Detect if an optional argument (which may include None) is provided

Question

Is there a better way than the following example to detect if an optional argument has been provided? Even if None is provided, we want to return that instead of raising the exception.

Expected results:

>>> get_value('key-exists-not')  # default not provided
KeyError
>>> get_value('key-exists-not', None)  # should support arg
None
>>> get_value('key-exists-not', default=None)  # as well as kwarg
None
>>> get_value('key-exists-not', default='')
''

Is this the best/recommended way to do it?

def get_value(key, default='somerubbishthatwillneverbeused'):
    try:
        return _data[key]
    except KeyError:
        if default != 'somerubbishthatwillneverbeused':
            return default
        raise

That's true, it was a tricky one to search for because it's a bit different to a standard "default value". — Michael Waterfall, Jun 03 '13 at 10:04

score 6 · Accepted Answer · answered Jun 03 '13 at 10:00

6

Use a sentinel value; it's a private singleton that you test agains with is:

_sentinel = object()
def get_value(key, default=_sentinel):
    if default is _sentinel:
         # no other value provided

answered Jun 03 '13 at 10:00

Martijn Pieters

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score 4 · Answer 2 · answered Jun 03 '13 at 10:00

4

Two options

def foobar (**kwargs):
    if 'some_key' in kwargs:
        # it was provided

And the other

default = object()
def foobar (some_key=default):
    if some_key is default:
        # some_key not specified

answered Jun 03 '13 at 10:00

Blubber

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I like the second option more, because it doesn't break the function signature. – kirelagin Jun 03 '13 at 10:02
Agreed. Saw that one when I was looking through cherrypy's source, it's a nice solution. – Blubber Jun 03 '13 at 12:07

Detect if an optional argument (which may include None) is provided

2 Answers2