Is there a more pythonic way of exploding a list over a function's arguments?

Question

def foo(a, b, c):
 print a+b+c

i = [1,2,3]

Is there a way to call foo(i) without explicit indexing on i? Trying to avoid foo(i[0], i[1], i[2])

Solution 1

Yes, use foo(*i):

>>> foo(*i)
6

You can also use * in function definition: def foo(*vargs) puts all non-keyword arguments into a tuple called vargs. and the use of **, for eg., def foo(**kargs), will put all keyword arguments into a dictionary called kargs:

>>> def foo(*vargs, **kargs):
        print vargs
        print kargs

>>> foo(1, 2, 3, a="A", b="B")
(1, 2, 3)
{'a': 'A', 'b': 'B'}

Solution 2

Yes, Python supports that:

foo(*i)

See the documentation on Unpacking Argument Lists. Works with anything iterable. With two stars ** it works for dicts and named arguments.

def bar(a, b, c): 
    return a * b * c
j = {'a': 5, 'b': 3, 'c': 2}

bar(**j)