重新组织嵌套的 `dict`

huangapple
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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：

  1. a = {
  2.  (0.0, 0.0): {'a': [25, 29, nan]},
  3.  (0.0, 2.0): {'a': [25, 29, nan], 'b': [25, 35, 31.0]},
  4.  (0.0, 4.0): {'b': [25, 35, 31.0]},
  5.  (2.0, 0.0): {'a': [25, 29, nan], 'c': [25, 26, 29.0]},
  6.  (2.0, 1.5): {'a': [25, 29, nan], 'c': [25, 26, 29.0]},
  7.  (2.0, 2.0): {'a': [25, 29, nan], 'b': [25, 35, 31.0]},
  8.  (2.0, 4.0): {'b': [25, 35, 31.0]},
  9.  (3.0, 3.0): {'d': [25, 31, 32.0]},
  10.  (3.0, 5.0): {'d': [25, 31, 32.0]},
  11.  (5.0, 0.0): {'c': [25, 26, 29.0]},
  12.  (5.0, 1.5): {'c': [25, 26, 29.0]},
  13.  (5.0, 3.0): {'d': [25, 31, 32.0]},
  14.  (5.0, 5.0): {'d': [25, 31, 32.0]},
  15.  (6.0, 1.0): {'e': [25, 28, 30.0]},
  16.  (6.0, 3.0): {'e': [25, 28, 30.0]},
  17.  (8.0, 1.0): {'e': [25, 28, 30.0]},
  18.  (8.0, 3.0): {'e': [25, 28, 30.0]}
  19. }

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：

  1. {'a': {(0.0, 0.0): [[25, 29, nan]],
  2.        (0.0, 2.0): [[25, 29, nan], [25, 35, 31.0]],
  3.        (2.0, 0.0): [[25, 29, nan], [25, 26, 29.0]],
  4.        (2.0, 1.5): [[25, 29, nan], [25, 26, 29.0]],
  5.        (2.0, 2.0): [[25, 29, nan], [25, 35, 31.0]]},
  6.  'b': {(0.0, 2.0): [[25, 29, nan], [25, 35, 31.0]],
  7.        (0.0, 4.0): [[25, 35, 31.0]],
  8.        (2.0, 2.0): [[25, 29, nan], [25, 35, 31.0]],
  9.        (2.0, 4.0): [[25, 35, 31.0]]},
  10.  'c': {(2.0, 0.0): [[25, 29, nan], [25, 26, 29.0]],
  11.        (2.0, 1.5): [[25, 29, nan], [25, 26, 29.0]],
  12.        (5.0, 0.0): [[25, 26, 29.0]],
  13.        (5.0, 1.5): [[25, 26, 29.0]]},
  14.  'd': {(3.0, 3.0): [[25, 31, 32.0]],
  15.        (3.0, 5.0): [[25, 31, 32.0]],
  16.        (5.0, 3.0): [[25, 31, 32.0]],
  17.        (5.0, 5.0): [[25, 31, 32.0]]},
  18.  'e': {(6.0, 1.0): [[25, 28, 30.0]],
  19.        (6.0, 3.0): [[25, 28, 30.0]],
  20.        (8.0, 1.0): [[25, 28, 30.0]],
  21.        (8.0, 3.0): [[25, 28, 30.0]]}
  22. }

直觉告诉我使用pd.DataFrame和.groupby() [并剔除NaN单元格] 是正确的方法...

  1. df = pd.DataFrame(dict_vertices)
  2. print(df.head(2))
  3.              0.0               2.0                    ...  8.0       6.0
  4.              0.0               0.0               1.5  ...  1.0  3.0  3.0
  5. a  [25, 29, nan]     [25, 29, nan]     [25, 29, nan]  ...  NaN  NaN  NaN
  6. c            NaN  [[25, 26, 29.0]]  [[25, 26, 29.0]]  ...  NaN  NaN  NaN
  8. [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:

  1. a = {
  2.  (0.0, 0.0): {'a': [25, 29, nan]},
  3.  (0.0, 2.0): {'a': [25, 29, nan], 'b': [25, 35, 31.0]},
  4.  (0.0, 4.0): {'b': [25, 35, 31.0]},
  5.  (2.0, 0.0): {'a': [25, 29, nan], 'c': [25, 26, 29.0]},
  6.  (2.0, 1.5): {'a': [25, 29, nan], 'c': [25, 26, 29.0]},
  7.  (2.0, 2.0): {'a': [25, 29, nan], 'b': [25, 35, 31.0]},
  8.  (2.0, 4.0): {'b': [25, 35, 31.0]},
  9.  (3.0, 3.0): {'d': [25, 31, 32.0]},
  10.  (3.0, 5.0): {'d': [25, 31, 32.0]},
  11.  (5.0, 0.0): {'c': [25, 26, 29.0]},
  12.  (5.0, 1.5): {'c': [25, 26, 29.0]},
  13.  (5.0, 3.0): {'d': [25, 31, 32.0]},
  14.  (5.0, 5.0): {'d': [25, 31, 32.0]},
  15.  (6.0, 1.0): {'e': [25, 28, 30.0]},
  16.  (6.0, 3.0): {'e': [25, 28, 30.0]},
  17.  (8.0, 1.0): {'e': [25, 28, 30.0]},
  18.  (8.0, 3.0): {'e': [25, 28, 30.0]}
  19. }

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:

