CODEWITHSUNDEEP
r
nathanielng
nathanielng

Reputation: 1725

Combining a list of named vectors without mangling the names

How do I combine a list of named vectors? I need to split a vector of integers (with characters for names) for use with parallel::parSapply() and combine them back again. Example code:

text <- 1:26
names(text) <- letters
n <- 4
text <- split(text, cut(1:length(text),breaks=n,labels=1:n))
# text <- parSapply(..., text, ...) would go here in the actual code

However, the names get mangled when I use unlist to convert the data back into a named vector:

> unlist(text)
1.a 1.b 1.c 1.d 1.e 1.f 1.g 2.h 2.i 2.j 2.k 2.l 2.m 3.n 3.o 3.p 3.q 3.r 3.s 4.t 4.u 4.v 
  1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20  21  22 
4.w 4.x 4.y 4.z 
 23  24  25  26

What I'm looking for is the following result (except that it should work with any value of n):

> c(text[[1]],text[[2]],text[[3]],text[[4]])
 a  b  c  d  e  f  g  h  i  j  k  l  m  n  o  p  q  r  s  t  u  v  w  x  y  z 
 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Upvotes: 3

Views: 687

Answers (2)

Ricky
Ricky

Reputation: 4686

You can remove the list elements names first then there won't be compound naming happening.

> names(text) <- NULL
> do.call(c, text)
 a  b  c  d  e  f  g  h  i  j  k  l  m  n  o  p  q  r  s  t  u  v  w  x  y  z 
 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Same as

> unlist(text)
 a  b  c  d  e  f  g  h  i  j  k  l  m  n  o  p  q  r  s  t  u  v  w  x  y  z 
 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Or as @RichardScriven pointed out in the comment, you can do it as follows without removing the name in the source variable: do.call("c", c(text, use.names = FALSE))

Upvotes: 2

akrun
akrun

Reputation: 887501

One option without changing the structure of 'text' would be to change the names of the vector (unlist(text)) with the names of onjects within the list elements.

setNames(unlist(text), unlist(sapply(text, names)))
#  a  b  c  d  e  f  g  h  i  j  k  l  m  n  o  p  q  r  s  t  u  v  w  x  y  z 
#  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Or if it is okay to remove the names of the 'text' object, set the names of 'text' to NULL and then unlist

unlist(setNames(text, NULL))
#  a  b  c  d  e  f  g  h  i  j  k  l  m  n  o  p  q  r  s  t  u  v  w  x  y  z 
# 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Upvotes: 4

Related Questions