CODEWITHSUNDEEP
python
itzy
itzy

Reputation: 11755

Call another function and optionally keep default arguments

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?

Upvotes: 13

Views: 3677

Answers (6)

Dunes
Dunes

Reputation: 40703

Using FunctionType from types, you can just take a function and create a new one specifying the defaults at runtime. You can put all this in a decorator so that at the point of where you write your code it will keep things tidy, whilst still giving the reader a clue about what you are trying to accomplish. It also allows the exact same call signature for funB as funA -- all arguments can be positional, or all arguments can be keywords, or any valid mix thereof, and any arguments with default values are optional. Should play nice with positional arguments (*args) and keyword arguments (**kwargs) too.

import inspect
from types import FunctionType

def copy_defaults(source_function):
    def decorator(destination_function):
        """Creates a wrapper for the destination function with the exact same 
        signature as source_function (including defaults)."""

        # check signature matches
        src_sig = inspect.signature(source_function)
        dst_sig = inspect.signature(destination_function)
        if list(src_sig.parameters) != list(dst_sig.parameters):
            raise ValueError("src func and dst func do not having matching " \
                "parameter names / order")

        return FunctionType(
            destination_function.__code__,
            destination_function.__globals__,
            destination_function.__name__,
            source_function.__defaults__, # use defaults from src
            destination_function.__closure__
        )
    return decorator

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

@copy_defaults(funA)
def funB(x, a, b):
    """this is fun B"""
    return funA(x, a, b)

assert funA(1, 2) == funB(1, 2)
assert funB.__name__ == "funB"
assert funB.__doc__ == "this is fun B"

Upvotes: 2

falsetru
falsetru

Reputation: 369044

Using inspect.getargspec, you can get the default values (fourth item of the returned tuple = defaults):

import inspect

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

# inspect.getargspec(funA) =>
#     ArgSpec(args=['x', 'a', 'b'], varargs=None, keywords=None, defaults=(1,))
def funcB(x, a, b=inspect.getargspec(funA)[3][0]):
    return funA(x, a, b)

OR (in Python 2.7+)

def funcB(x, a, b=inspect.getargspec(funA).defaults[0]):
    return funA(x, a, b)

In Python 3.5+, it's recommend to use inspect.signature instead:

def funcB(x, a, b=inspect.signature(funA).parameters['b'].default):
    return funA(x, a, b)

Upvotes: 2

Olivier Pellier-Cuit
Olivier Pellier-Cuit

Reputation: 1279

You can also use: 

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

def funB(x, a, b=None):
   return funA(*filter(lambda o: o is not None, [x, a, b]))

Version which will not fail if x or a are None:

def funB(x, a, b=None):
    return funA(*([x, a]+filter(lambda o: o is not None, [b])))

Upvotes: 0

Mad Physicist
Mad Physicist

Reputation: 114240

The way you did it is fine. Another way is for funB to have the same defaults as funA, so you can pass the same parameters right through. E.g., if you do def funB(x, a, b=1), then you can always call return funA(x, a, b) just like that.

For simple cases, the above will work fine. For more complex cases, you may want to use *args and **kwargs (explained here and here). Specifically, you can pass in all your keyword arguments as a dictionary (conventionally called kwargs). In this case, each function would set its own independent defaults, and you would just pass the whole dictionary through:

def funA(x, a, **kwargs):
   b = kwargs.get("b", 1)
   return a+b*x

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

If kwargs is empty when passed to funB (b is not specified), it will be set to the default in funA by the statement b = kwargs.get("b", 1). If b is specified, it will be passed through as-is. Note that in funB, you can access b with its own, independent default value and still get the behavior you are looking for.

While this may seem like overkill for your example, extracting a couple of arguments at the beginning of a function is not a big deal if the function is complex enough. It also gives you a lot more flexibility (such as avoiding many of the common gotchas).

Upvotes: 3

LeartS
LeartS

Reputation: 2896

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)

Upvotes: 16

k4ppa
k4ppa

Reputation: 4667

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()

Upvotes: 7

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