python - Conditionally passing arbitrary number of default named arguments to a function

Question

Is it possible to pass arbitrary number of named default arguments to a Python function conditionally ?

For eg. there's a function:

def func(arg, arg2='', arg3='def')

Now logic is that I have a condition which determines if arg3 needs to be passed, I can do it like this:

if condition == True:
    func('arg', arg2='arg2', arg3='some value')
else:
    func('arg', arg2='arg2')

Question is, can I have a shorthand like:

func('arg', 'arg2', 'some value' if condition == True else # nothing so default gets picked
)

Answer 1

That wouldn't be valid Python syntax you have to have something after else. What is done normally is:

func('arg', 'arg2', 'some value' if condition else None)

and function definition is changed accordingly:

def func(arg, arg2='', arg3=None):
    arg3 = 'def' if arg3 is None else arg3