How to check if object (variable) is defined in R?

Question

I'd like to check if some variable is defined in R - without getting an error. How can I do this?

My attempts (not successful):

> is.na(ooxx)
Error: object 'ooxx' not found
> is.finite(ooxx)
Error: object 'ooxx' not found

Thanks!

score 545 · Accepted Answer · answered Feb 20 '12 at 21:51

545

You want exists():

R> exists("somethingUnknown")
[1] FALSE
R> somethingUnknown <- 42
R> exists("somethingUnknown")
[1] TRUE
R>

answered Feb 20 '12 at 21:51

Dirk Eddelbuettel

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@Gavin & Dirk, you're so nice to each other :) Only solution is that you toss the coin (Bernoulli with p=0.5 :-)) who will get the accept! :-) – Tomas Feb 20 '12 at 22:00
36

@tim if you are inside a function, missing() is what you want. – CousinCocaine Jan 27 '14 at 14:31
2

Might get a bit trickier if checking for list elements: http://stackoverflow.com/q/7719741 – Tomas Sep 20 '14 at 11:58
6

what about for what the op wanted - using the variable name, not in quotes? – tim Jun 13 '15 at 13:46
@tim `exists(quote(ooxx))` – Julien Dec 13 '22 at 09:26

score 128 · Answer 2 · answered Feb 20 '12 at 21:50

128

See ?exists, for some definition of "...is defined". E.g.

> exists("foo")
[1] FALSE
> foo <- 1:10
> exists("foo")
[1] TRUE

answered Feb 20 '12 at 21:50

Gavin Simpson

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@DirkEddelbuettel Well, if you will use ridiculously long object names ;-) – Gavin Simpson Feb 20 '12 at 21:54
3

heh. Happens to me all the time when I am testing out examples before posting, Gavin or Josh have already answered it. – Maiasaura Feb 20 '12 at 22:17

score 69 · Answer 3 · answered Mar 09 '13 at 21:56

69

if you are inside a function, missing() is what you want.

exchequer = function(x) {
    if(missing(x)){
        message("x is missing… :-(")
    }
}

exchequer()
x is missing… :-(

answered Mar 09 '13 at 21:56

tim

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`missing` only works for function arguments, however. You can't do `foo <- function(x) {missing(x); missing(y)}` or you will get `foo(1)` `> Error in missing(y) : 'missing' can only be used for arguments`. – Dannid Feb 11 '19 at 17:48

score 61 · Answer 4 · edited Feb 03 '16 at 03:37

As others have pointed out, you're looking for exists. Keep in mind that using exists with names used by R's base packages would return true regardless of whether you defined the variable:

> exists("data")
[1] TRUE

To get around this (as pointed out by Bazz; see ?exists), use the inherits argument:

> exists("data", inherits = FALSE)
[1] FALSE

foo <- TRUE
> exists("foo", inherits = FALSE)
[1] TRUE

Of course, if you wanted to search the name spaces of attached packages, this would also fall short:

> exists("data.table")
[1] FALSE
require(data.table)
> exists("data.table", inherits = FALSE)
[1] FALSE
> exists("data.table")
[1] TRUE

The only thing I can think of to get around this -- to search in attached packages but not in base packages -- is the following:

any(sapply(1:(which(search() == "tools:rstudio") - 1L),
           function(pp) exists(_object_name_, where = pp, inherits = FALSE)))

Compare replacing _object_name_ with "data.table" (TRUE) vs. "var" (FALSE)

(of course, if you're not on RStudio, I think the first automatically attached environment is "package:stats")

Playing around, using argument `inherits = FALSE` seems to isolate things in the global environment. Does that sound right? — CJB, Jan 07 '16 at 12:49
This comment should be higher up, since I use variable name "data", just using exist gave me some trouble initially. — mzm, May 11 '16 at 14:44

score 38 · Answer 5 · edited Dec 16 '19 at 12:09

38

If you don't want to use quotes, you can use deparse(substitute()) trick which I found in the example section of ?substitute:

is.defined <- function(sym) {
  sym <- deparse(substitute(sym))
  env <- parent.frame()
  exists(sym, env)
}

is.defined(a)
# FALSE
a <- 10
is.defined(a)
# TRUE

edited Dec 16 '19 at 12:09

dpel

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answered Apr 17 '17 at 06:32

Nirmal

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you can also `force` or evaluate it in the function like this: `is.defined <- function(sym) class(try(sym, TRUE))!='try-error'` – chinsoon12 Oct 04 '17 at 00:49
2

this answer may be a tad more complicated...but this is the ideal answer if you don't want to contend with characters vs var-names – Gregg H Jul 12 '20 at 16:37
1

I have bumped into an issue that needs exactly your solution. When you wanted to use something that's not `NULL` for quosure (`enquo(x)` if x is not null), you need a good way to check if the expression passed in is not null. Your solution does exactly that. thanks – stucash Nov 17 '21 at 16:49

score 4 · Answer 6 · answered May 18 '21 at 15:24

4

If you don't mind using quotes, you can use:

exists("x")

If you don't want to use quotes you can use:

exists(deparse(substitute(x)))

answered May 18 '21 at 15:24

Sebastian Tacora

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This answer combines previous answers. Not sure if this is adding anything to the conversation so far. – Nirmal Nov 15 '21 at 13:29

score 1 · Answer 7 · answered Aug 19 '19 at 10:10

There may be situations in which you do not exactly know the name of the variable you are looking for, like when an array of results have been created by a queuing system. These can possibly be addressed with "ls" and its argument "pattern" that expects a regular expression.

The "exists" function could be reimplemented that way as

exists <-function(variablename) {
   #print(ls(env=globalenv()))
   return(1==length(ls(pattern=paste("^",variablename,"$",sep=""),env=globalenv())))
}

While preparing this answer, I was a bit surprised about the need for the need of the specification of the environment when invoking ls() from within a function. So, thank you for that, stackoverflow! There is also an "all.names" attribute that I should have set to true but have omitted.

How to check if object (variable) is defined in R?

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