Multiple optional arguments python

Question

So I have a function with several optional arguments like so:

def func1(arg1, arg2, optarg1=None, optarg2=None, optarg3=None):

Optarg1 & optarg2 are usually used together and if these 2 args are specified then optarg3 is not used. By contrast, if optarg3 is specified then optarg1 & optarg2 are not used. If it were one optional argument it'd be easy for the function to "know" which argument to use:

if optarg1 != None:
    do something
else:
    do something else 

My question is how to I "tell" the function which optional argument to use when there's multiple optional arguments and not all of them are always specified? Is parsing the arguments with **kwargs the way to go?

Answer 1

**kwargs is used to let Python functions take an arbitrary number of keyword arguments and then ** unpacks a dictionary of keyword arguments. Learn More here

def print_keyword_args(**kwargs):
    # kwargs is a dict of the keyword args passed to the function
    print kwargs
    if("optarg1" in kwargs and "optarg2" in kwargs):
        print "Who needs optarg3!"
        print kwargs['optarg1'], kwargs['optarg2']
    if("optarg3" in kwargs):
        print "Who needs optarg1, optarg2!!"
        print kwargs['optarg3']

print_keyword_args(optarg1="John", optarg2="Doe")
# {'optarg1': 'John', 'optarg2': 'Doe'}
# Who needs optarg3!
# John Doe
print_keyword_args(optarg3="Maxwell")
# {'optarg3': 'Maxwell'}
# Who needs optarg1, optarg2!!
# Maxwell
print_keyword_args(optarg1="John", optarg3="Duh!")
# {'optarg1': 'John', 'optarg3': 'Duh!'}
# Who needs optarg1, optarg2!!
# Duh!