How to check if object (variable) is defined in R?
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!
r
asked Feb 20 '12 at 21:46
TMSTMS
33.5k37170303
add a comment |
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!
r
asked Feb 20 '12 at 21:46
TMSTMS
33.5k37170303
add a comment |
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!
r
asked Feb 20 '12 at 21:46
TMSTMS
33.5k37170303
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!
r
r
asked Feb 20 '12 at 21:46
TMSTMS
33.5k37170303
asked Feb 20 '12 at 21:46
TMSTMS
33.5k37170303
edited Jun 1 '16 at 14:16
TMS
asked Feb 20 '12 at 21:46
TMSTMS
33.5k37170303
asked Feb 20 '12 at 21:46
TMSTMS
33.5k37170303
asked Feb 20 '12 at 21:46
TMSTMS
33.5k37170303
33.5k37170303
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
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 EddelbuettelDirk Eddelbuettel
277k37513602
1
@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! :-)
– TMS
Feb 20 '12 at 22:00
24
@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: stackoverflow.com/q/7719741
– TMS
Sep 20 '14 at 11:58
2
what about for what the op wanted - using the variable name, not in quotes?
– tim
Jun 13 '15 at 13:46
add a comment |
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 SimpsonGavin Simpson
135k19306383
7
You win by 52 seconds :)
– Dirk Eddelbuettel
Feb 20 '12 at 21:51
6
@DirkEddelbuettel Well, if you will use ridiculously long object names ;-)
– Gavin Simpson
Feb 20 '12 at 21:54
2
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
add a comment |
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 9 '13 at 21:56
timtim
1,8651733
add a comment |
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")
edited Feb 3 '16 at 3:37
MichaelChirico
20k861111
answered Sep 24 '15 at 16:19
Santiago BaldrichSantiago Baldrich
538510
1
Playing around, using argumentinherits = FALSEseems to isolate things in the global environment. Does that sound right?
– CJB
Jan 7 '16 at 12:49
@Bazz you're correct; I've edited this into the answer.
– MichaelChirico
Feb 3 '16 at 3:21
1
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
add a comment |
If you don't want to use quotes, you can use deparse(substitute()) trick which I found in 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
answered Apr 17 '17 at 6:32
NirmalNirmal
19115
I wish people knew how sick this was. Good work.
– Carl Boneri
Jun 27 '17 at 18:36
you can alsoforceor evaluate it in the function like this:is.defined <- function(sym) class(try(sym, TRUE))!='try-error'
– chinsoon12
Oct 4 '17 at 0:49
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 EddelbuettelDirk Eddelbuettel
277k37513602
1
@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! :-)
– TMS
Feb 20 '12 at 22:00
24
@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: stackoverflow.com/q/7719741
– TMS
Sep 20 '14 at 11:58
2
what about for what the op wanted - using the variable name, not in quotes?
– tim
Jun 13 '15 at 13:46
add a comment |
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 EddelbuettelDirk Eddelbuettel
277k37513602
1
@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! :-)
– TMS
Feb 20 '12 at 22:00
24
@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: stackoverflow.com/q/7719741
– TMS
Sep 20 '14 at 11:58
2
what about for what the op wanted - using the variable name, not in quotes?
– tim
Jun 13 '15 at 13:46
add a comment |
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 EddelbuettelDirk Eddelbuettel
277k37513602
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 EddelbuettelDirk Eddelbuettel
277k37513602
answered Feb 20 '12 at 21:51
Dirk EddelbuettelDirk Eddelbuettel
277k37513602
answered Feb 20 '12 at 21:51
Dirk EddelbuettelDirk Eddelbuettel
277k37513602
answered Feb 20 '12 at 21:51
Dirk EddelbuettelDirk Eddelbuettel
277k37513602
277k37513602
1
@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! :-)
– TMS
Feb 20 '12 at 22:00
24
@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: stackoverflow.com/q/7719741
– TMS
Sep 20 '14 at 11:58
2
what about for what the op wanted - using the variable name, not in quotes?
– tim
Jun 13 '15 at 13:46
add a comment |
1
@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! :-)
– TMS
Feb 20 '12 at 22:00
24
@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: stackoverflow.com/q/7719741
– TMS
Sep 20 '14 at 11:58
2
what about for what the op wanted - using the variable name, not in quotes?
– tim
Jun 13 '15 at 13:46
1
1
@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! :-)
– TMS
Feb 20 '12 at 22:00
@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! :-)
– TMS
Feb 20 '12 at 22:00
24
24
@tim if you are inside a function, missing() is what you want.
