Question

我有如下字典：

每个键都有一个与之关联的字典。

dict_sample = {'a': {'d0': '1', 'd1': '2', 'd2': '3'}, 'b': {'d0': '1'}, 'c': {'d1': '1'}}

我需要如下输出：

output_dict = {'d0': {'a': 1, 'b': 1}, 'd1': {'a': 2, 'c': 1}, 'd2': {'a': 3}}

我非常感谢在pythonic方法上实现这一目标的任何帮助。谢谢！

Answer 1

我相信这会产生所需的输出

>>> from collections import defaultdict
>>> d = defaultdict(dict)
>>>
>>> dict_sample = {'a': {'d0': '1', 'd1': '2', 'd2': '3'}, 'b': {'d0': '1'}, 'c': {'d1': '1'}}
>>>
>>> for key, value in dict_sample.items():
...     for k, v in value.items():
...         d[k][key] = v
...
>>> d
defaultdict(<class 'dict'>, {'d0': {'a': '1', 'b': '1'}, 'd1': {'a': '2', 'c': '1'}, 'd2': {'a': '3'}})

Answer 2

您可以在具有嵌套循环的新字典上使用dict.setdefault：

d = {}
# for each key and sub-dict in the main dict
for k1, s in dict_sample.items():
    # for each key and value in the sub-dict
    for k2, v in s.items():
        # this is equivalent to d[k2][k1] = int(v), except that when k2 is not yet in d,
        # setdefault will initialize d[k2] with {} (a new dict)
        d.setdefault(k2, {})[k1] = int(v)

d将变为：

{'d0': {'a': 1, 'b': 1}, 'd1': {'a': 2, 'c': 1}, 'd2': {'a': 3}}

交换嵌套字典的键

2 个答案: