Python function parameter *args**kwargs usage instance

The code is as follows:

#coding =utf8
__author__ = ' Administrator '

# *args and **kwargs can be used when the parameters of the function are not deterministic. *args no key value, **kwargs has a key value

def Fun_var_args (Farg, *args):
print ' args: ', farg
For value in args:
print ' Another arg: ', value

# *args can be used as a list or tuple that can accommodate multiple variables
Fun_var_args (1, ' both ', 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 and Myarg3 are considered key, feeling that **kwargs can be used to accommodate multiple keys and value dictionary
Fun_var_kwargs (1, myarg1= ' both ', myarg2=3, Myarg3=none)
Output
#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 ', ' both ', None] # definition List
Fun_args (*myargs)

Output
#arg1:1
#arg2:
#arg3: None

Mykwargs = {' arg1 ': ' 1 ', ' arg2 ': ' Both ', ' Arg3 ': None} # defines the dictionary type
Fun_args (**mykwargs)

Output
#arg1:1
#arg2:
#arg3: None

# both have
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 ': $, ' 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}

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