– CousinCocaine
Jan 27 '14 at 14:31
@tim if you are inside a function, missing() is what you want.
– CousinCocaine
Jan 27 '14 at 14:31
2
2
Might get a bit trickier if checking for list elements: stackoverflow.com/q/7719741
– TMS
Sep 20 '14 at 11:58
Might get a bit trickier if checking for list elements: stackoverflow.com/q/7719741
– TMS
Sep 20 '14 at 11:58
2
2
what about for what the op wanted - using the variable name, not in quotes?
– tim
Jun 13 '15 at 13:46
what about for what the op wanted - using the variable name, not in quotes?
– tim
Jun 13 '15 at 13:46
add a comment |
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 SimpsonGavin Simpson
135k19306383
7
You win by 52 seconds :)
– Dirk Eddelbuettel
Feb 20 '12 at 21:51
6
@DirkEddelbuettel Well, if you will use ridiculously long object names ;-)
– Gavin Simpson
Feb 20 '12 at 21:54
2
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
add a comment |
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 SimpsonGavin Simpson
135k19306383
7
You win by 52 seconds :)
– Dirk Eddelbuettel
Feb 20 '12 at 21:51
6
@DirkEddelbuettel Well, if you will use ridiculously long object names ;-)
– Gavin Simpson
Feb 20 '12 at 21:54
2
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
add a comment |
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 SimpsonGavin Simpson
135k19306383
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 SimpsonGavin Simpson
135k19306383
answered Feb 20 '12 at 21:50
Gavin SimpsonGavin Simpson
135k19306383
answered Feb 20 '12 at 21:50
Gavin SimpsonGavin Simpson
135k19306383
answered Feb 20 '12 at 21:50
Gavin SimpsonGavin Simpson
135k19306383
135k19306383
7
You win by 52 seconds :)
– Dirk Eddelbuettel
Feb 20 '12 at 21:51
6
@DirkEddelbuettel Well, if you will use ridiculously long object names ;-)
– Gavin Simpson
Feb 20 '12 at 21:54
2
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
add a comment |
7
You win by 52 seconds :)
– Dirk Eddelbuettel
Feb 20 '12 at 21:51
6
@DirkEddelbuettel Well, if you will use ridiculously long object names ;-)
– Gavin Simpson
Feb 20 '12 at 21:54
2
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
7
7
You win by 52 seconds :)
– Dirk Eddelbuettel
Feb 20 '12 at 21:51
You win by 52 seconds :)
– Dirk Eddelbuettel
Feb 20 '12 at 21:51
6
6
@DirkEddelbuettel Well, if you will use ridiculously long object names ;-)
– Gavin Simpson
Feb 20 '12 at 21:54
@DirkEddelbuettel Well, if you will use ridiculously long object names ;-)
– Gavin Simpson
Feb 20 '12 at 21:54
2
2
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
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
add a comment |
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 9 '13 at 21:56
timtim
1,8651733
add a comment |
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 9 '13 at 21:56
timtim
1,8651733
add a comment |
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 9 '13 at 21:56
timtim
1,8651733
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 9 '13 at 21:56
timtim
1,8651733
answered Mar 9 '13 at 21:56
timtim
1,8651733
answered Mar 9 '13 at 21:56
timtim
1,8651733
answered Mar 9 '13 at 21:56
timtim
1,8651733
1,8651733
add a comment |
add a comment |
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")
edited Feb 3 '16 at 3:37
MichaelChirico
20k861111
answered Sep 24 '15 at 16:19
Santiago BaldrichSantiago Baldrich
538510
1
Playing around, using argumentinherits = FALSEseems to isolate things in the global environment. Does that sound right?
– CJB
Jan 7 '16 at 12:49
@Bazz you're correct; I've edited this into the answer.
– MichaelChirico
Feb 3 '16 at 3:21
1
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
add a comment |
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")
edited Feb 3 '16 at 3:37
MichaelChirico
20k861111
answered Sep 24 '15 at 16:19
Santiago BaldrichSantiago Baldrich
538510
1
Playing around, using argumentinherits = FALSEseems to isolate things in the global environment. Does that sound right?
– CJB
Jan 7 '16 at 12:49
@Bazz you're correct; I've edited this into the answer.
