Get value at list/array index or "None" if out of range in Python

If you are dealing with small lists, you do not need to add an if statement or something of the sorts. An easy solution is to transform the list into a dict. Then you can use dict.get:

table = dict(enumerate(the_list))
return table.get(i)

You can even set another default value than None, using the second argument to dict.get. For example, use table.get(i, 'unknown') to return 'unknown' if the index is out of range.

Note that this method does not work with negative indices.

I find list slices good for this:

>>> x = [1, 2, 3]
>>> a = x [1:2]
>>> a
[2]
>>> b = x [4:5]
>>> b
[]

So, always access x[i:i+1], if you want x[i]. You'll get a list with the required element if it exists. Otherwise, you get an empty list.

Try:

try:
    return the_list[i]
except IndexError:
    return None

Or, one liner:

l[i] if i < len(l) else None

Example:

>>> l=list(range(5))
>>> i=6
>>> print(l[i] if i < len(l) else None)
None
>>> i=2
>>> print(l[i] if i < len(l) else None)
2