Data frame to nested list

Question

I have a dataframe which I read from a .csv file and looks like this:

              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")

I want to transform it into a list of lists, where, for example:

myList$developer

would give me a list containing all developers, and then

myList$developer$john

would give me a list of phone numbers associated with developers named John. Is there any simple way of doing it?

If you're curious as to why I'd want to do something like that: the actual data frame I'm working with is huge, so finding a specific entry by 4 parameters (in this example I can find a specific entry with 2 parameters: job, name) takes way too much time using filter. I think that the hash table structure of a nested list might take a lot of time to build, but would be searchable in O(1), which definitely works for me. If I'm wrong and you have a better way of doing it I'd love to hear it too.

`lapply(split(df,df$job), function(x) split(x$phone_number,x$name))` would do it. — Andrew Gustar, Sep 25 '17 at 14:36

Jaap · Accepted Answer · 2017-09-25T14:41:45.040

You can use a double split with lapply and the drop = TRUE-parameter for that. Using drop = TRUE will drop levels that do not occur, thus preventing the creation of empty list elements.

Using:

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

Or in one go:

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

gives:

> 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

Used data:

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"))

score 6 · Answer 2 · answered Sep 25 '17 at 14:42

the actual data frame I'm working with is huge, so finding a specific entry by 4 parameters (in this example I can find a specific entry with 2 parameters: job, name) takes way too much time using filter. I think that the hash table structure of a nested list might take a lot of time to build, but would be searchable in O(1), which definitely works for me. If I'm wrong and you have a better way of doing it I'd love to hear it too.

Apparently name lookup behaves like O(n), not O(1).

A possibly better way would be to use data.table, which uses binary search.

library(data.table)
setDT(DT, key = c("job", "name"))

get_phones = function(..., d = DT) d[list(...), phone]

Example usage

get_phones("developer", "john")
# [1] 654

get_phones("administrative")
# [1] 651 984 325 654 357

See vignette("datatable-keys-fast-subset") or the (possibly outdated) copy online.

not marking as solution, because it is not a solution to the question in the title, but it was a better fit for my needs. Thanks a lot! — shayelk, Sep 26 '17 at 09:27

Data frame to nested list

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