– MichaelChirico
Feb 3 '16 at 3:21
1
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
add a comment |
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")
edited Feb 3 '16 at 3:37
MichaelChirico
20k861111
answered Sep 24 '15 at 16:19
Santiago BaldrichSantiago Baldrich
538510
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")
edited Feb 3 '16 at 3:37
MichaelChirico
20k861111
answered Sep 24 '15 at 16:19
Santiago BaldrichSantiago Baldrich
538510
edited Feb 3 '16 at 3:37
MichaelChirico
20k861111
edited Feb 3 '16 at 3:37
MichaelChirico
20k861111
edited Feb 3 '16 at 3:37
MichaelChirico
20k861111
20k861111
answered Sep 24 '15 at 16:19
Santiago BaldrichSantiago Baldrich
538510
answered Sep 24 '15 at 16:19
Santiago BaldrichSantiago Baldrich
538510
answered Sep 24 '15 at 16:19
Santiago BaldrichSantiago Baldrich
538510
538510
1
Playing around, using argumentinherits = FALSEseems to isolate things in the global environment. Does that sound right?
– CJB
Jan 7 '16 at 12:49
@Bazz you're correct; I've edited this into the answer.
– MichaelChirico
Feb 3 '16 at 3:21
1
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
add a comment |
1
Playing around, using argumentinherits = FALSEseems to isolate things in the global environment. Does that sound right?
– CJB
Jan 7 '16 at 12:49
@Bazz you're correct; I've edited this into the answer.
– MichaelChirico
Feb 3 '16 at 3:21
1
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
1
1
Playing around, using argument
inherits = FALSE seems to isolate things in the global environment. Does that sound right?– CJB
Jan 7 '16 at 12:49
Playing around, using argument
inherits = FALSE seems to isolate things in the global environment. Does that sound right?– CJB
Jan 7 '16 at 12:49
@Bazz you're correct; I've edited this into the answer.
– MichaelChirico
Feb 3 '16 at 3:21
@Bazz you're correct; I've edited this into the answer.
– MichaelChirico
Feb 3 '16 at 3:21
1
1
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
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
add a comment |
If you don't want to use quotes, you can use deparse(substitute()) trick which I found in 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
answered Apr 17 '17 at 6:32
NirmalNirmal
19115
I wish people knew how sick this was. Good work.
– Carl Boneri
Jun 27 '17 at 18:36
you can alsoforceor evaluate it in the function like this:is.defined <- function(sym) class(try(sym, TRUE))!='try-error'
– chinsoon12
Oct 4 '17 at 0:49
add a comment |
If you don't want to use quotes, you can use deparse(substitute()) trick which I found in 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
answered Apr 17 '17 at 6:32
NirmalNirmal
19115
I wish people knew how sick this was. Good work.
– Carl Boneri
Jun 27 '17 at 18:36
you can alsoforceor evaluate it in the function like this:is.defined <- function(sym) class(try(sym, TRUE))!='try-error'
– chinsoon12
Oct 4 '17 at 0:49
add a comment |
If you don't want to use quotes, you can use deparse(substitute()) trick which I found in 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
answered Apr 17 '17 at 6:32
NirmalNirmal
19115
If you don't want to use quotes, you can use deparse(substitute()) trick which I found in 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
answered Apr 17 '17 at 6:32
NirmalNirmal
19115
edited Apr 17 '17 at 8:40
answered Apr 17 '17 at 6:32
NirmalNirmal
19115
answered Apr 17 '17 at 6:32
NirmalNirmal
19115
answered Apr 17 '17 at 6:32
NirmalNirmal
19115
19115
I wish people knew how sick this was. Good work.
– Carl Boneri
Jun 27 '17 at 18:36
you can alsoforceor evaluate it in the function like this:is.defined <- function(sym) class(try(sym, TRUE))!='try-error'
– chinsoon12
Oct 4 '17 at 0:49
add a comment |
I wish people knew how sick this was. Good work.
– Carl Boneri
Jun 27 '17 at 18:36
you can alsoforceor evaluate it in the function like this:is.defined <- function(sym) class(try(sym, TRUE))!='try-error'
– chinsoon12
Oct 4 '17 at 0:49
I wish people knew how sick this was. Good work.
– Carl Boneri
Jun 27 '17 at 18:36
I wish people knew how sick this was. Good work.
– Carl Boneri
Jun 27 '17 at 18:36
you can also
force or evaluate it in the function like this: is.defined <- function(sym) class(try(sym, TRUE))!='try-error'– chinsoon12
Oct 4 '17 at 0:49
you can also
force or evaluate it in the function like this: is.defined <- function(sym) class(try(sym, TRUE))!='try-error'– chinsoon12
Oct 4 '17 at 0:49
add a comment |
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