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

Is there clean way to get the value at a list index or None if the index is out or range in Python?

The obvious way to do it would be this:

if len(the_list) > i:
    return the_list[i]
else:
    return None

However, the verbosity reduces code readability. Is there a clean, simple, one-liner that can be used instead?

Solution 1:

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

Solution 2:

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.

Solution 3:

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.

Solution 4:

For your purposes you can exclude the else part as None is return by default if a given condition is not met.

def return_ele(x, i):
    if len(x) > i: return x[i]

Result

>>> x = [2,3,4]
>>> b = return_ele(x, 2)
>>> b
4
>>> b = return_ele(x, 5)
>>> b
>>> type(b)
<type 'NoneType'>