How to test if list element exists?

Question

Problem I would like to test if an element of a list exists, here is an example ``` foo <- list(a=1) exists('foo') TRUE #foo does exist exists('foo$a') FALSE #suggests that foo$a does not exist foo$a [1] 1 #but it does exist ``` In this example, I know that `foo$a` exists, but the test returns `FALSE`. I looked in `?exists` and have found that `with(foo, exists('a')` returns `TRUE`, but do not understand why `exists('foo$a')` returns `FALSE`. Questions - Why does `exists('foo$a')` return `FALSE`? - Is use of `with(...)` the preferred approach?

Accepted Answer

This is actually a bit trickier than you'd think. Since a list can actually (with some effort) contain NULL elements, it might not be enough to check `is.null(foo$a)`. A more stringent test might be to check that the name is actually defined in the list:

```
foo <- list(a=42, b=NULL)
foo

is.null(foo[["a"]]) # FALSE
is.null(foo[["b"]]) # TRUE, but the element "exists"...
is.null(foo[["c"]]) # TRUE

"a" %in% names(foo) # TRUE
"b" %in% names(foo) # TRUE
"c" %in% names(foo) # FALSE
```

...and `foo[["a"]]` is safer than `foo$a`, since the latter uses partial matching and thus might also match a longer name:

```
x <- list(abc=4)
x$a  # 4, since it partially matches abc
x[["a"]] # NULL, no match
```

[UPDATE] So, back to the question why `exists('foo$a')` doesn't work. The `exists` function only checks if a variable exists in an environment, not if parts of a object exist. The string `"foo$a"` is interpreted literary: Is there a variable called "foo$a"? ...and the answer is `FALSE`...

```
foo <- list(a=42, b=NULL) # variable "foo" with element "a"
"bar$a" <- 42   # A variable actually called "bar$a"...
ls() # will include "foo" and "bar$a" 
exists("foo$a") # FALSE 
exists("bar$a") # TRUE
```

How to test if list element exists?

Solution

Problem