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

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

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

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

(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

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)

… 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 > 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):
    "Returns the nth item or a default value"
    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  

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
>>> 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

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

A cheap solution is to really make a dict with enumerate and use .get() as usual, like

 dict(enumerate(l)).get(7, my_default)

After reading through the answers, I’m going to use:

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

Althought this is not a one-liner solution, you can define a function with a default value like so:

def get_val(myList, idx, default=5):
    try:
        return myList[idx]
    except IndexError:
        return default

For a small index, such as when parsing up to k arguments, I’d build a new list of length k, with added elements set to d, as follows:

def fill(l, k, d): 
    return l + (k - len(l)) * [d]

Typical usage:

N = 2
arg_one, arg_two = fill("first_word and the rest".split(maxsplit=N - 1), N, None)
# arg_one == "first_word"
# arg_two == "and the rest"

Same example, with a short list:

arg_one, arg_two = fill("one_word_only".split(maxsplit=N - 1), N, None)
# arg_one == "one_word_only"
# arg_two == None

With unpacking:

b, = a[n:n+1] or [default]