Call another function and optionally keep default arguments

Question

I have a function with one optional argument, like this:

def funA(x, a, b=1):
   return a+b*x

I want to write a new function that calls funA and also has an optional argument, but if no argument is passed, I want to keep the default in funA.

I was thinking something like this:

def funB(x, a, b=None):
   if b:
     return funA(x, a, b)
   else:
     return funA(x, a)

Is there a more pythonic way of doing this?

Answer 1

I would replace if b with if b is not None, so that if you pass b=0 (or any other "falsy" value) as argument to funB it will be passed to funA.

Apart from that it seems pretty pythonic to me: clear and explicit. (albeit maybe a bit useless, depending on what you're trying to do!)

A little more cryptic way that relies on calling funB with the correct keyword arguments (e.g. funB(3, 2, b=4):

def funB(x, a, **kwargs):
    return funA(x, a, **kwargs)

Answer 2

def funA(x, a, b=1):
    return a+b*x

def funB(x, a, b=1):     
   return funA(x, a, b)

Make the default value of b=1 in funB() and then pass it always to funA()