在Python中从其他列表/字典的嵌套字典中获取键层次结构

w8rqjzmb  于 2023-01-12  发布在  Python
关注(0)|答案(3)|浏览(137)

我有这样一个输入dict：

  1. input={'boo': 'its', 'soo': 'your', 'roo': 'choice', 'qoo': 'this', 'fizz': 'is', 'buzz': 'very', 'yoyo': 'rambling', 'wazzw': 'lorem', 'bnn': 'ipsum', 'cc': [{'boo': 'fill', 'soo': 'ing', 'roo': 'in', 'qoo': 'the', 'fizz': 'words', 'buzz': 'here', 'yoyo': 'we', 'wazzw': 'go', 'nummm': 2, 'bsdfff': 3, 'hgdjgkk': 4, 'opu': 1, 'mnb': True}, {'boo': 'again', 'soo': 'loop', 'roo': 'de', 'qoo': 'loop', 'fizz': 'wowzers', 'buzz': 'try', 'yoyo': 'again', 'wazzw': 'how', 'nummm': 1, 'bsdfff': 7, 'hgdjgkk': 0, 'opu': 1, 'mnb': True}], 'soos': ['ya'], 'tyu': 'doin', 'dddd3': 'today'}

使用python内置库如何获得每个键的层次结构（点分隔）。即：

  1. expected_output=['boo','soo','roo','qoo','fizz','buzz','yoyo','wazzw','bnn','cc','cc.boo','cc.soo','cc.roo','cc.qoo','cc.fizz','cc.buzz','cc.yoyo','cc.wazzw','cc.nummm','cc.bsdfff','cc.hgdjgkk','cc.opu','cc.mnb','soos','tyu','dddd3']

第一次尝试未处理列表：

  1. def getKeys(object, prev_key = None, keys = []):
  2. if type(object) != type({}):
  3.     keys.append(prev_key)
  4.     return keys
  5. new_keys = []
  6. for k, v in object.items():
  7.     if prev_key != None:
  8.         new_key = "{}.{}".format(prev_key, k)
  9.     else:
  10.         new_key = k
  11.     new_keys.extend(getKeys(v, new_key, []))
  12. return new_keys
python

3条答案

按热度按时间
bvpmtnay

bvpmtnay1#

使用递归生成器：

  1. def hierarchy(d, prefix=None):
  2.     if isinstance(d, dict):
  3.         for k, v in d.items():
  4.             prefix2 = f'{prefix}.{k}' if prefix else k
  5.             yield prefix2
  6.             if isinstance(v, list):
  7.                 seen = set()
  8.                 for x in v:
  9.                     if isinstance(x, dict):
  10.                         yield from hierarchy({k: v for k, v in x.items()
  11.                                               if k not in seen},
  12.                                              prefix=prefix2)
  13.                         seen.update(x.keys())
  14.                     else:
  15.                         yield from hierarchy(x, prefix=prefix2)
  16.             elif isinstance(v, dict):
  17.                 yield from hierarchy(v, prefix=prefix2)
  18.                 
  19. out = list(hierarchy(inpt))
  21. # validation
  22. assert out == expected_output

输出：

  1. ['boo', 'soo', 'roo', 'qoo', 'fizz', 'buzz', 'yoyo', 'wazzw', 'bnn',
  2.  'cc', 'cc.boo', 'cc.soo', 'cc.roo', 'cc.qoo', 'cc.fizz', 'cc.buzz',
  3.  'cc.yoyo', 'cc.wazzw', 'cc.nummm', 'cc.bsdfff', 'cc.hgdjgkk', 'cc.opu', 'cc.mnb',
  4.  'soos', 'tyu', 'dddd3']

不同示例：

  1. list(hierarchy({'l1': {'l2': {'l3': 'test', 'l4': [['abc'], {'l5': 'def'}]}}}))
  2. # ['l1', 'l1.l2', 'l1.l2.l3', 'l1.l2.l4', 'l1.l2.l4.l5']
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u3r8eeie

u3r8eeie2#

莫兹韦回答的修正https://www.mycompiler.io/view/6LB7k4TVOuj

  1. # Includes $ for root node, and [] where access is through an array
  3. def hierarchy(struct, path=None):
  4.     if isinstance(struct, dict):
  5.         path = path if path else '$'
  6.         return set(
  7.             child_path
  8.                 for key, obj   in struct.items()
  9.                 for child_path in hierarchy(obj, f'{path}.{key}')
  10.         ).union(
  11.             [path]
  12.         )
  13.     elif isinstance(struct, list):
  14.         path = f'{path}[]' if path else '$[]'
  15.         return set(
  16.             child_path
  17.                 for obj        in struct
  18.                 for child_path in hierarchy(obj, path)
  19.         ).union(
  20.             [path]
  21.         )
  22.     else:
  23.         return [path]

或者...

  1. from itertools import chain
  3. # Excludes those $ and [] markers
  5. def hierarchy2(d):
  6.     if isinstance(d, dict):
  7.         return set(
  8.             f'{k}.{x}' if x else k
  9.                 for k,v in d.items()
  10.                 for x in chain([''], hierarchy2(v))
  11.         )
  12.     elif isinstance(d, list):
  13.         return set(
  14.             v
  15.                 for l in d
  16.                 for v in hierarchy2(l)
  17.                     if v
  18.         )
  19.     else:
  20.         return set()
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db2dz4w8

db2dz4w83#

要处理子列表，您可以迭代地检查每个子项是否是一个字典，如果是，则递归地将子字典的键路径追加到当前键：

  1. def get_keys(d):
  2.     keys = []
  3.     for key, value in d.items():
  4.         if isinstance(value, list):
  5.             for obj in value:
  6.                 if isinstance(obj, dict):
  7.                     for path in get_keys(obj):
  8.                         keys.append(f'{key}.{path}')
  9.                 else:
  10.                     keys.append(key)
  11.         else:
  12.             keys.append(key)
  13.     return keys

因此，对于给定示例input，get_keys(input)将返回：

  1. ['boo', 'soo', 'roo', 'qoo', 'fizz', 'buzz', 'yoyo', 'wazzw', 'bnn', 'cc.boo', 'cc.soo', 'cc.roo', 'cc.qoo', 'cc.fizz', 'cc.buzz', 'cc.yoyo', 'cc.wazzw', 'cc.nummm', 'cc.bsdfff', 'cc.hgdjgkk', 'cc.opu', 'cc.mnb', 'cc.boo', 'cc.soo', 'cc.roo', 'cc.qoo', 'cc.fizz', 'cc.buzz', 'cc.yoyo', 'cc.wazzw', 'cc.nummm', 'cc.bsdfff', 'cc.hgdjgkk', 'cc.opu', 'cc.mnb', 'soos', 'tyu', 'dddd3']

演示：https://replit.com/@blhsing/OpenGoldenIntegrationtesting

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