*args和**kwargs是什么意思?
def foo(x, y, *args):
def bar(x, y, **kwargs):
*args和**kwargs是什么意思?
def foo(x, y, *args):
def bar(x, y, **kwargs):
当前回答
除了函数调用之外,*args和**kwargs在类层次结构中也很有用,并且还可以避免在Python中编写__init__方法。类似的用法可以在Django代码等框架中看到。
例如
def __init__(self, *args, **kwargs):
for attribute_name, value in zip(self._expected_attributes, args):
setattr(self, attribute_name, value)
if kwargs.has_key(attribute_name):
kwargs.pop(attribute_name)
for attribute_name in kwargs.viewkeys():
setattr(self, attribute_name, kwargs[attribute_name])
子类可以是
class RetailItem(Item):
_expected_attributes = Item._expected_attributes + ['name', 'price', 'category', 'country_of_origin']
class FoodItem(RetailItem):
_expected_attributes = RetailItem._expected_attributes + ['expiry_date']
然后将子类实例化为
food_item = FoodItem(name = 'Jam',
price = 12.0,
category = 'Foods',
country_of_origin = 'US',
expiry_date = datetime.datetime.now())
此外,具有仅对该子类实例有意义的新属性的子类可以调用基类__init__来卸载属性设置。这是通过*args和**kwargs完成的。kwargs主要用于使用命名参数使代码可读。例如
class ElectronicAccessories(RetailItem):
_expected_attributes = RetailItem._expected_attributes + ['specifications']
# Depend on args and kwargs to populate the data as needed.
def __init__(self, specifications = None, *args, **kwargs):
self.specifications = specifications # Rest of attributes will make sense to parent class.
super(ElectronicAccessories, self).__init__(*args, **kwargs)
其可以被初始化为
usb_key = ElectronicAccessories(name = 'Sandisk',
price = '$6.00',
category = 'Electronics',
country_of_origin = 'CN',
specifications = '4GB USB 2.0/USB 3.0')
完整的代码在这里
其他回答
在Python 3.5中,您还可以在列表、字典、元组和集合显示(有时也称为文字)中使用此语法。参见PEP 488:其他解包概括。
>>> (0, *range(1, 4), 5, *range(6, 8))
(0, 1, 2, 3, 5, 6, 7)
>>> [0, *range(1, 4), 5, *range(6, 8)]
[0, 1, 2, 3, 5, 6, 7]
>>> {0, *range(1, 4), 5, *range(6, 8)}
{0, 1, 2, 3, 5, 6, 7}
>>> d = {'one': 1, 'two': 2, 'three': 3}
>>> e = {'six': 6, 'seven': 7}
>>> {'zero': 0, **d, 'five': 5, **e}
{'five': 5, 'seven': 7, 'two': 2, 'one': 1, 'three': 3, 'six': 6, 'zero': 0}
它还允许在单个函数调用中解包多个可迭代项。
>>> range(*[1, 10], *[2])
range(1, 10, 2)
(感谢mgilson提供PEP链接。)
根据尼克的回答。。。
def foo(param1, *param2):
print(param1)
print(param2)
def bar(param1, **param2):
print(param1)
print(param2)
def three_params(param1, *param2, **param3):
print(param1)
print(param2)
print(param3)
foo(1, 2, 3, 4, 5)
print("\n")
bar(1, a=2, b=3)
print("\n")
three_params(1, 2, 3, 4, s=5)
输出:
1
(2, 3, 4, 5)
1
{'a': 2, 'b': 3}
1
(2, 3, 4)
{'s': 5}
基本上,任何数量的位置参数都可以使用*args,任何命名参数(或kwargs又名关键字参数)都可以使用**kwargs。
*表示以元组形式接收变量参数
**表示接收变量参数作为字典
使用方式如下:
1) 单个*
def foo(*args):
for arg in args:
print(arg)
foo("two", 3)
输出:
two
3
2) 现在**
def bar(**kwargs):
for key in kwargs:
print(key, kwargs[key])
bar(dic1="two", dic2=3)
输出:
dic1 two
dic2 3
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'
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