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

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

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

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

(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

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