嵌套和转义 JSON 有效负载到扁平字典 - python

Question

我正在寻找解决我面临的问题的任何建议。这似乎是一个简单的问题，但几天后试图找到答案 - 我认为它不再是了。 我正在接收以下类似 JSON 格式的数据（StringType），并且需要将其转换为平面键值对字典。这是一个有效载荷示例：

s = """{"status": "active", "name": "{\"first\": \"John\", \"last\": \"Smith\"}", "street_address": "100 \"Y\" Street"}"""

所需的输出应如下所示：

{'status': 'active', 'name_first': 'John', 'name_last': 'Smith', 'street_address': '100 "Y" Street'}

问题是我找不到将原始字符串转换为字典的方法。如果我能做到，展平部分工作得很好。

import json
import collections
import ast


#############################################################
# Flatten complex structure into a flat dictionary
#############################################################
def flatten_dictionary(dictionary, parent_key=False, separator='_', value_to_str=True):
    """
    Turn a nested complex json into a flattened dictionary
    :param dictionary: The dictionary to flatten
    :param parent_key: The string to prepend to dictionary's keys
    :param separator: The string used to separate flattened keys
    :param value_to_str: Force all returned values to string type
    :return: A flattened dictionary
    """

    items = []

    for key, value in dictionary.items():
        new_key = str(parent_key) + separator + key if parent_key else key

        try:
            value = json.loads(value)
        except BaseException:
            value = value

        if isinstance(value, collections.MutableMapping):
            if not value.items():
                items.append((new_key,None))
            else:
                items.extend(flatten_dictionary(value, new_key, separator).items())
        elif isinstance(value, list):
            if len(value):
                for k, v in enumerate(value):
                    items.extend(flatten_dictionary({str(k): (str(v) if value_to_str else v)}, new_key).items())
            else:
                items.append((new_key,None))
        else:
            items.append((new_key, (str(value) if value_to_str else value)))

    return dict(items)


# Data sample; sting and dictionary
s = """{"status": "active", "name": "{\"first\": \"John\", \"last\": \"Smith\"}", "street_address": "100 \"Y\" Street"}"""
d =    {"status": "active", "name": "{\"first\": \"John\", \"last\": \"Smith\"}", "street_address": "100 \"Y\" Street"}

# Works for dictionary type
print(flatten_dictionary(d))

# Doesn't work for string type, for any of the below methods
e = eval(s)
# a = ast.literal_eval(s)
# j = json.loads(s)

Answer 1

试试：

import json
import re


def jsonify(s):
  s = s.replace('"{','{').replace('}"','}')
  s = re.sub(r'street_address":\s+"(.+)"(.+)"(.+)"', r'street_address": "\1\2\3"',s)
  return json.loads(s)

如果您必须在 Y 周围保留引号，请尝试：

def jsonify(s):
  s = s.replace('"{','{').replace('}"','}')
  search = re.search(r'street_address":\s+"(.+)"(.+)"(.+)"',s)
  if search:
    s = re.sub(r'street_address":\s+"(.+)"(.+)"(.+)"', r'street_address": "\1\2\3"',s)
    dict_version = json.loads(s)
    dict_version['street_address'] = dict_version['street_address'].replace(search.group(2),'"'+search.group(2)+'"')
  return dict_version

更普遍的尝试：

def jsonify(s):
  pattern = r'(?<=[,}])\s*"(.[^\{\}:,]+?)":\s+"([^\{\}:,]+?)"([^\{\}:,]+?)"([^\{\}:,]+?)"([,\}])'
  s = s.replace('"{','{').replace('}"','}')
  search = re.search(pattern,s)
  matches = []
  if search:
    matches = re.findall(pattern,s)
    s = re.sub(pattern, r'"\1": "\2\3\4"\5',s)
  dict_version = json.loads(s)
  for match in matches:
    dict_version[match[0]] = dict_version[match[0]].replace(match[2],'"'+match[2]+'"')
  return dict_version