#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}