复制代码 代码如下:

coding=utf8

author = 'Administrator'

当函数的参数不确定时，可以使用*args和*kwargs。args没有key值，**kwargs有key值

def fun_var_args(farg, *args):
print 'args:', farg
for value in args:
print 'another arg:',value

*args可以当作可容纳多个变量组成的list或tuple

fun_var_args(1, 'two', 3, None)

args: 1

another arg: two

another arg: 3

another arg: None

def fun_var_kwargs(farg, **kwargs):
print 'args:',farg
for key in kwargs:
print 'another keyword arg:%s:%s' % (key, kwargs[key])

myarg1,myarg2和myarg3被视为key， 感觉**kwargs可以当作容纳多个key和value的dictionary

fun_var_kwargs(1, myarg1='two', myarg2=3, myarg3=None)

输出：

args: 1

another keyword arg:myarg1:two

another keyword arg:myarg2:3

another keyword arg:myarg3:None

def fun_args(arg1, arg2, arg3):
print 'arg1:', arg1
print 'arg2:', arg2
print 'arg3:', arg3

myargs = ['1', 'two', None] # 定义列表
fun_args(*myargs)

输出：

arg1: 1

arg2: two

arg3: None

mykwargs = {'arg1': '1', 'arg2': 'two', 'arg3': None} # 定义字典类型
fun_args(**mykwargs)

输出：

arg1: 1

arg2: two

arg3: None

两者都有

def fun_args_kwargs(*args, **kwargs):
print 'args:', args
print 'kwargs:', kwargs

args = [1, 2, 3, 4]
kwargs = {'name': 'BeginMan', 'age': 22}
fun_args_kwargs(args,kwargs)

args: ([1, 2, 3, 4], {'age': 22, 'name': 'BeginMan'})

kwargs: {}

fun_args_kwargs(1,2,3,a=100)

args: (1, 2, 3)

kwargs: {'a': 100}

fun_args_kwargs(*(1,2,3,4),**{'a':None})

args: (1, 2, 3, 4)

kwargs: {'a': None}