Python 删除嵌套键和特定值的键

huangapple
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英文:

Python remove nested keys and a certain value's keys

问题

  1. 需要删除任何键的任何级别中具有值为 null 的值
  2. 需要删除名为 key_2.key_c 的任何链接键
    应该得到以下结果块。

原始 JSON

  1. {
  2.   "key_1": {
  3.     "key_a": 111111
  4.   },
  5.   "key_2": {
  6.     "key_a": "value",
  7.     "key_b": null,
  8.     "key_c": {
  9.       "key_c_a": {
  10.         "key_c_b": "value"
  11.       }
  12.     },
  13.     "key_d": [
  14.       {
  15.         "key_c": "value"
  16.       }
  17.     ]
  18.   }
  19. }

结果

  1. {
  2.   "key_1": {
  3.     "key_a": 111111
  4.   },
  5.   "key_2": {
  6.     "key_a": "value",
  7.     "key_d": [
  8.       {
  9.         "key_c": "value"
  10.       }
  11.     ]
  12.   }
  13. }
英文:

1.Need to remove any value in any level of the keys that has a value null
2.Need to remove any chained key that is named key_2.key_c
Should result in the outcome block below.

Original json

  1. {
  2. "key_1": {
  3.     "key_a": 111111},
  4. "key_2": {
  5.     "key_a": "value",
  6.     "key_b": null,
  7.     "key_c": {
  8.         "key_c_a": {
  9.             "key_c_b": "value"}
  10.             },
  11.     "key_d": [{"key_c": "value"}],
  12. }

Outcome

  1. {
  2. "key_1": {
  3.     "key_a": 111111},
  4. "key_2": {
  5.     "key_a": "value",
  6.     "key_d": [{"key_c": "value"}],
  7. }

答案1

得分: 1

你可以通过递归遍历输入的JSON对象并过滤掉不需要的值来实现这一目标：

  1. import json
  3. def filter_json(obj):
  4.     if isinstance(obj, dict):
  5.         new_obj = {}
  6.         for k, v in obj.items():
  7.             if v is None:
  8.                 continue
  9.             if k == "key_2":
  10.                 new_obj[k] = filter_json({k2: v2 for k2, v2 in v.items() if k2 != "key_c"})
  11.             else:
  12.                 new_obj[k] = filter_json(v)
  13.         return new_obj
  14.     elif isinstance obj, list:
  15.         return [filter_json(elem) for elem in obj]
  16.     else:
  17.         return obj

使用：

  1. json_str = '''
  2. {
  3.   "key_1": {
  4.     "key_a": 111111
  5.   },
  6.   "key_2": {
  7.     "key_a": "value",
  8.     "key_b": null,
  9.     "key_c": {
  10.       "key_c_a": {
  11.         "key_c_b": "value"
  12.       }
  13.     },
  14.     "key_d": [
  15.       {
  16.         "key_c": "value"
  17.       }
  18.     ]
  19.   }
  20. }
  21. '''
  23. json_obj = json.loads(json_str)
  24. filtered_obj = filter_json(json_obj)
英文:

You can achieve this by recursively traversing the input JSON object and filtering out the unwanted values:

  1. import json
  3. def filter_json(obj):
  4.     if isinstance(obj, dict):
  5.         new_obj = {}
  6.         for k, v in obj.items():
  7.             if v is None:
  8.                 continue
  9.             if k == "key_2":
  10.                 new_obj[k] = filter_json({k2: v2 for k2, v2 in v.items() if k2 != "key_c"})
  11.             else:
  12.                 new_obj[k] = filter_json(v)
  13.         return new_obj
  14.     elif isinstance(obj, list):
  15.         return [filter_json(elem) for elem in obj]
  16.     else:
  17.         return obj

Usage

  1. json_str = '''
  2. {
  3.   "key_1": {
  4.     "key_a": 111111
  5.   },
  6.   "key_2": {
  7.     "key_a": "value",
  8.     "key_b": null,
  9.     "key_c": {
  10.       "key_c_a": {
  11.         "key_c_b": "value"
  12.       }
  13.     },
  14.     "key_d": [
  15.       {
  16.         "key_c": "value"
  17.       }
  18.     ]
  19.   }
  20. }
  21. '''
  23. json_obj = json.loads(json_str)
  24. filtered_obj = filter_json(json_obj)

huangapple
  • 本文由 发表于 2023年2月18日 00:19:34
  • 转载请务必保留本文链接：https://go.coder-hub.com/75486784.html
  • dictionary
  • json
  • python
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