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

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. — EliasK93
– EliasK93, Commented Jul 30, 2022 at 14:02
not know why question with 1 answer closed, any way what i looking for to not repeat the same code generated list and get value or no value if empty list *[1] *[] — Mahmoud Magdy
– Mahmoud Magdy, Commented Apr 28, 2024 at 14:27

Mouse · Accepted Answer · 2022-07-30 14:06:02Z

12

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

answered Jul 30, 2022 at 14:03

Mouse

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2 Comments

You could shorten len(arr) > 0 to just arr because an empty array is False and non-empty True.

Ah even better - and something I didn't know - thanks :-)