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

Python 2.6+

next(iter(your_list), None) 

If your_list can be None:

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

Python 2.4

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 

Answer 2

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]