Can lists be created that name themselves based on input object names?

问题

It would be very helpful to me to be able to create an R list object without having to specify the names of each element. For example:

a1 <- 1
a2 <- 20
a3 <- 1:20

b <- list(a1,a2,a3, inherit.name=TRUE)
> b

[[a1]]
[1] 1

[[a2]]
[1] 20

[[a3]]
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20

This would be ideal. Any suggestions?

回答1:

Coincidentally, I just wrote this function. It looks a lot like @joran's solution, but it tries not to stomp on already-named arguments.

namedList <- function(...) {
    L <- list(...)
    snm <- sapply(substitute(list(...)),deparse)[-1]
    if (is.null(nm <- names(L))) nm <- snm
    if (any(nonames <- nm=="")) nm[nonames] <- snm[nonames]
    setNames(L,nm)
}
## TESTING:
a <- b <- c <- 1
namedList(a,b,c)
namedList(a,b,d=c)
namedList(e=a,f=b,d=c)

Copied from comments: if you want something from a CRAN package, you can use Hmisc::llist:

Hmisc::llist(a, b, c, d=a, labels = FALSE)

The only apparent difference is that the individual vectors also have names in this case.

回答2:

The tidyverse package tibble has a function that can do this as well. Try out tibble::lst

tibble::lst(a1, a2, a3)
# $a1
#  [1] 1
#
# $a2
#  [1] 20
# 
# $a3
#  [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20

回答3:

A random idea:

a1<-1
a2<-20
a3<-1:20

my_list <- function(...){
    names <- as.list(substitute(list(...)))[-1L]
    result <- list(...)
    names(result) <- names
    result
}

> my_list(a1,a2,a3)
$a1
[1] 1

$a2
[1] 20

$a3
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20

(The idea is stolen from the code in data.frame.)

回答4:

Another idea ,

 sapply(ls(pattern='^a[0-9]'), get)
$a1
[1] 1

$a2
[1] 20

$a3
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20

来源：https://stackoverflow.com/questions/16951080/can-lists-be-created-that-name-themselves-based-on-input-object-names

标签

list

names