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  lt - 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

User · Answer

Here is a performance comparison of the proposed methods in other answers.

> foo <- sapply(letters, function(x){runif(5)}, simplify = FALSE)
> microbenchmark::microbenchmark('k' %in% names(foo), 
                                 is.null(foo[['k']]), 
                                 exists('k', where = foo))
Unit: nanoseconds
                     expr  min   lq    mean median   uq   max neval cld
      "k" %in% names(foo)  467  933 1064.31    934  934 10730   100  a 
      is.null(foo[["k"]])    0    0  168.50      1  467  3266   100  a 
 exists("k", where = foo) 6532 6998 7940.78   7232 7465 56917   100   b

If you are planing to use the list as a fast dictionary accessed many times, then the is.null approach might be the only viable option. I assume it is O(1), while the %in% approach is O(n)?

User · 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  lt - 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  lt - 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  lt - list a 42  b NULL    variable  foo  with element  a   bar a   lt - 42     A variable actually called  bar a     ls     will include  foo  and  bar a   exists  foo a     FALSE  exists  bar a     TRUE

User · Answer

One solution that hasn t come up yet is using length  which successfully handles NULL  As far as I can tell  all values except NULL have a length greater than 0   x  lt - list 4  -1  NULL  NA  Inf  -Inf  NaN  T  x   0  y       z   c 1 2 3   lapply x  function el  print length el     1  1  1  1  1  0  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  3   Thus we could make a simple function that works with both named and numbered indices   element exists  lt - function var  element      tryCatch       if length var  element     gt  -1        return T       error   function e        return F           If the element doesn t exist  it causes an out-of-bounds condition caught by the tryCatch block

User · Answer

A slight modified version of  salient salamander   if one wants to check on full path  this can be used   Element Exists Check   function  full index path      tryCatch       len element   length full index path      exists indicator   ifelse len element  gt  0  T  F        return exists indicator       error   function e        return F

User · Answer

The best way to check for named elements is to use exist    however the above answers are not using the function properly   You need to use the where argument to check for the variable within the list   foo  lt - list a 42  b NULL   exists  a   where foo   TRUE exists  b   where foo   TRUE exists  c   where foo   FALSE

User · Answer

Use purrr  has element to check against the value of a list element    gt  x  lt - list c 1  2   c 3  4    gt  purrr  has element x  c 3  4    1  TRUE  gt  purrr  has element x  c 3  5    1  FALSE

User · Answer

rlang  has name   can do this too   foo   list a   1  bb   NULL  rlang  has name foo   a      TRUE rlang  has name foo   b      FALSE  No partial matching rlang  has name foo   bb      TRUE  Handles NULL correctly rlang  has name foo   c      FALSE   As you can see  it inherently handles all the cases that  Tommy showed how to handle using base R and works for lists with unnamed items  I would still recommend exists  bb   where   foo  as proposed in another answer for readability  but has name is an alternative if you have unnamed items

[r] How to test if list element exists?

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