Suppose I have
label <- 'My val'
and I try to create the list
Output <- list(
label = pi
)
I get that the name of the first (only) object in the list is "label" but I want "My val".
I understand I can do
names(Output) <- label
But the list is quite long and I'd rather name it in the list function.
Another option is lst from dplyr
library(dplyr)
lst(!!label := pi)
#$`My val`
#[1] 3.141593
No, you can't reference variables for names when using the list() function to create a list. It will just interpret any variable name as name for the entry. But after constructing the list, you can change the names:
label <- 'My val'
Output <- list(pi)
names(Output)=label
Maybe this could help you, but it would've been much better if you could share a more detailed sample as I thought there might be more variable names involved:
label <- 'My val'
Output <- list(
label = pi
)
Output |>
setNames(label)
$`My val`
[1] 3.141593
data.tables substitute2() might help:
library(data.table)
labels <- list(l1 = 'My val', l2 = 'Other label')
Output <- eval(data.table::substitute2(
list( l1 = pi,
l2 = exp(1)),
env = labels
) )