  1. {'a': {(0.0, 0.0): [[25, 29, nan]],
  2.        (0.0, 2.0): [[25, 29, nan], [25, 35, 31.0]],
  3.        (2.0, 0.0): [[25, 29, nan], [25, 26, 29.0]],
  4.        (2.0, 1.5): [[25, 29, nan], [25, 26, 29.0]],
  5.        (2.0, 2.0): [[25, 29, nan], [25, 35, 31.0]]},
  6.  'b': {(0.0, 2.0): [[25, 29, nan], [25, 35, 31.0]],
  7.        (0.0, 4.0): [[25, 35, 31.0]],
  8.        (2.0, 2.0): [[25, 29, nan], [25, 35, 31.0]],
  9.        (2.0, 4.0): [[25, 35, 31.0]]},
  10.  'c': {(2.0, 0.0): [[25, 29, nan], [25, 26, 29.0]],
  11.        (2.0, 1.5): [[25, 29, nan], [25, 26, 29.0]],
  12.        (5.0, 0.0): [[25, 26, 29.0]],
  13.        (5.0, 1.5): [[25, 26, 29.0]]},
  14.  'd': {(3.0, 3.0): [[25, 31, 32.0]],
  15.        (3.0, 5.0): [[25, 31, 32.0]],
  16.        (5.0, 3.0): [[25, 31, 32.0]],
  17.        (5.0, 5.0): [[25, 31, 32.0]]},
  18.  'e': {(6.0, 1.0): [[25, 28, 30.0]],
  19.        (6.0, 3.0): [[25, 28, 30.0]],
  20.        (8.0, 1.0): [[25, 28, 30.0]],
  21.        (8.0, 3.0): [[25, 28, 30.0]]}
  22. }

Intuition tells me pd.DataFrame with a .groupby() [and cull the NaN cells] would be the way to go...

  1. df = pd.DataFrame(dict_vertices)
  2. print(df.head(2))
  3.              0.0               2.0                    ...  8.0       6.0
  4.              0.0               0.0               1.5  ...  1.0  3.0  3.0
  5. a  [25, 29, nan]     [25, 29, nan]     [25, 29, nan]  ...  NaN  NaN  NaN
  6. c            NaN  [[25, 26, 29.0]]  [[25, 26, 29.0]]  ...  NaN  NaN  NaN
  8. [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：

  1. out = {}
  3. for k1, d in a.items():
  4.     for k2 in d:
  5.         out.setdefault(k2, {})[k1] = list(d.values())

输出如下：

  1. {
  2.     'a': {
  3.         (0.0, 0.0): [[25, 29, nan]],
  4.         (0.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]],
  5.         (2.0, 0.0): [[25, 29, nan], [[25, 26, 29.0]]],
  6.         (2.0, 1.5): [[25, 29, nan], [[25, 26, 29.0]]],
  7.         (2.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]]
  8.     },
  9.     'b': {
  10.         (0.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]],
  11.         (0.0, 4.0): [[25, 35, 31.0]],
  12.         (2.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]],
  13.         (2.0, 4.0): [[25, 35, 31.0]]
  14.     },
  15.     'c': {
  16.         (2.0, 0.0): [[25, 29, nan], [[25, 26, 29.0]]],
  17.         (2.0, 1.5): [[25, 29, nan], [[25, 26, 29.0]]],
  18.         (5.0, 0.0): [[25, 26, 29.0]],
  19.         (5.0, 1.5): [[25, 26, 29.0]]
  20.     },
  21.     'd': {
  22.         (3.0, 3.0): [[25, 31, 32.0]],
  23.         (3.0, 5.0): [[25, 31, 32.0]],
  24.         (5.0, 3.0): [[25, 31, 32.0]],
  25.         (5.0, 5.0): [[25, 31, 32.0]]
  26.     },
  27.     'e': {
  28.         (6.0, 1.0): [[25, 28, 30.0]],
  29.         (6.0, 3.0): [[25, 28, 30.0]],
  30.         (8.0, 1.0): [[25, 28, 30.0]],
  31.         (8.0, 3.0): [[25, 28, 30.0]]
  32.     }
  33. }
英文:

You can use:

  1. out = {}
  3. for k1, d in a.items():
  4.     for k2 in d:
  5.         out.setdefault(k2, {})[k1] = list(d.values())

Output:

  1. {'a': {(0.0, 0.0): [[25, 29, nan]],
  2.        (0.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]],
  3.        (2.0, 0.0): [[25, 29, nan], [[25, 26, 29.0]]],
  4.        (2.0, 1.5): [[25, 29, nan], [[25, 26, 29.0]]],
  5.        (2.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]]},
  6.  'b': {(0.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]],
  7.        (0.0, 4.0): [[25, 35, 31.0]],
  8.        (2.0, 2.0): [[25, 29, nan], [[25, 35, 31.0]]],
  9.        (2.0, 4.0): [[25, 35, 31.0]]},
  10.  'c': {(2.0, 0.0): [[25, 29, nan], [[25, 26, 29.0]]],
  11.        (2.0, 1.5): [[25, 29, nan], [[25, 26, 29.0]]],
  12.        (5.0, 0.0): [[25, 26, 29.0]],
  13.        (5.0, 1.5): [[25, 26, 29.0]]},
  14.  'd': {(3.0, 3.0): [[25, 31, 32.0]],
  15.        (3.0, 5.0): [[25, 31, 32.0]], 
  16.        (5.0, 3.0): [[25, 31, 32.0]],
  17.        (5.0, 5.0): [[25, 31, 32.0]]},
  18.  'e': {(6.0, 1.0): [[25, 28, 30.0]],
  19.        (6.0, 3.0): [[25, 28, 30.0]],
  20.        (8.0, 1.0): [[25, 28, 30.0]],
  21.        (8.0, 3.0): [[25, 28, 30.0]]},
  22. }

huangapple
  • 本文由 发表于 2023年6月2日 00:21:40
  • 转载请务必保留本文链接：https://go.coder-hub.com/76383903.html
  • dictionary
  • group-by
  • pandas
  • python
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