Getting the keyword arguments actually passed to a Python method

Question

I\'m dreaming of a Python method with explicit keyword args:

def func(a=None, b=None, c=None):
    for arg, val in magic_arg_dict.items():   # Where do I get         


        

Answer 1

Magic is not the answer:

def funky(a=None, b=None, c=None):
    for name, value in [('a', a), ('b', b), ('c', c)]:
        print name, value

Answer 2

Here is the easiest and simplest way:

def func(a=None, b=None, c=None):
    args = locals().copy()
    print args

func(2, "egg")

This give the output: {'a': 2, 'c': None, 'b': 'egg'}. The reason args should be a copy of the locals dictionary is that dictionaries are mutable, so if you created any local variables in this function args would contain all of the local variables and their values, not just the arguments.

More documentation on the built-in locals function here.