Unpack a list in Python?

问题

I think 'unpack' might be the wrong vocabulary here - apologies because I'm sure this is a duplicate question.

My question is pretty simple: in a function that expects a list of items, how can I pass a Python list item without getting an error?

my_list = ['red', 'blue', 'orange']
function_that_needs_strings('red', 'blue', 'orange') # works!
function_that_needs_strings(my_list) # breaks!

Surely there must be a way to expand the list, and pass the function 'red','blue','orange' on the hoof?

回答1:

function_that_needs_strings(*my_list) # works!

You can read all about it here.

回答2:

Yes, you can use the *args (splat) syntax:

function_that_needs_strings(*my_list)

where my_list can be any iterable; Python will loop over the given object and use each element as a separate argument to the function.

See the call expression documentation.

There is a keyword-parameter equivalent as well, using two stars:

kwargs = {'foo': 'bar', 'spam': 'ham'}
f(**kwargs)

and there is equivalent syntax for specifying catch-all arguments in a function signature:

def func(*args, **kw):
    # args now holds positional arguments, kw keyword arguments

回答3:

Since Python 3.5 you can unpack unlimited amount of lists.

PEP 448 - Additional Unpacking Generalizations

So this will work:

a = ['1', '2', '3', '4']
b = ['5', '6']
function_that_needs_strings(*a, *b)

来源：https://stackoverflow.com/questions/3480184/unpack-a-list-in-python