How to force one keyword in python function

Question

I am implementing a function with three keywords. The default value for each keyword is None, but I need to force the user to pass at least one keyword. The reason why I want to use keywords is that the keyword names a, b and c are descriptive, and will help the user to figure out what does he need to pass to method. How do I achieve my task?

def method(a=None, b=None, c=None):

    if a!=None:
        func_a(a)
    elif b!=None:
        func_b(b)
    elif c!=None:
        func_c(c)
    else:
        raise MyError('Don\'t be silly, user - please!')

In the above example, assume of course that a, b and c have different attributes. The obvious solution would be:

def method(x):
    if is_instance(x, A):
        func_a(x)
    elif is_instance(x, B):
        func_b(x)
    [...]

But the problem is that as I said I want to use the keyword names a, b and c to help the user understand what he does need to pass to method!

Is there a more pythonic way to achieve the result?

Which Python version is this? In Python 3.4 you can use single-dispatch generic functions for this: http://legacy.python.org/dev/peps/pep-0443/ — Simeon Visser, Apr 10 '14 at 17:25
@SimeonVisser Thanks for the link, that's very interesting! Unfortunately I am stuck with python 2.7. I think I want overloading, that's it! — astabada, Apr 10 '14 at 17:39
@SimeonVisser Thanks for the link, too. Everyday you learn something new. — Hyperboreus, Apr 10 '14 at 17:51
[related, but with correlated arguments](http://stackoverflow.com/q/28604741/321973) — Tobias Kienzler, Mar 11 '15 at 10:13

score 2 · Answer 1 · answered Apr 10 '14 at 17:26

2

You could use all() to raise an error early:

def foo(a=None, b=None, c=None):
    if all(x is None for x in (a, b, c)):
        raise ValueError('You need to set at least *one* of a, b, or c')

    if a is not None:
        func_a(a)
    # etc.

answered Apr 10 '14 at 17:26

Martijn Pieters

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score 0 · Answer 2 · answered Apr 10 '14 at 17:35

You can use decorator, simulating the Contract programming paradigm.

def check_params(func):
    def wrapper(*args, **kwargs):
        a = kwargs.get('a', None)
        b = kwargs.get('b', None)
        c = kwargs.get('c', None)
        if (a == b == c == None):
            raise Exception("Set one of a, b or c is mandatory.")
        else:
            return func(*args, **kwargs)
    return wrapper


@check_params
def foo(a=None, b=None, c=None):
    print("Ok")


foo(a=4)  # This will print ok.
foo()     # This will rise an exception.

Interesting approach, though you could generalize the decorator using [`inspect.getargspec`](http://stackoverflow.com/a/218709/321973) and [`all`](http://stackoverflow.com/a/22994928/321973) — Tobias Kienzler, Mar 10 '15 at 12:13

score 0 · Answer 3 · edited May 23 '17 at 12:10

Note that a call such as method(b="something", a="otherthing") would return func_a(a), and not func_b(b), which the user might expect. In fact, it would be better to make sure one and only one keyword is not None (see e.g. here), for which it would probably make more sense for the user to simply directly call the respective methods (though you might want to call them method_from_a etc. then).

score -2 · Accepted Answer · answered Apr 10 '14 at 17:29

-2

Try this expression to evaluate that at least one argument is passed

if not (a or b or c):
    raise MyError

answered Apr 10 '14 at 17:29

Trimax

2,413
7
35
59

1

This fails if `a`, `b` or `c` can assume a false, but valid value (e.g. `0` or any empty container). – Bakuriu Apr 10 '14 at 17:48

How to force one keyword in python function

4 Answers4