names() function in R Language is used to get or set the name of an Object. This function takes object i.e. vector, matrix or data frame as argument along with the value that is to be assigned as name to the object. The length of the value vector passed must be exactly equal to the length of the object to be named.
Definition: The names R function returns or sets the names of a data object.
Objects in R For example, objects are assigned a value using <-, consider the following piece of code where 'a1' and 'a2' are the variables where the numeric value is assigned to it. An object can be assigned a set of numbers where 'a3' is the object containing a vector of different numbers.
ls() function in R lists the objects in your environment as a vector. In this tutorial, we will learn the use of ls() functions with examples.
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.
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