Python: define arguments of a function which include the arguments of the other function

Question

Please let me know if there are duplicated ones (I believe there are) and I will remove the post. I am just not sure about the jargon for this question.

When writing a new function augmented_func which contains the other function already defined original_func , I am trying to avoid defining arguments that are already in original_func :

def original_func(a1, a2 ,a3):
    print(a1, a2, a3)

def augmented_func(b, a1, a2, a3):
    print(original_func(a1, a2, a3), b)

How do I avoid writing a 's when defining augmented_func , and make it simpler especially when the number of arguments for original_func is more than, say, three in the example?

Answer 1

Use * notation for an arbitrary number of positional arguments:

def augmented_func(b, *args):
    print(original_func(*args), b)

augmented_func('b', 1, 2, 3)

Or use keyword arguments with ** :

def augmented_func(b, **kwargs):
    print(original_func(**kwargs), b)

augmented_func('b', a1=1, a2=2, a3=3)

Answer 2

I think what you looking for is something like

def original_func(a1, a2 ,a3):
    print(a1, a2, a3)

def augmented_func(b, *args):
    print(original_func(*args), b)

you can call it then as augmented_func(1, 2, 3, 4)

*args and **kwargs is a good way for passing the variable number of arguments and passing them to other internal function calls.

Python docs for further info.