Pythonic solution for conditional arguments passing

Going by the now-deleted comments to the question that the check is meant to be for the variables being None rather than being falsey, change func so that it handles the arguments being None:

def func(a=None, b=None):
   if a is None:
      a = 0
   if b is None:
      b = 10

And then just call func(a, b) every time.

You can add a decorator that would eliminate None arguments:

def skip_nones(fun):
    def _(*args, **kwargs):
        for a, v in zip(fun.__code__.co_varnames, args):
            if v is not None:
                kwargs[a] = v
        return fun(**kwargs)
    return _

@skip_nones
def func(a=10, b=20):
    print a, b

func(None, None) # 10 20
func(11, None)   # 11 20
func(None, 33)   # 10 33

to solve your specific question I would do:

args = {'a' : a, 'b' : b}
for varName, varVal in args.items():
    if not varVal:
        del args[varName]
f(**args)

But the most pythonic way would be to use None as the default value in your function:

f(a=None, b=None):
    a = 10 if a is None else a
    ...

and just call f(a, b)

By default, all methods in Python take variable arguments.

When you define an argument in the signature of the method, you explicity make it required. In your snippet, what you are doing is giving it a default value - not making them optional.

Consider the following:

>>> def func(a,b):
...    a = a if a else 0
...    b = b if b else 10
...    return a+b
...
>>> a = b = None
>>> func(a,b)
10
>>> a = 5
>>> b = 2
>>> func(a,b)
7
>>> func()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: func() takes exactly 2 arguments (0 given)

In this snippet, both a and b are required, since I didn't define any default values. They are not optional.

However in your snippet, because you have given them defaults in the method signature they look like they are optional, but in fact they just have defaults assigned to them.

>>> def func(a=0,b=10):
...    return a+b
...
>>> func()
10

You can confirm this by checking the argument list inside the body of the method:

>>> def func(a=b,b=10):
...    print locals().keys()
...
>>> func()
['a', 'b']

One way to have your function accept any number of arguments (in other words, making them all optional):

>>> def func(*args,**kwargs):
...    print len(args),args
...    print len(kwargs),kwargs
...
>>> func("hello","world",a=5,b=10)
2 ('hello', 'world')
2 {'a': 5, 'b': 10}
>>> func()
0 ()
0 {}
>>> func(1,2)
2 (1, 2)
0 {}
>>> func(a)
1 (5,)
0 {}
>>> func(foo="hello")
0 ()
1 {'foo': 'hello'}

Why not pass that logic to the function?

def func(a, b):
    a = a or 0
    b = b or 10
    return a + b

To answer your literal question:

func(a or 0, b or 10) 

Edit: See comment why this doesn't answer the question.

You can use the ternary if-then operator to pass conditional arguements into functions https://www.geeksforgeeks.org/ternary-operator-in-python/

For example, if you want to check if a key actually exists before passing it you can do something like:

 
    def func(arg):
        print(arg)

    kwargs = {'name':'Adam')

    func( kwargs['name'] if 'name' in kwargs.keys() else '' )       #Expected Output: 'Adam'
    func( kwargs['notakey'] if 'notakey' in kwargs.keys() else '' ) #Expected Output: ''
 

This might work:

def f(**kwargs):
    a = get(kwargs, 0)
    b = get(kwargs, 10)
    return a + b