Pass a list to a function to act as multiple arguments [duplicate]

Question

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? I think this is called 'unpacking'.

wjandrea · Accepted Answer · 2022-11-26 19:13:41Z

372

function_that_needs_strings(*my_list) # works!

wjandrea

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answered Aug 13, 2010 at 19:40

Jochen Ritzel

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Comments

Community · Accepted Answer · 2017-05-23 12:26:32Z

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.

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

vishes_shell · Accepted Answer · 2016-09-04 19:11:35Z

22

Since Python 3.5 you can unpack unlimited amount of lists.

So this will work:

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

answered Sep 4, 2016 at 19:11

vishes_shell

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how can I do the same with python 2.7 or 3.4?

@answerSeeker not efficient, but function_that_needs_strings(*(a+b))

wow, can't believe You couldn't before that.

@answerSeeker more efficient: function_that_needs_strings(*itertools.chain(a, b))