Getting the keyword arguments actually passed to a Python method

Asked 2009-09-11T03:21:12.753

I'm dreaming of a Python method with explicit keyword args:

def func(a=None, b=None, c=None):
    for arg, val in magic_arg_dict.items():   # Where do I get the magic?
        print '%s: %s' % (arg, val)

I want to get a dictionary of only those arguments the caller actually passed into the method, just like **kwargs, but I don't want the caller to be able to pass any old random args, unlike **kwargs.

>>> func(b=2)
b: 2
>>> func(a=3, c=5)
a: 3
c: 5

So: is there such an incantation? In my case, I happen to be able to compare each argument against its default to find the ones that are different, but this is kind of inelegant and gets tedious when you have nine arguments. For bonus points, provide an incantation that can tell me even when the caller passes in a keyword argument assigned its default value:

>>> func(a=None)
a: None

Tricksy!

Edit: The (lexical) function signature has to remain intact. It's part of a public API, and the primary worth of the explicit keyword args lies in their documentary value. Just to make things interesting. :)

python arguments keyword

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3 Answers

I was inspired by lost-theory's decorator goodness, and after playing about with it for a bit came up with this:

def actual_kwargs():
    """
    Decorator that provides the wrapped function with an attribute 'actual_kwargs'
    containing just those keyword arguments actually passed in to the function.
    """
    def decorator(function):
        def inner(*args, **kwargs):
            inner.actual_kwargs = kwargs
            return function(*args, **kwargs)
        return inner
    return decorator


if __name__ == "__main__":

    @actual_kwargs()
    def func(msg, a=None, b=False, c='', d=0):
        print msg
        for arg, val in sorted(func.actual_kwargs.iteritems()):
            print '  %s: %s' % (arg, val)

    func("I'm only passing a", a='a')
    func("Here's b and c", b=True, c='c')
    func("All defaults", a=None, b=False, c='', d=0)
    func("Nothin'")
    try:
        func("Invalid kwarg", e="bogon")
    except TypeError, err:
        print 'Invalid kwarg\n  %s' % err

Which prints this:

I'm only passing a
  a: a
Here's b and c
  b: True
  c: c
All defaults
  a: None
  b: False
  c: 
  d: 0
Nothin'
Invalid kwarg
  func() got an unexpected keyword argument 'e'

I'm happy with this. A more flexible approach is to pass the name of the attribute you want to use to the decorator, instead of hard-coding it to 'actual_kwargs', but this is the simplest approach that illustrates the solution.

Mmm, Python is tasty.

answered 2009-09-11T06:25:43.097

One possibility:

def f(**kw):
  acceptable_names = set('a', 'b', 'c')
  if not (set(kw) <= acceptable_names):
    raise WhateverYouWantException(whatever)
  ...proceed...

IOW, it's very easy to check that the passed-in names are within the acceptable set and otherwise raise whatever you'd want Python to raise (TypeError, I guess;-). Pretty easy to turn into a decorator, btw.

Another possibility:

_sentinel = object():
def f(a=_sentinel, b=_sentinel, c=_sentinel):
   ...proceed with checks `is _sentinel`...

by making a unique object _sentinel you remove the risk that the caller might be accidentally passing None (or other non-unique default values the caller could possibly pass). This is all object() is good for, btw: an extremely-lightweight, unique sentinel that cannot possibly be accidentally confused with any other object (when you check with the is operator).

Either solution is preferable for slightly different problems.

answered 2009-09-11T03:31:08.570

Magic is not the answer:

def funky(a=None, b=None, c=None):
    for name, value in [('a', a), ('b', b), ('c', c)]:
        print name, value

answered 2009-09-11T06:53:28.173

Getting the keyword arguments actually passed to a Python method

3 Answers

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