deparse(substitute(x)) in lapply?

问题

I would like use a function that uses the standard deparse(substitute(x)) trick within lapply. Unfortunately I just get the argument of the loop back. Here's my completely useless reproducible example:

# some test data
a <- 5
b <- 6 
li <- list(a1=a,b2=b)

# my test function
tf <- function(obj){
nm <- deparse(substitute(obj))
res <- list(myName=nm)
res
}

tf(a)
#returns
$myName
[1] "a"

which is fine. If I use lapply I either get [[1L]] or the x argument of an anonymous function.

lapply(li,function(x) tf(x))
# returns
$a1
$a1$myName
[1] "x"


$b2
$b2$myName
[1] "x"

Is there any way to obtain the following?

$a1
$a1$myName
[1] "a1"


$b2
$b2$myName
[1] "b1"

If there's anything more general on deparse(substitute(x)) and lapply I'd also eager to know.

EDIT: The problem as opposed to using an anonymous function that accepts multiple arguments and can thus use the name of the object and the object itself does not work because, the tf function will only accept one argument. So this does not work here...

回答1:

A possible solution :

lapply(li, function(x) {
  call1 <-  sys.call(1)
  call1[[1]] <- as.name("names")
  call1 <- call1[1:2]
  nm <- eval(call1)
  y <- deparse(substitute(x))
  y <- gsub("\\D", "", y)
  y <- as.numeric(y)
  list(myname=nm[y])
})

来源：https://stackoverflow.com/questions/18508790/deparsesubstitutex-in-lapply