In R, how to get an object's name after it is sent to a function?

The old deparse-substitute trick:

a<-data.frame(x=1:10,y=1:10)
test<-function(z){
   mean.x<-mean(z$x)
   nm <-deparse(substitute(z))
   print(nm)
   return(mean.x)}
 
 test(a)
#[1] "a"   ... this is the side-effect of the print() call
#          ... you could have done something useful with that character value
#[1] 5.5   ... this is the result of the function call

Edit: Ran it with the new test-object

Note: this will not succeed inside a local function when a set of list items are passed from the first argument to lapply (and it also fails when an object is passed from a list given to a for-loop.) You would be able to extract the ".Names"-attribute and the order of processing from the structure result, if it were a named vector that were being processed.

> lapply( list(a=4,b=5), function(x) {nm <- deparse(substitute(x)); strsplit(nm, '\\[')} )
$a      # This "a" and the next one in the print output are put in after processing
$a[[1]]
[1] "X"    ""     "1L]]"  # Notice that there was no "a"


$b
$b[[1]]
[1] "X"    ""     "2L]]"

> lapply( c(a=4,b=5), function(x) {nm <- deparse(substitute(x)); strsplit(nm, '\\[')} )
$a
$a[[1]]   # but it's theoretically possible to extract when its an atomic vector
[1] "structure(c(4, 5), .Names = c(\"a\", \"b\"))" ""                                            
[3] "1L]]"                                        


$b
$b[[1]]
[1] "structure(c(4, 5), .Names = c(\"a\", \"b\"))" ""                                            
[3] "2L]]"  

deparse(quote(var))

My intuitive understanding In which the quote freeze the var or expression from evaluation and the deparse function which is the inverse of parse function makes that freezed symbol back to String

Note that for print methods the behavior can be different.

print.foo=function(x){ print(deparse(substitute(x))) }
test = list(a=1, b=2)
class(test)="foo"
#this shows "test" as expected
print(test)

#this (just typing 'test' on the R command line)
test
#shows
#"structure(list(a = 1, b = 2), .Names = c(\"a\", \"b\"), class = \"foo\")"

Other comments I've seen on forums suggests that the last behavior is unavoidable. This is unfortunate if you are writing print methods for packages.