Accesing variable name passed as argument inside apply - r

Question

I made almost the same question in another post, but asking just for column name, and received a perfect solution for that need. Now what I need is the variable full name. I reformulate here.
I use 'deparse(substitute(x))' from inside my function to get variable name passed as parameter. It works great... but not with 'lapply'
myfun <- function(x)
{
return(deparse(substitute(x)))
}
a <- c(1,2,3)
b <- c(4,5,5)
df<-data.frame(a,b)
myfun(df$a)
[1] "df$a"
but, with 'lapply'...
lapply(df, myfun)
$a
[1] "X[[i]]"
$b
[1] "X[[i]]"
How can I get the variable name inside 'lapply'?
Thanks

Answer 1

When you pass a data frame to lapply, it iterates through the columns by numerical indexing using the double square bracket, not name indexing using the $ accessor. It is equivalent to using the following loop:
X <- df
result <- list()
for(i in seq_along(X)) {
result[[i]] <- myfun(X[[i]])
}
names(result) <- names(X)
result
#> $a
#> [1] "X[[i]]"
#>
#> $b
#> [1] "X[[i]]"
So a simple deparse(substitute(x)) will not work inside lapply. You are not recovering the column name, but rather would need to reconstruct it from the call stack. This is full of caveats and gotchas, but a (relatively) simple approach would be:
myfun <- function(x) {
stack <- lapply(sys.calls(), function(x) sapply(as.list(x), deparse))
if(stack[[length(stack)]][1] == 'myfun') {
return(stack[[length(stack)]][2])
}
if(stack[[length(stack)]][1] == 'FUN') {
return(paste0(stack[[length(stack) - 1]][2], '$',
eval(quote(names(X)[i]), parent.frame())))
}
deparse(substitute(x))
}
This means your function will still work if called directly:
myfun(df$a)
#> [1] "df$a"
But will also work within lapply
lapply(df, myfun)
#> $a
#> [1] "df$a"
#>
#> $b
#> [1] "df$b"
lapply(iris, myfun)
#> $Sepal.Length
#> [1] "iris$Sepal.Length"
#>
#> $Sepal.Width
#> [1] "iris$Sepal.Width"
#>
#> $Petal.Length
#> [1] "iris$Petal.Length"
#>
#> $Petal.Width
#> [1] "iris$Petal.Width"
#>
#> $Species
#> [1] "iris$Species"
It is specifically written to cover direct use or use within lapply. If you wanted to expand its use to work within other functional calls like Map or the various purrr mapping functions, then these would have to be covered specifically by their own if clauses.

Answer 2

Here is another solution, its a bit verbose and Allen's solution is much better:
myfun <- function(x) {
pf <- parent.frame()
x_nm <- deparse(substitute(x))
frame_n <- sys.nframe()
apply <- FALSE
while(frame_n > 0) {
cl <- as.list(sys.call(frame_n))
if (grepl("apply", cl[[1]])) {
x_obj <- cl[[2]]
apply <- TRUE
break
}
frame_n <- frame_n - 1L
}
if (apply) {
idx <- parent.frame()$i[]
obj <- get(x_obj, envir = pf)
if (!is.null(names(obj)[idx])) {
nm_or_idx <- names(obj)[idx]
} else {
nm_or_idx <- idx
}
x_nm <- paste0(x_obj, '$', nm_or_idx)
}
return(x_nm)
}
myfun(df$a)
#> [1] "df$a"
lapply(df, myfun)
#> $a
#> [1] "df$a"
#>
#> $b
#> [1] "df$b"
Created on 2023-02-09 by the reprex package (v2.0.1)

Answer 3

We can define a character string 'col_name'to take the name of the data frame column in the function. For example, if col_name is "a", df[[col_name]] extracts "a" column from data frame.Then we can use the paste() function to concatenate the string 'df$' and 'col_name':
myfun <- function(col_name) {
col <- df[[col_name]]
return(paste("df$", col_name, sep = ""))
}
lapply(colnames(df), myfun)
output
[[1]]
[1] "df$a"
[[2]]
[1] "df$b"
If we would like to assign any data we could do the assignment and then run lapply for example:
df <- iris
lapply(colnames(df), myfun)
output
[[1]]
[1] "df$Sepal.Length"
[[2]]
[1] "df$Sepal.Width"
[[3]]
[1] "df$Petal.Length"
[[4]]
[1] "df$Petal.Width"
[[5]]
[1] "df$Species"
I hope this could helps.