A detailed description of Python's *args and **kwargs

*args represents any number of nameless arguments, which is a tuple;**kwargs representing the keyword argument, 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 results are 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 mutable parameters in Python.

Note: When using both *args and **kwargs, the parameter column must be *args before **kwargs, such as foo (a=1, b= ' 2 ', c=3, a ', 1, None,) and will prompt syntax error "SyntaxError: Non-keyword arg after keyword arg ".

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 =  1args =  (2,3,4) Kwargs =  {' A ': 1, ' C ': 3, ' B ': 2} # # #

1 assigned to the param1, the rest of the 2,3,4 to *args, the others are assigned to the **kwargs

There is also a very nice way to create a dictionary:

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

Results:

{' A ': 1, ' B ': 2, ' C ': 3}

In fact Python has the Dict class, you can use Dict (a=1,b=2,c=3) to create a dictionary.

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