How to use a single named parameter after *args

How can I use single **kwargs in my program? I need it to set a port number in an application uses sockets. I have to use **kwargs because I use *args as well so I can't just add port=xxxx in my function arguments.

def myFunction(*args, **kwargs):
   ... 
   socket.connect(("xx.xx.xx.xx", kwargs.values()))
   ... 

myFunction("something", "other", port = xxxx)

I'm now using kwargs.values() to set the port number, but it isn't good idea I think.

I don't want kwargs, I thought that I can't just add port=xxxx in declaration parameters in my function while I have *args

Python 3

In Python 3 it's possible to define a paramter after *args:

>>> def func(*args, port):
...     print(port)
...     print(args)
... 
>>> func('something', 'args', port='8000')
8000
('something', 'args')

However this will generate a syntax error in Python 2.

To adapt this to your code you can use:

#python3 only
def func(*args, port):
    socket.connect(('xx.xx.xx.xx', port))

A more natural solution: using the **kwargs itself (Python 2 and 3)

>>> def func(*args, **kwargs):
...     print(args)
...     print(kwargs)
... 
>>> func('something', 'other', port='8000')
('something', 'other')
{'port': '8000'}

As you can see, you can pull the port parameter if it is provided, in the kwargs dict

def func(*args, **kwargs):
   socket.connect(('xx.xx.xx.xx', kwargs['port']))

If you want to have a default value like @wovano suggested :

def func(*args, **kwargs):
   socket.connect(('xx.xx.xx.xx', kwargs.get('port', default_port)))

Call it using:

func('something', 'other', port='8000')

Or using the default port value:

func('something', 'other')