forcing keyword arguments in python 2.7

Question

I know I can use * to force all keyword arguments to a function/method to be "named".

If I have

def abc(a, *, x=10, z=30):
    pass

then the following all work

abc(5)
abc(8, x=12)
abc(9, z=31)
abc(x=17, a=4)

even if I change the function signature to def abc(a, *, x=10, y=20, z=30), and

abc(7, 13)

throws an error.

This is extremely important because, I can use the logical place, which will help maintenance over time, without being forced to use the end position based on history.

But * is not valid in Python 2.7, and abc(a, *args, x=10, z=30) (which I tried) doesn't work either.

Is there a way to force the use of x=12 in Python 2.7? Or another way of saying: make abc(7, 13) be invalid on Python 2.7.

Answer 1

One way of doing this is by adding a dummy keyword argument that never gets a valid positional value (so don't check for None):

_dummy = object()

def abc(a, dummy_kw=_dummy, x=10, z=30):
    if dummy_kw is not _dummy:
        raise TypeError("abc() takes 1 positional argument but at least 2 were given")

That will prohibit abc(7, 13) and allow all the others. It works on Python 2 and Python 3, so it is useful when you have code that needs to run on both.

Originally I used:

 def _dummy():
    pass

but as @mata pointed out _dummy=object() works as well, and cleaner. Essentially any unique memory location that is not used in another way will work.

Answer 2

What about the following:

def abc(a, **kwargs):
    # Get arguments from kwargs otherwise use default values
    x = kwargs.pop('x', 10) 
    z = kwargs.pop('z', 30) 

    if not kwargs: # if kwargs is not empty
        print 'extra parameters passed'

    pass

This allows to force the use of kwargs and still have default values.

pop removes the key from kwargs, once you use it. This is potentially very useful as you can check if the user gave extra parameters that do not belong to the function and in this case you can throw an error (for example).