读取的数据框，如下所示:

              job    name `phone number`
            <chr>   <chr>          <int>
 1      developer    john            654
 2      developer    mike            321
 3      developer  albert            987
 4        manager    dana            741
 5        manager     guy            852
 6        manager    anna            936
 7      developer     dan            951
 8      developer   shean            841
 9 administrative  rebeca            357
10 administrative  krissy            984
11 administrative   hilma            651
12 administrative    otis            325
13 administrative   piper            654
14        manager   mendy            984
15        manager corliss            321

DT = structure(list(job = c("developer", "developer", "developer", 
"manager", "manager", "manager", "developer", "developer", "administrative", 
"administrative", "administrative", "administrative", "administrative", 
"manager", "manager"), name = c("john", "mike", "albert", "dana", 
"guy", "anna", "dan", "shean", "rebeca", "krissy", "hilma", "otis", 
"piper", "mendy", "corliss"), phone = c(654L, 321L, 987L, 741L, 
852L, 936L, 951L, 841L, 357L, 984L, 651L, 325L, 654L, 984L, 321L
)), .Names = c("job", "name", "phone"), row.names = c(NA, -15L
), class = "data.frame")

我想将其转换为列表列表，例如:

myList$developer

会给我一个包含所有开发人员的列表，然后

myList$developer$john

会给我一个与名为 John 的开发人员相关联的电话号码列表。有什么简单的方法吗？

如果您对我为什么想做这样的事情感到好奇:我正在使用的实际数据框很大，因此可以通过 4 个参数找到特定条目(在本例中，我可以找到具有 2 个参数的特定条目)参数:工作，名称)使用过滤器花费太多时间。我认为嵌套列表的哈希表结构可能需要很多时间来构建，但可以在 O(1) 中进行搜索，这对我来说绝对有用。
如果我错了，而您有更好的方法，我也很想听听。

Answer 1

您可以使用带有 split 和 lapply 参数的双 drop = TRUE 。使用 drop = TRUE 将删除不会发生的级别，从而防止创建空列表元素。

使用:

l <- split(dat, dat$job, drop = TRUE)
nestedlist <- lapply(l, function(x) split(x, x[['name']], drop = TRUE))

或者一口气:

nestedlist <- lapply(split(dat, dat$job, drop = TRUE),
                     function(x) split(x, x[['name']], drop = TRUE))

给出:

> nestedlist
$administrative
$administrative$hilma
              job  name phonenumber
11 administrative hilma         651

$administrative$krissy
              job   name phonenumber
10 administrative krissy         984

$administrative$otis
              job name phonenumber
12 administrative otis         325

$administrative$piper
              job  name phonenumber
13 administrative piper         654

$administrative$rebeca
             job   name phonenumber
9 administrative rebeca         357


$developer
$developer$albert
        job   name phonenumber
3 developer albert         987

$developer$dan
        job name phonenumber
7 developer  dan         951

$developer$john
        job name phonenumber
1 developer john         654

$developer$mike
        job name phonenumber
2 developer mike         321

$developer$shean
        job  name phonenumber
8 developer shean         841


$manager
$manager$anna
      job name phonenumber
6 manager anna         936

$manager$corliss
       job    name phonenumber
15 manager corliss         321

$manager$dana
      job name phonenumber
4 manager dana         741

$manager$guy
      job name phonenumber
5 manager  guy         852

$manager$mendy
       job  name phonenumber
14 manager mendy         984

使用数据:

dat <- structure(list(job = c("developer", "developer", "developer", "manager", "manager", "manager", "developer", "developer", "administrative", "administrative", "administrative", "administrative", "administrative", "manager", "manager"),
                      name = c("john", "mike", "albert", "dana", "guy", "anna", "dan", "shean", "rebeca", "krissy", "hilma", "otis", "piper", "mendy", "corliss"),
                      phonenumber = c(654L, 321L, 987L, 741L, 852L, 936L, 951L, 841L, 357L, 984L, 651L, 325L, 654L, 984L, 321L)),
                 .Names = c("job", "name", "phonenumber"), class = "data.frame", row.names = c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15"))