Get a function argument’s default value?

Python3.x

In a python3.x world, you should probably use a Signature object:

import inspect

def get_default_args(func):
    signature = inspect.signature(func)
    return {
        k: v.default
        for k, v in signature.parameters.items()
        if v.default is not inspect.Parameter.empty
    }

Python2.x (old answer)

The args/defaults can be combined as:

import inspect
a = inspect.getargspec(eat_dog)
zip(a.args[-len(a.defaults):],a.defaults)

Here a.args[-len(a.defaults):] are the arguments with defaults values and obviously a.defaults are the corresponding default values.

You could even pass the output of zip to the dict constructor and create a mapping suitable for keyword unpacking.

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looking at the docs, this solution will only work on python2.6 or newer since I assume that inspect.getargspec returns a named tuple. Earlier versions returned a regular tuple, but it would be very easy to modify accordingly. Here’s a version which works with older (and newer) versions:

import inspect
def get_default_args(func):
    """
    returns a dictionary of arg_name:default_values for the input function
    """
    args, varargs, keywords, defaults = inspect.getargspec(func)
    return dict(zip(args[-len(defaults):], defaults))

Come to think of it:

    return dict(zip(reversed(args), reversed(defaults)))

would also work and may be more intuitive to some people.

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