How to get first element of list or `None` when list is empty?

Question

I can do it like this but there has to be better way:

arr = []
if len(arr) > 0:
    first_or_None = arr[0]
else:
    first_or_None = None

If I just do arr[0] I get IndexError. Is there something where I can give default argument?

You can use: `first_or_none = arr[0] if arr else None` (`if arr` is just a shortened way to write `if len(arr) > 0`). Indexing the first item needs index `0`. — ewz93, Jul 30 '22 at 14:02

Mouse · Accepted Answer · 2022-07-30T14:06:02.973

4

I think the example you give is absolutely fine - it is very readable and would not suffer performance issues.

You could use the ternary operator python equivalent, if you really want it to be shorter code:

last_or_none = arr[0] if len(arr) > 0 else None

edited Jul 30 '22 at 14:06

answered Jul 30 '22 at 14:03

Mouse

2

You could shorten `len(arr) > 0` to just `arr` because an empty array is False and non-empty True. – tdelaney Jul 30 '22 at 14:14
Ah even better - and something I didn't know - thanks :-) – Mouse Jul 30 '22 at 14:17

1 Answers1