*args和**kwargs是什么意思?

def foo(x, y, *args):
def bar(x, y, **kwargs):

当前回答

这个示例将帮助您立即记住Python中的*args、**kwargs甚至super和继承。

class base(object):
    def __init__(self, base_param):
        self.base_param = base_param


class child1(base): # inherited from base class
    def __init__(self, child_param, *args) # *args for non-keyword args
        self.child_param = child_param
        super(child1, self).__init__(*args) # call __init__ of the base class and initialize it with a NON-KEYWORD arg

class child2(base):
    def __init__(self, child_param, **kwargs):
        self.child_param = child_param
        super(child2, self).__init__(**kwargs) # call __init__ of the base class and initialize it with a KEYWORD arg

c1 = child1(1,0)
c2 = child2(1,base_param=0)
print c1.base_param # 0
print c1.child_param # 1
print c2.base_param # 0
print c2.child_param # 1

其他回答

在函数中同时使用两者的一个好例子是:

>>> def foo(*arg,**kwargs):
...     print arg
...     print kwargs
>>>
>>> a = (1, 2, 3)
>>> b = {'aa': 11, 'bb': 22}
>>>
>>>
>>> foo(*a,**b)
(1, 2, 3)
{'aa': 11, 'bb': 22}
>>>
>>>
>>> foo(a,**b) 
((1, 2, 3),)
{'aa': 11, 'bb': 22}
>>>
>>>
>>> foo(a,b) 
((1, 2, 3), {'aa': 11, 'bb': 22})
{}
>>>
>>>
>>> foo(a,*b)
((1, 2, 3), 'aa', 'bb')
{}

带*Args和**kwargs的“无限”Args

*args和**kwargs只是向函数输入无限字符的一种方式,例如:


def print_all(*args, **kwargs):
    print(args) # print any number of arguments like: "print_all("foo", "bar")"
    print(kwargs.get("to_print")) # print the value of the keyworded argument "to_print"


# example:
print_all("Hello", "World", to_print="!")
# will print:
"""
('Hello', 'World')
!
"""

还值得注意的是,在调用函数时也可以使用*和**。这是一个快捷方式,允许您直接使用列表/元组或字典将多个参数传递给函数。例如,如果您具有以下功能:

def foo(x,y,z):
    print("x=" + str(x))
    print("y=" + str(y))
    print("z=" + str(z))

您可以执行以下操作:

>>> mylist = [1,2,3]
>>> foo(*mylist)
x=1
y=2
z=3

>>> mydict = {'x':1,'y':2,'z':3}
>>> foo(**mydict)
x=1
y=2
z=3

>>> mytuple = (1, 2, 3)
>>> foo(*mytuple)
x=1
y=2
z=3

注意:mydict中的键必须与函数foo的参数完全相同。否则将抛出TypeError:

>>> mydict = {'x':1,'y':2,'z':3,'badnews':9}
>>> foo(**mydict)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() got an unexpected keyword argument 'badnews'

此表便于在函数构造和函数调用中使用*和**:

            In function construction         In function call
=======================================================================
          |  def f(*args):                 |  def f(a, b):
*args     |      for arg in args:          |      return a + b
          |          print(arg)            |  args = (1, 2)
          |  f(1, 2)                       |  f(*args)
----------|--------------------------------|---------------------------
          |  def f(a, b):                  |  def f(a, b):
**kwargs  |      return a + b              |      return a + b
          |  def g(**kwargs):              |  kwargs = dict(a=1, b=2)
          |      return f(**kwargs)        |  f(**kwargs)
          |  g(a=1, b=2)                   |
-----------------------------------------------------------------------

这真的只是用来概括洛林·霍希斯坦的答案,但我觉得它很有用。

相关地:在Python 3中扩展了星形/splat运算符的用法

上下文

python 3.x使用打开包装**与字符串格式一起使用

与字符串格式一起使用

除了本主题中的答案之外,还有一个其他地方没有提到的细节。这扩展了Brad Solomon的答案

使用python str.format时,使用**解包也很有用。

这有点类似于使用python f-string f-string所做的操作,但增加了声明dict以保存变量的开销(f-string不需要dict)。

快速示例

  ## init vars
  ddvars = dict()
  ddcalc = dict()
  pass
  ddvars['fname']     = 'Huomer'
  ddvars['lname']     = 'Huimpson'
  ddvars['motto']     = 'I love donuts!'
  ddvars['age']       = 33
  pass
  ddcalc['ydiff']     = 5
  ddcalc['ycalc']     = ddvars['age'] + ddcalc['ydiff']
  pass
  vdemo = []

  ## ********************
  ## single unpack supported in py 2.7
  vdemo.append('''
  Hello {fname} {lname}!

  Today you are {age} years old!

  We love your motto "{motto}" and we agree with you!
  '''.format(**ddvars)) 
  pass

  ## ********************
  ## multiple unpack supported in py 3.x
  vdemo.append('''
  Hello {fname} {lname}!

  In {ydiff} years you will be {ycalc} years old!
  '''.format(**ddvars,**ddcalc)) 
  pass

  ## ********************
  print(vdemo[-1])