英文:
Reorganize nested `dict`
问题
这个问题与此处相关。我想要重新组织以下嵌套的dict：
a = {
(0.0, 0.0): {'a': [25, 29, nan]},
(0.0, 2.0): {'a': [25, 29, nan], 'b': [25, 35, 31.0]},
(0.0, 4.0): {'b': [25, 35, 31.0]},
(2.0, 0.0): {'a': [25, 29, nan], 'c': [25, 26, 29.0]},
(2.0, 1.5): {'a': [25, 29, nan], 'c': [25, 26, 29.0]},
(2.0, 2.0): {'a': [25, 29, nan], 'b': [25, 35, 31.0]},
(2.0, 4.0): {'b': [25, 35, 31.0]},
(3.0, 3.0): {'d': [25, 31, 32.0]},
(3.0, 5.0): {'d': [25, 31, 32.0]},
(5.0, 0.0): {'c': [25, 26, 29.0]},
(5.0, 1.5): {'c': [25, 26, 29.0]},
(5.0, 3.0): {'d': [25, 31, 32.0]},
(5.0, 5.0): {'d': [25, 31, 32.0]},
(6.0, 1.0): {'e': [25, 28, 30.0]},
(6.0, 3.0): {'e': [25, 28, 30.0]},
(8.0, 1.0): {'e': [25, 28, 30.0]},
(8.0, 3.0): {'e': [25, 28, 30.0]}
}
我想要交换内部和外部的key。一些外部的key会重复，value 应该变成一个list的lists。结果应该是：
{'a': {(0.0, 0.0): [[25, 29, nan]],
(0.0, 2.0): [[25, 29, nan], [25, 35, 31.0]],
(2.0, 0.0): [[25, 29, nan], [25, 26, 29.0]],
(2.0, 1.5): [[25, 29, nan], [25, 26, 29.0]],
(2.0, 2.0): [[25, 29, nan], [25, 35, 31.0]]},
'b': {(0.0, 2.0): [[25, 29, nan], [25, 35, 31.0]],
(0.0, 4.0): [[25, 35, 31.0]],
(2.0, 2.0): [[25, 29, nan], [25, 35, 31.0]],
(2.0, 4.0): [[25, 35, 31.0]]},
'c': {(2.0, 0.0): [[25, 29, nan], [25, 26, 29.0]],
(2.0, 1.5): [[25, 29, nan], [25, 26, 29.0]],
(5.0, 0.0): [[25, 26, 29.0]],
(5.0, 1.5): [[25, 26, 29.0]]},
'd': {(3.0, 3.0): [[25, 31, 32.0]],
(3.0, 5.0): [[25, 31, 32.0]],
(5.0, 3.0): [[25, 31, 32.0]],
(5.0, 5.0): [[25, 31, 32.0]]},
'e': {(6.0, 1.0): [[25, 28, 30.0]],
(6.0, 3.0): [[25, 28, 30.0]],
(8.0, 1.0): [[25, 28, 30.0]],
(8.0, 3.0): [[25, 28, 30.0]]}
}
直觉告诉我使用pd.DataFrame和.groupby() [并剔除NaN单元格] 是正确的方法...
df = pd.DataFrame(dict_vertices)
print(df.head(2))
0.0 2.0 ... 8.0 6.0
0.0 0.0 1.5 ... 1.0 3.0 3.0
a [25, 29, nan] [25, 29, nan] [25, 29, nan] ... NaN NaN NaN
c NaN [[25, 26, 29.0]] [[25, 26, 29.0]] ... NaN NaN NaN
[2 rows x 17 columns]
...但我不确定。如何重新组织以下嵌套的dict，其中value跟随外部的key？
英文:
This question is connected to [-> here].
I would like to reorganize the following nested dict please:
a = {
(0.0, 0.0): {'a': [25, 29, nan]},
(0.0, 2.0): {'a': [25, 29, nan], 'b': [25, 35, 31.0]},
(0.0, 4.0): {'b': [25, 35, 31.0]},
(2.0, 0.0): {'a': [25, 29, nan], 'c': [25, 26, 29.0]},
(2.0, 1.5): {'a': [25, 29, nan], 'c': [25, 26, 29.0]},
(2.0, 2.0): {'a': [25, 29, nan], 'b': [25, 35, 31.0]},
(2.0, 4.0): {'b': [25, 35, 31.0]},
(3.0, 3.0): {'d': [25, 31, 32.0]},
(3.0, 5.0): {'d': [25, 31, 32.0]},
(5.0, 0.0): {'c': [25, 26, 29.0]},
(5.0, 1.5): {'c': [25, 26, 29.0]},
(5.0, 3.0): {'d': [25, 31, 32.0]},
(5.0, 5.0): {'d': [25, 31, 32.0]},
(6.0, 1.0): {'e': [25, 28, 30.0]},
(6.0, 3.0): {'e': [25, 28, 30.0]},
(8.0, 1.0): {'e': [25, 28, 30.0]},
(8.0, 3.0): {'e': [25, 28, 30.0]}
}
I want to swap the inner and outer keys.
