递归遍历 Python 中的参差不齐的 JSON 层次结构 3 以执行部分​​叶节点删除

Recursively Traversing Ragged JSON Hierarchy in Python 3 to perform a partial leaf node deletion

所以，我有一个 JSON 块，看起来有点像这样：

[
{
    "children": [
        {
            "address": "123 Main Street",
            "class": "blarg",
            "children": [
                {
                    "children": [
                        {
                            "children": [
                                {
                                    "y_type": "string",
                                    "x_type": "002",
                                    "002_id": 2222,
                                    "updated_at": "2018-03-29T13:47:42.986Z",
                                    "field020": "AAA",
                                    "field030": "DDD"
                                },
                                {
                                    "y_type": "bool",
                                    "x_type": "007",
                                    "007_id": 2222,
                                    "updated_at": "2018-03-29T13:47:42.986Z",
                                    "field1": True,
                                    "field2": True
                                }
                            ],
                            "create_at": "2018-03-29T13:45:20.875Z",
                            "x_id": "3e0e1b44-ac0d-4bf7-985e-11d74b8be323",
                            "junk_field": {},
                            "x_type": "000",
                            "timezone": "America/New_York",
                            "update_at": "2018-03-29T13:45:20.875Z"
                        },
                        {
                            "sibling": [1,2,3]
                        }
                    ]
                }
            ]
        }
    ]
}]

我需要做的是找到 "children" 个属于 x_type“007”的叶节点，并删除与该数据块关联的字段 1 条目。我在尝试隔离仅与正确类型的叶节点（子节点，而不是兄弟节点）关联的整个字典时遇到问题，以便我可以检查它是否正确 x_type 并执行删除操作。

我不确定要从我拼凑的递归函数中传递什么样的值 to/back。我以前从未在 Python 中做过递归，更不用说针对参差不齐的层次结构 JSON，所以我可以使用一些 help/guidance 来了解 use/google 的方法。如果你能给我任何帮助，让我朝着正确的方向前进，我将不胜感激！！

您可以通过递归使用字典解包：

def d_filter(d):
  return {**({a:b for a, b in d.items() if d.get('x_type') != '007' or a != 'field1'}), \
   'children':list(map(d_filter, d.get('children', [])))} 

new_data = list(map(d_filter, data))

import json
print(json.dumps(new_data, indent=4))

输出：

[
  {
    "children": [
        {
            "address": "123 Main Street",
            "class": "blarg",
            "children": [
                {
                    "children": [
                        {
                            "children": [
                                {
                                    "y_type": "string",
                                    "x_type": "002",
                                    "002_id": 2222,
                                    "updated_at": "2018-03-29T13:47:42.986Z",
                                    "field020": "AAA",
                                    "field030": "DDD",
                                    "children": []
                                },
                                {
                                    "y_type": "bool",
                                    "x_type": "007",
                                    "007_id": 2222,
                                    "updated_at": "2018-03-29T13:47:42.986Z",
                                    "field2": true,
                                    "children": []
                                }
                            ],
                            "create_at": "2018-03-29T13:45:20.875Z",
                            "x_id": "3e0e1b44-ac0d-4bf7-985e-11d74b8be323",
                            "junk_field": {},
                            "x_type": "000",
                            "timezone": "America/New_York",
                            "update_at": "2018-03-29T13:45:20.875Z"
                        },
                        {
                            "sibling": [
                                1,
                                2,
                                3
                            ],
                            "children": []
                        }
                    ]
                }
            ]
        }
     ]
   }
]