Naming a list of vectors by the vectors names

Question

I have 3 different vectors, and i would like to combine then into a list of vectors. But i would like to maintain the names of the vectors in the lists names. Below is the way i've been doing it.

A<-seq(from=1, to=10, by=1)
B<-rep(11,10)
C<-seq(from=50, to=100, by=10)

ABC<-list(A,B,C)

This is the part i would like streamlined

names(ABC)<-c("A", "B", "C")

Thank you.

If you're doing this frequently, I suppose you could define a utility function for it, like: ```named_list <- function(...) `names<-`(list(...), as.character(substitute(list(...))[-1]))``` Then it's just `named_list(A, B, C)` — Tim Goodman, Apr 03 '17 at 23:38
Probably not worth it unless you have something more like: `list(A=A,B=B,C=C,D=D,E=E,F=F,G=G,H=H,...,Z=Z)` :) — Tim Goodman, Apr 03 '17 at 23:44
If you happen to have tibble/tidyverse loaded anyway, `tibble::lst(A, B, C)` will automatically set names. — alistaire, Apr 03 '17 at 23:44
Inspired by the output of `dput`: `ABC <- structure(list(A, B, C), names = c("A", "B", "C"))` — mt1022, Apr 04 '17 at 01:16
@TimGoodman This is more of what i am looking for. Still a bit confused on when to use substitute vs. quote. — MadmanLee, Apr 04 '17 at 01:51
@MadmanLee A key difference it that if the input to `substitute` contains formal parameters for the function you're in, it will replace them with the actual (unevaluated) arguments from outside the function. So in my comment above, `substitute(list(...))` returned the call `list(A, B, C)`, whereas `quote(list(...))` would have returned the call `list(...)`. — Tim Goodman, Apr 04 '17 at 17:29

score 0 · Answer 1 · answered Apr 13 '17 at 21:29

Alrighty, using @TimGoodman 's suggestion i created a function that does this for me. Wishing list had this as an optional function.

named.list<-function(...){
    bob<-list(...)
    names(bob)<-as.character(substitute((...)))[-1]
    return(bob)
}

Thank you!

Naming a list of vectors by the vectors names

1 Answers1