是否有更多pythonic方法在函数的参数上爆炸列表？

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])

有没有办法在没有显式索引的情况下调用foo(i)?试图避免foo(i [0],i [1],i [2])

3 个解决方案

#1

Yes, use foo(*i):

是的,使用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:

您还可以在函数定义中使用*:def foo(* vargs)将所有非关键字参数放入名为vargs的元组中。并且使用**,例如,def foo(** kargs),将所有关键字参数放入名为kargs的字典中:

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

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

#2

It's called argument unpacking and there's an operator for it.

它被称为参数解包,并且有一个运算符。

foo(*i)

By the way, it works with any iterable, not just with lists.

顺便说一句,它适用于任何可迭代的,而不仅仅是列表。

#3

Yes, Python supports that:

是的,Python支持:

foo(*i)

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

请参阅有关解压缩参数列表的文档。适用于任何可迭代的东西。有两颗星**它适用于dicts和命名参数。

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

bar(**j)

#1