不插入缺失值的 Python defaultdict

Question

So the defaultdict documentation mentions that, if an item is missing, the value returned by default_factory "is inserted in the dictionary for the key, and returned." That's great most of the time, but what I actually want in this case is for the value to be returned but not inserted into the defaultdict.

I figured I could probably subclass defaultdict and override... I guess __missing__ ? Not sure. What's the best way to go about this?

Thanks in advance.

Answer 1

You can subclass dict and implement __missing__ :

class missingdict(dict):
    def __missing__(self, key):
        return 'default'  # note, does *not* set self[key]

Demo:

>>> d = missingdict()
>>> d['foo']
'default'
>>> d
{}

You could subclass defaultdict too, you'd get the factory handling plus copy and pickle support thrown in:

from collections import defaultdict

class missingdict(defaultdict):
    def __missing__(self, key):
        return self.default_factory() 

Demo:

>>> from collections import defaultdict
>>> class missingdict(defaultdict):
...     def __missing__(self, key):
...         return self.default_factory() 
... 
>>> d = missingdict(list)
>>> d['foo']
[]
>>> d
defaultdict(<type 'list'>, {})

but, as you can see, the __repr__ does lie about its name.

Answer 2

It's even simpler than subclassing. While dict[key] adds the key, dict.get(key, default=None) will NOT add the key to the dictionary:

import collections

d = collections.defaultdict(int)

assert d['test1'] == 0

print(d)
# defaultdict(<class 'int'>, {'test1': 0})

assert d.get('test2', 0) == 0
# for a more generic version, use `d.default_factory()` instead of 0

print(d)
# defaultdict(<class 'int'>, {'test1': 0})
# => did NOT insert a key for 'test2' =)