合并命名向量列表而无需修改名称

Question

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 

Answer 1

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 

Answer 2

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))