*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
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])