如何将defaultdicts的defaultdict转换为dicts的dict?

问题描述

使用此答案，我创建了defaultdict的defaultdict.现在，我想将深层嵌套的dict对象变成普通的python dict.

Using this answer, I created a defaultdict of defaultdicts. Now, I'd like to turn that deeply nested dict object back into an ordinary python dict.

from collections import defaultdict

factory = lambda: defaultdict(factory)
defdict = factory()
defdict['one']['two']['three']['four'] = 5

# defaultdict(<function <lambda> at 0x10886f0c8>, {
#             'one': defaultdict(<function <lambda> at 0x10886f0c8>, {
#                 'two': defaultdict(<function <lambda> at 0x10886f0c8>, {
#                     'three': defaultdict(<function <lambda> at 0x10886f0c8>, {
#                         'four': 5})})})})

我认为这不是正确的解决方案:

I assume this is not the right solution:

import json

regdict = json.loads(json.dumps(defdict))

# {u'one': {u'two': {u'three': {u'four': 5}}}}

此外，此答案是不充分的，因为它不会对嵌套的字典进行递归.

Also, this answer is inadequate since it does not recurse on the nested dict(s).

推荐答案

您可以遍历树，将每个defaultdict实例替换为由dict理解产生的dict:

You can recurse over the tree, replacing each defaultdict instance with a dict produced by a dict comprehension:

def default_to_regular(d):
    if isinstance(d, defaultdict):
        d = {k: default_to_regular(v) for k, v in d.items()}
    return d

演示:

>>> from collections import defaultdict
>>> factory = lambda: defaultdict(factory)
>>> defdict = factory()
>>> defdict['one']['two']['three']['four'] = 5
>>> defdict
defaultdict(<function <lambda> at 0x103098ed8>, {'one': defaultdict(<function <lambda> at 0x103098ed8>, {'two': defaultdict(<function <lambda> at 0x103098ed8>, {'three': defaultdict(<function <lambda> at 0x103098ed8>, {'four': 5})})})})
>>> default_to_regular(defdict)
{'one': {'two': {'three': {'four': 5}}}}

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