0

我正在尝试通过使用熊猫中的内置函数和堆栈上可用的一些解决方案将多嵌套 JSON 转换为平面，能够正常化直到第一级或扁平化创建不需要的结果或修改它们而不给出期望输出，任何见解，欢迎。

示例 JSON：

{
  "Records": [
    {
      "Name": "Student1",
      "Result": "Pass",
      "Marks": [
        {
          "Sub1": "50",
          "Sub2": "40",
          "YOP": [
            {
              "prim": "2010",
              "sch": "abc"
            },
            {
              "prim": "2010",
              "sch": "abc"
            }
          ]
        }
      ]
    },
    {
      "Name": "Stu2",
      "Result": "Pass",
      "Marks": [
        {
          "Sub1": "33",
          "Sub2": "33",
          "YOP": [
            {
              "prim": "2010",
              "sch": "def"
            },
            {
              "high": "2010",
              "sch": "abc"
            }
          ]
        }
      ]
    }
  ]
}

当前代码

from itertools import chain, starmap
import json
from itertools import islice
from pandas.io.json import json_normalize



from collections import MutableMapping
crumbs = True

def flatten(dictionary, parent_key=False, separator='.'):
    """
    Turn a nested dictionary 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
    :return: A flattened dictionary
    """

    items = []
    for key, value in dictionary.items():
        if crumbs: print('checking:',key)
        new_key = str(parent_key) + separator + key if parent_key else key
        if isinstance(value, MutableMapping):
            if crumbs: print(new_key,': dict found')
            if not value.items():
                if crumbs: print('Adding key-value pair:',new_key,None)
                items.append((new_key,None))
            else:
                items.extend(flatten(value, new_key, separator).items())
        elif isinstance(value, list):
            if crumbs: print(new_key,': list found')
            if len(value):
                for k, v in enumerate(value):
                    items.extend(flatten({str(k): v}, new_key).items())
            else:
                if crumbs: print('Adding key-value pair:',new_key,None)
                items.append((new_key,None))
        else:
            if crumbs: print('Adding key-value pair:',new_key,value)
            items.append((new_key, value))
    return dict(items)

    
def main():
  with open("aaa.json", "r") as f:
    data = json.loads(f.read())
  print(type(data))
  flat = flatten(data)
  print(flat)


if __name__ == '__main__':
    main()

输出

{
  "Records.0.Name": "Student1",
  "Records.0.Result": "Pass",
  "Records.0.Marks.0.Sub1": "50",
  "Records.0.Marks.0.Sub2": "40",
  "Records.0.Marks.0.YOP.0.prim": "2010",
  "Records.0.Marks.0.YOP.0.sch": "abc",
  "Records.0.Marks.0.YOP.1.high": "2012",
  "Records.0.Marks.0.YOP.1.sch": "abc",
  "Records.1.Name": "Stu2",
  "Records.1.Result": "Pass",
  "Records.1.Marks.0.Sub1": "33",
  "Records.1.Marks.0.Sub2": "33",
  "Records.1.Marks.0.YOP.0.prim": "210",
  "Records.1.Marks.0.YOP.0.sch": "def",
  "Records.1.Marks.0.YOP.1.high": "999",
  "Records.1.Marks.0.YOP.1.sch": "abc"
}

使用此代码，知道如何删除数字并将其分开

结束预期

{
  "Records.Name": "Student1",
  "Records.Result": "Pass",
  "Records.Marks.Sub1": "50",
  "Records.Marks.Sub2": "40",
  "Records.Marks.YOP.prim": "2010",
  "Records.Marks.YOP.sch": "abc",
  "Records.Marks.YOP.high": "2012",
  "Records.Marks.YOP.sch": "abc",
},

{
  "Records.Name": "Stu2",
  "Records.Result": "Pass",
  "Records.Marks.Sub1": "33",
  "Records.Marks.Sub2": "33",
  "Records.Marks.YOP.prim": "210",
  "Records.Marks.YOP.sch": "def",
  "Records.Marks.YOP.high": "999",
  "Records.Marks.YOP.sch": "abc"
}
4

1 回答 1

0 
from flatten_json import flatten

records = flatten(json[0])

输入：

json = [{
  "Records": [
    {
      "Name": "Student1",
      "Result": "Pass",
      "Marks": [
        {
          "Sub1": "50",
          "Sub2": "40",
          "YOP": [
            {
              "prim": "2010",
              "sch": "abc"
            },
            {
              "prim": "2010",
              "sch": "abc"
            }
          ]
        }
      ]
    },
    {
      "Name": "Stu2",
      "Result": "Pass",
      "Marks": [
        {
          "Sub1": "33",
          "Sub2": "33",
          "YOP": [
            {
              "prim": "2010",
              "sch": "def"
            },
            {
              "high": "2010",
              "sch": "abc"
            }
          ]
        }
      ]
    }
  ]
}]

输出：

{'Records_0_Name': 'Student1',
 'Records_0_Result': 'Pass',
 'Records_0_Marks_0_Sub1': '50',
 'Records_0_Marks_0_Sub2': '40',
 'Records_0_Marks_0_YOP_0_prim': '2010',
 'Records_0_Marks_0_YOP_0_sch': 'abc',
 'Records_0_Marks_0_YOP_1_prim': '2010',
 'Records_0_Marks_0_YOP_1_sch': 'abc',
 'Records_1_Name': 'Stu2',
 'Records_1_Result': 'Pass',
 'Records_1_Marks_0_Sub1': '33',
 'Records_1_Marks_0_Sub2': '33',
 'Records_1_Marks_0_YOP_0_prim': '2010',
 'Records_1_Marks_0_YOP_0_sch': 'def',
 'Records_1_Marks_0_YOP_1_high': '2010',
 'Records_1_Marks_0_YOP_1_sch': 'abc'}

更新：

您正在寻找的结果：

records = [flatten(record, “.”) for record in json[0]['Records']]

输出：

[{'Name': 'Student1',
  'Result': 'Pass',
  'Marks_0_Sub1': '50',
  'Marks_0_Sub2': '40',
  'Marks_0_YOP_0_prim': '2010',
  'Marks_0_YOP_0_sch': 'abc',
  'Marks_0_YOP_1_prim': '2010',
  'Marks_0_YOP_1_sch': 'abc'},
 {'Name': 'Stu2',
  'Result': 'Pass',
  'Marks_0_Sub1': '33',
  'Marks_0_Sub2': '33',
  'Marks_0_YOP_0_prim': '2010',
  'Marks_0_YOP_0_sch': 'def',
  'Marks_0_YOP_1_high': '2010',
  'Marks_0_YOP_1_sch': 'abc'}]
于 2022-01-15T18:49:17.043 回答