Python idiom to return first item or None

Python 2.6+

next(iter(your_list), None)

If your_list can be None:

next(iter(your_list or []), None)

def get_first(iterable, default=None):
    if iterable:
        for item in iterable:
            return item
    return default

Example:

x = get_first(get_first_list())
if x:
    ...
y = get_first(get_second_list())
if y:
    ...

Another option is to inline the above function:

for x in get_first_list() or []:
    # process x
    break # process at most one item
for y in get_second_list() or []:
    # process y
    break

To avoid break you could write:

for x in yield_first(get_first_list()):
    x # process x
for y in yield_first(get_second_list()):
    y # process y

Where:

def yield_first(iterable):
    for item in iterable or []:
        yield item
        return

The best way is this:

a = get_list()
return a[0] if a else None

You could also do it in one line, but it's much harder for the programmer to read:

return (get_list()[:1] or [None])[0]

(get_list() or [None])[0]

That should work.

BTW I didn't use the variable list, because that overwrites the builtin list() function.

Edit: I had a slightly simpler, but wrong version here earlier.