Python 3.7 递归删除动态 JSON 字典/列表中不在列表中的名称的项目

Question

这是一个文件的示例，我想清除一些 key ，我想知道删除不在我的“保留列表”中的 key 的最佳方法是什么

{"address":"item/address","data":{"set1":{"sub_ref1":0,"sub_ref2":1550620800,"sub_ref3":false,"sub_ref4":false},"set2":[{"sub_ref1":1550534400,"sub_ref2":0,"sub_ref3":0.0,"sub_ref4":0.0,"sub_ref5":"D"}],"set3":[{"sub_ref1":1550534400,"sub_ref2":7,"sub_ref3":5}]},"info":"test","id":190800005945008523}
{"address":"item/address","data":{"set1":{"sub_ref1":0,"sub_ref2":1550620800,"sub_ref3":true,"sub_ref4":false},"set2":[{"sub_ref1":1550534400,"sub_ref2":0,"sub_ref3":0.0,"sub_ref4":0.0,"sub_ref5":"D"}],"set3":[{"sub_ref1":1550534400,"sub_ref2":8,"sub_ref3":6}]},"info":"test","id":190800005945008632}

我想保留的 key 引用如下:

address
data.set1.sub_ref2
data.set1.sub_ref4
data.set3
id

由于不同的文件具有复杂的 JSON 架构，最好的处理方式是什么？

我已经完成了检查键是否与文件中的键匹配并添加列表长度(如果路径中有)的部分。

Answer 1

我认为这可以满足您的需求:

import json

# Loads a JSON document keeping only the given keys
def load_json_with_keys(json_data, keys):
    # Load JSON document
    d = json.loads(json_data)
    # Make recursive call to delete unnecessary keys
    keys = [k.split('.') for k in keys]
    if isinstance(d, dict):
        _delete_extra_keys_rec(d, keys, [])
    if isinstance(d, list):
        for elem in d:
            _delete_extra_keys_rec(elem, keys, [])
    return d

# Recursively deletes unnecessary keys
def _delete_extra_keys_rec(d, keys, current):
    level = len(current) + 1
    # Iterates over list of keys
    # It is important to use list(...) to make a snapshot of the keys
    # before any deletion
    for k in list(d.keys()):
        # Add child key to current key
        current.append(k)
        # Look for a key to maintain matching the current partial key
        for ks in keys:
            if current != ks[:level]: continue
            # Maching key found - this child is kept
            # If the matching key is not complete
            if len(ks) > level:
                # Delete recursively in child
                child = d[k]
                if isinstance(child, dict):
                    _delete_extra_keys_rec(child, keys, current)
                if isinstance(child, list):
                    for elem in child:
                        _delete_extra_keys_rec(elem, keys, current)
            break
        else:
            # No matching key found - this child is deleted
            del d[k]
        # Remove child key
        current.pop()

dict1 = '{"address":"item/address","data":{"set1":{"sub_ref1":0,"sub_ref2":1550620800,"sub_ref3":false,"sub_ref4":false},"set2":[{"sub_ref1":1550534400,"sub_ref2":0,"sub_ref3":0.0,"sub_ref4":0.0,"sub_ref5":"D"}],"set3":[{"sub_ref1":1550534400,"sub_ref2":7,"sub_ref3":5}]},"info":"test","id":190800005945008523}'
dict2 = '{"address":"item/address","data":{"set1":{"sub_ref1":0,"sub_ref2":1550620800,"sub_ref3":true,"sub_ref4":false},"set2":[{"sub_ref1":1550534400,"sub_ref2":0,"sub_ref3":0.0,"sub_ref4":0.0,"sub_ref5":"D"}],"set3":[{"sub_ref1":1550534400,"sub_ref2":8,"sub_ref3":6}]},"info":"test","id":190800005945008632}'
keys = [
    'address',
    'data.set1.sub_ref2',
    'data.set1.sub_ref4',
    'data.set3',
    'id',
    'data.set2.sub_ref2',
]

print(load_json_with_keys(dict1, keys))
# {'address': 'item/address', 'data': {'set1': {'sub_ref2': 1550620800, 'sub_ref4': False}, 'set2': [{'sub_ref2': 0}], 'set3': [{'sub_ref1': 1550534400, 'sub_ref2': 7, 'sub_ref3': 5}]}, 'id': 190800005945008523}
print(load_json_with_keys(dict2, keys))
# {'address': 'item/address', 'data': {'set1': {'sub_ref2': 1550620800, 'sub_ref4': False}, 'set2': [{'sub_ref2': 0}], 'set3': [{'sub_ref1': 1550534400, 'sub_ref2': 8, 'sub_ref3': 6}]}, 'id': 190800005945008632}

有一些可能不受欢迎的极端情况。例如，在你的例子中，如果对象包含一个键为 data.set1 的字典，它不包含 sub_ref2 或 sub_ref4 键，那dict 仍然保留，即使它所属的任何完整键都没有完全匹配。根据具体情况，这可能是所需的行为，也可能不是。