Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]


Groups > comp.lang.python > #80177

Re: Dictionaries with variable default.

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)

Show all headers | View raw


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)

Back to comp.lang.python | Previous | Next — Next in thread | Find similar | Unroll thread


Thread

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

csiph-web