deparse(substitute()) 与 lapply

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)