[ACCEPTED]-Exposing `defaultdict` as a regular `dict`-defaultdict
defaultdict docs say for default_factory:
If the default_factory attribute 5 is None, this raises a KeyError exception 4 with the key as argument.
What if you just 3 set your defaultdict's default_factory to 2 None? E.g.,
>>> d = defaultdict(int)
>>> d['a'] += 1
>>> d
defaultdict(<type 'int'>, {'a': 1})
>>> d.default_factory = None
>>> d['b'] += 2
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
KeyError: 'b'
>>>
Not sure if this is the best approach, but 1 seems to work.
Once you have finished populating your defaultdict, you 4 can simply create a regular dict from it:
my_dict = dict(my_default_dict)
One 3 can optionally use the typing.Final type annotation.
If 2 the default dict is a recursive default 1 dict, see this answer which uses a recursive solution.
You could make a class that holds a reference 1 to your dict and prevent setitem()
from collections import Mapping
class MyDict(Mapping):
def __init__(self, d):
self.d = d;
def __getitem__(self, k):
return self.d[k]
def __iter__(self):
return self.__iter__()
def __setitem__(self, k, v):
if k not in self.d.keys():
raise KeyError
else:
self.d[k] = v
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