How to indicate that a function should use its default value of an optional argument without omitting it?

Question

I am trying to write code that does not repeat itself, following the DRY principle.

Consider a function call with many arguments, both mandatory and optional. In some cases, I would like to specify a value for an optional argument, whereas in other cases I would like to leave that value to its default value. To simplify my problem:

def func(a, b=2):
    print("b = {}".format(b))

avalue = 1
condition = 2
arg = None  # <-- Means: "use default" 
if condition == 1:
    arg = 3

func(avalue, b=arg)

Output:

b = None

Expected output:

b = 2

Thus, I am trying to avoid coding the function call twice like this:

if arg:
    func(avalue, b=arg)
else:
    func(avalue)

Is it possible in Python?

Answer 1

Use a dictionary, and only set the optional argument as a key-value pair in that if you want to specify it. Then apply the dictionary using the **kwargs call syntax:

avalue = 1
condition = 2

kwargs = {}
if condition == 1:
    kwargs['b'] = 3

func(avalue, **kwargs)

An empty dictionary (the condition != 1 case) leaves b set to the default value.