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Groups > comp.lang.python > #80177
| Date | 2014-11-04 14:21 +0100 |
|---|---|
| From | Antoon Pardon <antoon.pardon@rece.vub.ac.be> |
| Subject | Re: Dictionaries with variable default. |
| References | <54576150.6090002@rece.vub.ac.be> <CAPTjJmo451DeTNhJnQQNaWJPWMDeBvMoiKK=DQidqKDVedxDLw@mail.gmail.com> |
| Newsgroups | comp.lang.python |
| Message-ID | <mailman.15444.1415107297.18130.python-list@python.org> (permalink) |
Op 03-11-14 om 12:09 schreef Chris Angelico:
> On Mon, Nov 3, 2014 at 10:04 PM, Antoon Pardon
> <antoon.pardon@rece.vub.ac.be> wrote:
>> Is it possible to have a default dictionary where the default is dependant
>> on the key?
>>
>> I was hoping something like this might work:
>>>>> m = defaultdict(lambda key: key+1)
>> But it obviously doesn't:
>>>>> m[3]
>> Traceback (most recent call last):
>> File "<stdin>", line 1, in <module>
>> TypeError: <lambda>() takes exactly 1 argument (0 given)
> Yes, but not with the defaultdict. Instead, just implement __missing__:
>
> class KeyPlusOne(dict):
> def __missing__(self, key): return key+1
>
> ChrisA
So How should I call this:
class ...dict(dict):
def __init__(self, fun):
self.fun = fun
def __missing__(self, key):
return self.fun(key)
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Re: Dictionaries with variable default. Antoon Pardon <antoon.pardon@rece.vub.ac.be> - 2014-11-04 14:21 +0100 Re: Dictionaries with variable default. Cousin Stanley <cousinstanley@gmail.com> - 2014-11-04 07:13 -0700
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