在密钥中共享defaultdict中的Dict

Question

I have a dictionary inside a defaultdict. I noticed that the dictionary is being shared across keys and therefore it takes the values of the last write. How can I isolate those dictionaries?

>>> from collections import defaultdict
>>> defaults = [('a', 1), ('b', {})]
>>> dd = defaultdict(lambda: dict(defaults))
>>> dd[0]
{'a': 1, 'b': {}}
>>> dd[1]
{'a': 1, 'b': {}}
>>> dd[0]['b']['k'] = 'v'
>>> dd
defaultdict(<function <lambda> at 0x7f4b3688b398>, {0: {'a': 1, 'b': {'k': 'v'}}, 1:{'a': 1, 'b': {'k': 'v'}}})
>>> dd[1]['b']['k'] = 'v2'
>>> dd
defaultdict(<function <lambda> at 0x7f4b3688b398>, {0: {'a': 1, 'b': {'k': 'v2'}}, 1: {'a': 1, 'b': {'k': 'v2'}}})

Notice that v was set to v2 for both dictionaries. Why is that? and how to change this behavior without much performance overhead?

Answer 1

When you do dict(defaults) you're not copying the inner dictionary, just making another reference to it. So when you change that dictionary, you're going to see the change everywhere it's referenced.

You need deepcopy here to avoid the problem:

import copy
from collections import defaultdict
defaults = {'a': 1, 'b': {}}
dd = defaultdict(lambda: copy.deepcopy(defaults))

Or you need to not use the same inner mutable objects in successive calls by not repeatedly referencing defaults :

dd = defaultdict(lambda: {'a': 1, 'b': {}})

Answer 2

Your values all contain references to the same object from defaults : you rebuild the outer dict, but not the inner one. Just make a function that creates a new, separate object:

def builder():
    return {'a': 1, 'b': {}}
dd = defaultdict(builder)