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

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Chapters
00:00 How To Get The Nth Element Of A Python List Or A Default If Not Available
00:28 Accepted Answer Score 174
00:41 Answer 2 Score 75
00:53 Answer 3 Score 33
01:00 Answer 4 Score 53
01:22 Thank you

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Full question
https://stackoverflow.com/questions/2492...

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Tags
#python #list

#avk47

ACCEPTED ANSWER

Score 174

l[index] if index < len(l) else default

To support negative indices we can use:

l[index] if -len(l) <= index < len(l) else default

ANSWER 2

Score 75

try:
   a = b[n]
except IndexError:
   a = default

Edit: I removed the check for TypeError - probably better to let the caller handle this.

ANSWER 3

Score 53

(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 => len(a)

Here is an example of how this works with range(5)

>>> range(5)[3:4]
[3]
>>> range(5)[4:5]
[4]
>>> range(5)[5:6]
[]
>>> range(5)[6:7]
[]

And the full expression

>>> (range(5)[3:4]+[999])[0]
3
>>> (range(5)[4:5]+[999])[0]
4
>>> (range(5)[5:6]+[999])[0]
999
>>> (range(5)[6:7]+[999])[0]
999

ANSWER 4

Score 33

(L[n:n+1] or [somedefault])[0]