Count positional arguments in function signature

Question

I have the following function signature:

# my signature
def myfunction(x, s, c=None, d=None):
    # some irrelevant function body

I need the number of positional arguments. How can I return the number (2) of positional arguments (x & s). The number of keyword arguments is not relevant.

Answer 1

You can get the number of all arguments (using f.__code__.co_argcount), the number of keyword arguments (using f.__defaults__), and subtract the latter from the first:

def myfunction(x, s, c=None, d=None):
  pass

all_args = myfunction.__code__.co_argcount

if myfunction.__defaults__ is not None:  #  in case there are no kwargs
  kwargs = len(myfunction.__defaults__)
else:
  kwargs = 0

print(all_args - kwargs)

Output:

2

From the Docs:

__defaults__: A tuple containing default argument values for those arguments that have defaults, or None if no arguments have a default value.

and:

co_argcount is the number of positional arguments (including arguments with default values);