Get a function argument's default value?

Question

For this function

def eat_dog(name, should_digest=True):
    print "ate dog named %s. Digested, too? %" % (name, str(should_digest))

I want to, external to the function, read its arguments and any default values attached. So for this specific example, I want to know that name has no default value (i.e. that it is a required argument) and that True is the default value for should_digest.

I'm aware of inspect.getargspec(), which does give me information about arguments and default values, but I see no connection between the two:

ArgSpec(args=['name', 'should_digest'], varargs=None, keywords=None, defaults=(True,))

From this output how can I tell that True (in the defaults tuple) is the default value for should_digest?

Additionally, I'm aware of the "ask for forgiveness" model of approaching a problem, but unfortunately output from that error won't tell me the name of the missing argument:

>>> eat_dog()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: eat_dog() takes at least 1 argument (0 given)

To give context (why I want to do this), I'm exposing functions in a module over a JSON API. If the caller omits certain function arguments, I want to return a specific error that names the specific function argument that was omitted. If a client omits an argument, but there's a default provided in the function signature, I want to use that default.

Answer 1

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.

---

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.

---

Answer 2

Depending on exactly what you need, you might not need the inspect module since you can check the __defaults__ attribute of the function:

>>> eat_dog.__defaults__
(True,)
>>> eat_dog.__code__.co_argcount
2
>>> eat_dog.__code__.co_varnames
('name', 'should_digest')
>>> 
>>> eat_dog.__kwdefaults__
>>> eat_dog.__code__.co_kwonlyargcount
0 

Answer 3

You can use inspect module with its getargspec function:

inspect.getargspec(func)

Get the names and default values of a Python function’s arguments. A tuple of four things is returned: (args, varargs, keywords, defaults). args is a list of the argument names (it may contain nested lists). varargs and keywords are the names of the * and ** arguments or None. defaults is a tuple of default argument values or None if there are no default arguments; if this tuple has n elements, they correspond to the last n elements listed in args.

See mgilson's answer for exact code on how to retrieve argument names and their default values.

Answer 4

To those looking for a version to grab a specific default parameter with mgilson's answer.

value = signature(my_func).parameters['param_name'].default

Here's a full working version, done in Python 3.8.2

from inspect import signature

def my_func(a, b, c, param_name='apple'):
    pass

value = signature(my_func).parameters['param_name'].default

print(value == 'apple') # True

Answer 5

to take care of keyword-only args (and because defaults and kwonlydefaults can be None):

spec = inspect.getfullargspec(func)
defaults = dict(zip(spec.args[::-1], (spec.defaults or ())[::-1]))
defaults.update(spec.kwonlydefaults or {})

Answer 6

You can get this via some of the __dunder__ vars as mentioned by other posts. Putting that into a simple helper function can get you a dictionary of default values.

• .__code__.co_varnames: A tuple of all input variables
• .__defaults__: A tuple of the default values
  • It is worth noting that this tuple only incudes the default provided variables which must always be positioned last in the function arguments

You can use these two items to match the last n variables in the .__code__.co_varnames with all the items in the .__defaults__

EDIT Thanks to @griloHBG - Added if statement to prevent exceptions when no defaults are specified.

def my_fn(a, b=2, c='a'):
    pass

def get_defaults(fn):
    if fn.__defaults__==None:
        return {}
    return dict(zip(
        fn.__code__.co_varnames[-len(fn.__defaults__):],
        fn.__defaults__
    ))

print(get_defaults(my_fn))

Should give:

{'b': 2, 'c': 'a'}

Answer 7

In python, all the arguments with default value come after the arguments without default value. So the mapping should start from the end till you exhaust the default value list. Hence the logic:

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

gives the correctly mapped defaults.