CODEWITHSUNDEEP
rlist
user1322296
user1322296

Reputation: 566

list names as vector in list

What is the quickest way to get from the following list:

 set.seed(1)
 mylist <- list(1:2,1:3,1:4)
 names(mylist) <- sample(LETTERS,3)

 mylist
$J
[1] 1 2

$V
[1] 1 2 3

$Z
[1] 1 2 3 4

to a list with an extra vector containing the names of list i.e.

mylist
$J
[1] 1 2
[2] J J

$V
[1] 1 2 3
[2] V V V
$Z
[1] 1 2 3 4
[2] Z Z Z Z

the result could equally be a list of data frames. Some sort of 

lapply(mylist, function(x) { data.frame(x=x,y=names(x)})

Upvotes: 1

Views: 161

Answers (2)

Ricardo Saporta
Ricardo Saporta

Reputation: 55420

Personally, I like @Matthew Plourde's use of stack

Here's an option using mapply that preserves the same structure of the original list 

mapply(function(L, N) 
     {names(L) <- rep(N, length(L)); L},   mylist, names(mylist))

Results:

$G
G G 
1 2 

$J
J J J 
1 2 3 

$N
N N N N 
1 2 3 4

Upvotes: 4

Matthew Plourde
Matthew Plourde

Reputation: 44614

Another option:

df<-stack(mylist)
split(df, df$ind)

Maybe stack(mylist) is all you really need?

Upvotes: 10

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