List function parameters not equal to their default values

Question

Given a Python function definition of the form:

def foo(a=None, b=1, c='bar'):

How can I dynamically determine which parameters have been set to a value different than their default value? For instance:

foo(48)

I want to be able to dynamically identify that a was the only parameter that was set. The solution I'm seeking would continue to work even if I added additional parameters to the signature (i.e. I don't want to do a manual check if a == None, etc.).

==UPDATE==

To clarify my goal: For some users of function foo, I allow it to be executed regardless. If I identify that user is of user type/class/category Bar then I only want to allow it to succeed only if they called foo with an argument for parameter a. If they provided arguments (or at least, arguments not equal to the defaults) for any other parameter then an Exception should be raised. (I know which user is calling the function based on other global data).

Again, I could simply say if b != 1 or c != 'bar', but then I would have to update it every time foo's signature gets modified.

possible duplicate of [Get a function argument's default value?](http://stackoverflow.com/questions/12627118/get-a-function-arguments-default-value) — jonrsharpe, Oct 13 '14 at 10:24
I don't think this is a dupplicate. What if you call foo(None, 1, 'bar'), would this account for a call where parameters have been set to a value different than their default value? — Flavian Hautbois, Oct 13 '14 at 10:38
@Flav To keep it simple, I would say that `foo(None, 1, 'bar')` has zero parameters which are set different than their defaults. — ChaimKut, Oct 13 '14 at 10:50
@ChaimKut can you explain more about why you want to know if it's the default or not? There might be a simpler way to achieve what you want. — Ned Batchelder, Oct 13 '14 at 15:32

Flavian Hautbois · Accepted Answer · 2014-10-13T15:36:34.970

3

I think your best course of action is to use inspect.

import inspect

def foo(non_default_var, a=None, b=1, c='bar'):
    inspector = inspect.getargspec(foo)
    local_vars = inspector.args                      # = ['d', 'a', 'b', 'c']
    default_values = inspector.defaults              # = (None, 1, 'bar')
    default_vars = local_vars[-len(default_values):] # Since default values are at the end
    for var, default_value in zip(default_vars, list(default_values)): # Now iterate and compare
        print locals()[var] == default_value         # True if the variable is default

I guess this would work even better as a decorator.

The only drawback is that you cannot know if the user intended to use the default value, or entered this value explicitely in the function call.

edited Oct 13 '14 at 15:36

answered Oct 13 '14 at 10:48

Flavian Hautbois

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You could even use `is`. – Veedrac Oct 14 '14 at 16:00
a = 1; b = 1; a is b -> True. How would `is` improve the answer? – Flavian Hautbois Oct 14 '14 at 16:28
That's not guaranteed. Using `is` will be more likely (but definitely not guaranteed) to distinguish between `f('default')` and just `f()`. Of course, it depends what behaviour is actually wanted. – Veedrac Oct 14 '14 at 16:34

List function parameters not equal to their default values

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