Some outer keys will duplicate and the value should become a list of lists. The result should be:
{'a': {(0.0, 0.0): [[25, 29, nan]],
(0.0, 2.0): [[25, 29, nan], [25, 35, 31.0]],
(2.0, 0.0): [[25, 29, nan], [25, 26, 29.0]],
(2.0, 1.5): [[25, 29, nan], [25, 26, 29.0]],
(2.0, 2.0): [[25, 29, nan], [25, 35, 31.0]]},
'b': {(0.0, 2.0): [[25, 29, nan], [25, 35, 31.0]],
(0.0, 4.0): [[25, 35, 31.0]],
(2.0, 2.0): [[25, 29, nan], [25, 35, 31.0]],
(2.0, 4.0): [[25, 35, 31.0]]},
'c': {(2.0, 0.0): [[25, 29, nan], [25, 26, 29.0]],
(2.0, 1.5): [[25, 29, nan], [25, 26, 29.0]],
(5.0, 0.0): [[25, 26, 29.0]],
(5.0, 1.5): [[25, 26, 29.0]]},
'd': {(3.0, 3.0): [[25, 31, 32.0]],
(3.0, 5.0): [[25, 31, 32.0]],
(5.0, 3.0): [[25, 31, 32.0]],
(5.0, 5.0): [[25, 31, 32.0]]},
'e': {(6.0, 1.0): [[25, 28, 30.0]],
(6.0, 3.0): [[25, 28, 30.0]],
(8.0, 1.0): [[25, 28, 30.0]],
(8.0, 3.0): [[25, 28, 30.0]]}
}
Intuition tells me pd.DataFrame with a .groupby() [and cull the NaN cells] would be the way to go...
df = pd.DataFrame(dict_vertices)
print(df.head(2))
0.0 2.0 ... 8.0 6.0
0.0 0.0 1.5 ... 1.0 3.0 3.0
a [25, 29, nan] [25, 29, nan] [25, 29, nan] ... NaN NaN NaN
c NaN [[25, 26, 29.0]] [[25, 26, 29.0]] ... NaN NaN NaN
[2 rows x 17 columns]
...but I don't know.
How do I reorganize the following nested dict please; where the value follows the outer key?
答案1
得分: 1
你可以使用以下代码：
out = {}
for k1, d in a.items():
for k2 in d:
out.setdefault(k2, {})[k1] = list(d.values())
输出如下：
{
'a': {
(0.0, 0.0): [[25, 29, nan]],
(0.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]],
(2.0, 0.0): [[25, 29, nan], [[25, 26, 29.0]]],
(2.0, 1.5): [[25, 29, nan], [[25, 26, 29.0]]],
(2.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]]
},
'b': {
(0.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]],
(0.0, 4.0): [[25, 35, 31.0]],
(2.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]],
(2.0, 4.0): [[25, 35, 31.0]]
},
'c': {
(2.0, 0.0): [[25, 29, nan], [[25, 26, 29.0]]],
(2.0, 1.5): [[25, 29, nan], [[25, 26, 29.0]]],
(5.0, 0.0): [[25, 26, 29.0]],
(5.0, 1.5): [[25, 26, 29.0]]
},
'd': {
(3.0, 3.0): [[25, 31, 32.0]],
(3.0, 5.0): [[25, 31, 32.0]],
(5.0, 3.0): [[25, 31, 32.0]],
(5.0, 5.0): [[25, 31, 32.0]]
},
'e': {
(6.0, 1.0): [[25, 28, 30.0]],
(6.0, 3.0): [[25, 28, 30.0]],
(8.0, 1.0): [[25, 28, 30.0]],
(8.0, 3.0): [[25, 28, 30.0]]
}
}
英文:
You can use:
out = {}
for k1, d in a.items():
for k2 in d:
out.setdefault(k2, {})[k1] = list(d.values())
Output:
{'a': {(0.0, 0.0): [[25, 29, nan]],
(0.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]],
(2.0, 0.0): [[25, 29, nan], [[25, 26, 29.0]]],
(2.0, 1.5): [[25, 29, nan], [[25, 26, 29.0]]],
(2.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]]},
'b': {(0.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]],
(0.0, 4.0): [[25, 35, 31.0]],
(2.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]],
(2.0, 4.0): [[25, 35, 31.0]]},
'c': {(2.0, 0.0): [[25, 29, nan], [[25, 26, 29.0]]],
(2.0, 1.5): [[25, 29, nan], [[25, 26, 29.0]]],
(5.0, 0.0): [[25, 26, 29.0]],
(5.0, 1.5): [[25, 26, 29.0]]},
'd': {(3.0, 3.0): [[25, 31, 32.0]],
(3.0, 5.0): [[25, 31, 32.0]],
(5.0, 3.0): [[25, 31, 32.0]],
(5.0, 5.0): [[25, 31, 32.0]]},
'e': {(6.0, 1.0): [[25, 28, 30.0]],
(6.0, 3.0): [[25, 28, 30.0]],
(8.0, 1.0): [[25, 28, 30.0]],
(8.0, 3.0): [[25, 28, 30.0]]},
}
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