Detailed explanation of Python's *args and **kwargs

*args represents any multiple unnamed parameters, which is a tuple; **kwargs represents keyword parameters, which is a dict.

def fun(*args, **kwargs):
    print 'args = ', args
    print 'kwargs = ', kwargs
    print '###'
if __name__ == '__main__':
    foo(1,2,3,4)
    foo(a=1,b=2,c=3)
    foo(1,2,3,4, a=1,b=2,c=3)
    foo('a', 1, None, a=1, b='2', c=3)

The output result is as follows:

args =  (1, 2, 3, 4) 
kwargs =  {} 
###
args =  () 
kwargs =  {'a': 1, 'c': 3, 'b': 2} 
###
args =  (1, 2, 3, 4) 
kwargs =  {'a': 1, 'c': 3, 'b': 2} 
###
args =  ('a', 1, None) 
kwargs =  {'a': 1, 'c': 3, 'b': '2'} 
###

As you can see, these two are variable parameters in python.

Note: When using *args and **kwargs at the same time, the *args parameter must be listed before **kwargs, like foo(a=1, b='2', c=3, a', 1, None, ) If called in this way, the syntax error "SyntaxError: non-keyword arg after keyword arg" will be prompted.

def fun2(param1, *args, **kwargs):
    print 'param1 = ', param1
    print 'args = ', args
    print 'kwargs = ', kwargs
    print '###'
fun2(1, 2, 3, 4, a=1,b=2,c=3)

Output result:

param1 =  1
args =  (2,3,4)
kwargs =  {'a': 1, 'c': 3, 'b': 2} 
###

1 is assigned to param1, the remaining 2, 3, and 4 are assigned to *args, and the others are assigned to **kwargs

There is also a very beautiful usage, which is to create a dictionary:

def kw_dict(**kwargs):
    return kwargs
print kw_dict(a=1,b=2,c=3)

Result:

{'a':1, 'b':2, 'c':3}

In fact, there is a dict class in Python, use dict(a=1,b=2,c= 3) You can create a dictionary.