我知道星号在Python中的函数定义中是什么意思。
我经常看到使用参数调用函数的星号:
def foo(*args, **kwargs):
first_func(args, kwargs)
second_func(*args, **kwargs)
第一个和第二个函数调用有什么区别?
答案 0 :(得分:17)
args = [1,2,3]
func(*args) == func(1,2,3)
变量从列表(或任何其他序列类型)中解压缩为参数
func(args) == func([1,2,3])
传递一个列表
kwargs = dict(a=1,b=2,c=3)
func(kwargs) == func({'a':1, 'b':2, 'c':3})
传递了一个词典
func(**kwargs) == func(a=1,b=2,c=3)
(key,value)从dict(或任何其他映射类型)解压缩为命名参数
答案 1 :(得分:9)
不同之处在于参数如何传递给被调用的函数。当您使用*
时,参数会被解压缩(如果它们是列表或元组) - 另外,它们只是按原样传入。
以下是差异的一个例子:
>>> def add(a, b):
... print a + b
...
>>> add(*[2,3])
5
>>> add([2,3])
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: add() takes exactly 2 arguments (1 given)
>>> add(4, 5)
9
当我用*
作为参数前缀时,它实际上将列表解压缩为两个单独的参数,这些参数作为add
和a
传递给b
。没有它,它只是作为单个参数传递给列表。
字典和**
的情况也是如此,除了它们作为命名参数而不是有序参数传入。
>>> def show_two_stars(first, second='second', third='third'):
... print "first: " + str(first)
... print "second: " + str(second)
... print "third: " + str(third)
>>> show_two_stars('a', 'b', 'c')
first: a
second: b
third: c
>>> show_two_stars(**{'second': 'hey', 'first': 'you'})
first: you
second: hey
third: third
>>> show_two_stars({'second': 'hey', 'first': 'you'})
first: {'second': 'hey', 'first': 'you'}
second: second
third: third
答案 2 :(得分:0)
def fun1(*args):
""" This function accepts a non keyworded variable length argument as a parameter.
"""
print args
print len(args)
>>> a = []
>>> fun1(a)
([],)
1
# This clearly shows that, the empty list itself is passed as a first argument. Since *args now contains one empty list as its first argument, so the length is 1
>>> fun1(*a)
()
0
# Here the empty list is unwrapped (elements are brought out as separate variable length arguments) and passed to the function. Since there is no element inside, the length of *args is 0
>>>