Python idiom to return first item or None

Question

I'm sure there's a simpler way of doing this that's just not occurring to me.

I'm calling a bunch of methods that return a list. The list may be empty. If the list is non-empty, I want to return the first item; otherwise, I want to return None. This code works:

my_list = get_list()
if len(my_list) > 0: return my_list[0]
return None

It seems to me that there should be a simple one-line idiom for doing this, but for the life of me I can't think of it. Is there?

Edit:

The reason that I'm looking for a one-line expression here is not that I like incredibly terse code, but because I'm having to write a lot of code like this:

x = get_first_list()
if x:
    # do something with x[0]
    # inevitably forget the [0] part, and have a bug to fix
y = get_second_list()
if y:
    # do something with y[0]
    # inevitably forget the [0] part AGAIN, and have another bug to fix

What I'd like to be doing can certainly be accomplished with a function (and probably will be):

def first_item(list_or_none):
    if list_or_none: return list_or_none[0]

x = first_item(get_first_list())
if x:
    # do something with x
y = first_item(get_second_list())
if y:
    # do something with y

I posted the question because I'm frequently surprised by what simple expressions in Python can do, and I thought that writing a function was a silly thing to do if there was a simple expression could do the trick. But seeing these answers, it seems like a function is the simple solution.

Answer 1

The most python idiomatic way is to use the next() on a iterator since list is iterable. just like what @J.F.Sebastian put in the comment on Dec 13, 2011.

next(iter(the_list), None) This returns None if the_list is empty. see next() Python 2.6+

or if you know for sure the_list is not empty:

iter(the_list).next() see iterator.next() Python 2.2+

Answer 2

Using the and-or trick:

a = get_list()
return a and a[0] or None

Answer 3

Probably not the fastest solution, but nobody mentioned this option:

dict(enumerate(get_list())).get(0)

if get_list() can return None you can use:

dict(enumerate(get_list() or [])).get(0)

Advantages:

-one line

-you just call get_list() once

-easy to understand

Answer 4

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]

Answer 5

def head(iterable):
    try:
        return iter(iterable).next()
    except StopIteration:
        return None

print head(xrange(42, 1000)  # 42
print head([])               # None

BTW: I'd rework your general program flow into something like this:

lists = [
    ["first", "list"],
    ["second", "list"],
    ["third", "list"]
]

def do_something(element):
    if not element:
        return
    else:
        # do something
        pass

for li in lists:
    do_something(head(li))

(Avoiding repetition whenever possible)

Answer 6

Several people have suggested doing something like this:

list = get_list()
return list and list[0] or None

That works in many cases, but it will only work if list[0] is not equal to 0, False, or an empty string. If list[0] is 0, False, or an empty string, the method will incorrectly return None.

I've created this bug in my own code one too many times !