deparse(substitute()) with lapply - r

Question

I want to give labels to data frames using a combination of a small function in combination with lapply()
I have the following code:
df1 <- data.frame(c(1,2,3), c(3,4,5))
df2 <- data.frame(c(6,7,8), c(9,10,11))
f.generate.name <- function(x) {
x$name <- deparse(substitute(x))
return(x)
}
my_list <- list(df1, df2)
# This works fine.
f.generate.name(df1)
# This does not work.
lapply(my_list, f.generate.name)
which produces the following output
[[1]]
c.1..2..3. c.3..4..5. name
1 1 3 X[[i]]
2 2 4 X[[i]]
3 3 5 X[[i]]
[[2]]
c.6..7..8. c.9..10..11. name
1 6 9 X[[i]]
2 7 10 X[[i]]
3 8 11 X[[i]]
What I want instead is:
[[1]]
c.1..2..3. c.3..4..5. name
1 1 3 df1
2 2 4 df1
3 3 5 df1
[[2]]
c.6..7..8. c.9..10..11. name
1 6 9 df2
2 7 10 df2
3 8 11 df2
What is the best way of dong this without using loops? How can I tweak the lapply() function or the function that I created in order to achieve the desired result?

Answer 1

Base R
lapply() cannot iterate over more than one argument. You can use mapply() or its wrapper Map() in this case that always returns a list.
Map(f = function(x, y){
x$name <- y
x },
my_list,
names(my_list))
$df1
c.1..2..3. c.3..4..5. name
1 1 3 df1
2 2 4 df1
3 3 5 df1
$df2
c.6..7..8. c.9..10..11. name
1 6 9 df2
2 7 10 df2
3 8 11 df2
Tidyverse
In case you are open to a purrr solution, you can use imap(). It makes the names of the object conveniently available. There is no need to write a function then:
my_list <- list(df1 = df1, df2 = df2)
imap(my_list, ~{
.x$name <- .y
.x
})
$df1
c.1..2..3. c.3..4..5. name
1 1 3 df1
2 2 4 df1
3 3 5 df1
$df2
c.6..7..8. c.9..10..11. name
1 6 9 df2
2 7 10 df2
3 8 11 df2

Answer 2

Really the question is where do the names come from? An unnamed list like my_list in the question has lost the df1 and df2 names as we can see by looking at its internals:
dput(my_list) # no df1 or df2 seen
## list(structure(list(c.1..2..3. = c(1, 2, 3), c.3..4..5. = c(3,
## 4, 5)), class = "data.frame", row.names = c(NA, -3L)), structure(list(
## c.6..7..8. = c(6, 7, 8), c.9..10..11. = c(9, 10, 11)), class =
## "data.frame", row.names = c(NA,
## -3L)))
Thus either we need to create a named list in the first place or else provide a vector of names. We show both using only base R.
Named list
First create a named list of the data frames and then use Map as shown:
L <- mget(ls("^df")) # create named list
Map(data.frame, L, name = names(L))
Unnamed list
Alternately if all you have is an unnamed list then we can Map over that and a vector of names:
my_list <- list(df1, df2) # unnamed list as in question
Map(data.frame, my_list, name = c("df1", "df2"))
Pass individual data frames
Yet another approach is to pass the individual data frames instead of a list. Because we have not destroyed the original names by creating an unnamed list we can still retrieve them. On R 4.0 and later deparse1 could optionally be used in place of deparse in the code.
add_names <- function(...) {
mc <- match.call()
Map(data.frame, list(...), names = sapply(mc[-1], deparse))
}
add_names(df1, df2)