How to get the nth element of a python list or a default if not available

Question

I m looking for an equivalent in python of dictionary get key  default  for lists  Is there any one liner idiom to get the nth element of a list or a default value if not available   For example  given a list myList I would like to get myList 0   or 5 ifmyList is an empty list   Thanks

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Just discovered that     next iter myList   5    iter l  returns an iterator on myList  next   consumes the first element of the iterator  and raises a StopIteration error except if called with a default value  which is the case here  the second argument  5  This only works when you want the 1st element  which is the case in your example  but not in the text of you question  so     Additionally  it does not need to create temporary lists in memory and it works for any kind of iterable  even if it does not have a name  see Xiong Chiamiov s comment on gruszczy s answer

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A cheap solution is to really make a dict with enumerate and use  get   as usual  like   dict enumerate l   get 7  my default

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After reading through the answers  I m going to use   L n   or  somedefault   0

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a n    default   0    This is probably better as a gets larger    a n n 1   default   0    This works because if a n   is an empty list if n   gt  len a   Here is an example of how this works with range 5    gt  gt  gt  range 5  3 4   3   gt  gt  gt  range 5  4 5   4   gt  gt  gt  range 5  5 6      gt  gt  gt  range 5  6 7       And the full expression   gt  gt  gt   range 5  3 4   999   0  3  gt  gt  gt   range 5  4 5   999   0  4  gt  gt  gt   range 5  5 6   999   0  999  gt  gt  gt   range 5  6 7   999   0  999

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looking for an equivalent in python of dict get key  default  for lists  There is an itertools recipes that does this for general iterables   For convenience  you can  gt  pip install more itertools and import this third-party library that implements such recipes for you  Code import more itertools as mit   mit nth  1  2  3   0    1      mit nth     0  5    5       Detail Here is the implementation of the nth recipe  def nth iterable  n  default None        quot Returns the nth item or a default value quot      return next itertools islice iterable  n  None   default   Like dict get    this tool returns a default for missing indices   It applies to  general iterables  mit nth  0  1  2   1                                         tuple   1  mit nth range 3   1                                          range generator  py3    1  mit nth iter  0  1  2    1                                   list iterator    1

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L n n 1  or  somedefault   0

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Combining  Joachim s with the above  you could use next iter my list index index 1    default   Examples  next iter range 10  8 9    11  8  gt  gt  gt  next iter range 10  12 13    11  11   Or  maybe more clear  but without the len my list index  if my list index index   1  else default

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try     a   b n  except IndexError     a   default   Edit  I removed the check for TypeError - probably better to let the caller handle this

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l index  if index  lt  len l  else default   To support negative indices we can use   l index  if -len l   lt   index  lt  len l  else default

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Using Python 3 4 s contextlib suppress exceptions  to build a getitem   method similar to getattr     import contextlib  def getitem iterable  index  default None          Return iterable index  or default if IndexError is raised         with contextlib suppress IndexError           return iterable index      return default

[python] How to get the nth element of a python list or a default if not available

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