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将损坏的JSON转换为CSV python_Python_Json_List_Csv_For Loop - Fatal编程技术网

将损坏的JSON转换为CSV python

将损坏的JSON转换为CSV python,python,json,list,csv,for-loop,Python,Json,List,Csv,For Loop,我正在尝试获取格式不好的JSON并将其转换为CSV。下面是json的一个示例: [ { "fields": [ { "label": "starttime", "field": "starttime", "type": "integer" }, {

我正在尝试获取格式不好的JSON并将其转换为CSV。下面是json的一个示例:

[
  {
    "fields": [
      {
        "label": "starttime",
        "field": "starttime",
        "type": "integer"
      },
      {
        "label": "endTime",
        "field": "endTime",
        "type": "integer"
      },
      {
        "label": "duration",
        "field": "duration",
        "type": "integer"
      },
      {
        "label": "metrics",
        "field": "metrics",
        "type": "integer"
      },
      {
        "label": "email",
        "field": "email",
        "type": "string"
      },
     ]
    "results": [
      [
        15949132375,
        15949133139,
        763,
        7,
        "newemail@gmail.com"
      ],
      [
        15949132376,
        15949133140,
        764,
        8,
        "newemail1@gmail.com"
      ],
      [
        15949132377,
        15949133141,
        765,
        9,
        "newemail2@gmail.com"
      ],
      [
        15949132378,
        15949133142,
        766,
        10,
        "newemail3@gmail.com"
      ],
      [
        15949132379,
        15949133143,
        767,
        11,
        "newemail4@gmail.com"
      ],
      [
        15949132380,
        15949133144,
        768,
        12,
        "newemail5@gmail.com"
      ],
      [
        15949132381,
        15949133145,
        769,
        13,
        "newemail6@gmail.com"
      ],
      [
        15949132382,
        15949133146,
        770,
        14,
        "newemail7@gmail.com"
      ],
      [
        15949132383,
        15949133147,
        771,
        15,
        "newemail8@gmail.com"
      ],
    ]
  }
]
这是一个示例,因为结果有几千个条目。我需要尝试将上面的“字段”部分的值作为标题,将“结果”放入这些标题下面的行中,因为它们是与“字段”一起使用的数据

正如我所说的,json没有正确导出,使用键以正确的格式获取json不是一个选项。如何将此数据输入CSV?如果不是CSV,如何将其格式化为正确的JSON,使用“字段”标签作为结果中每个对应行的键

我试着通过熊猫来创建一个df。从每个数据集(字段、标签)和结果创建列表,但无法正确连接它们


请提供帮助?

可以读取json文件,并可以使用“re”模块对其进行评估,以分离字段和结果。然后可以将结果写入输出csv文件

建议代码:

import re
import csv
with open("sample.json") as fptr :
    fstr = fptr.read()
    #jsonStr = eval(fstr)
    fields = re.findall ('"fields": (\[.*?\])', fstr, re.M+re.S+re.I)
    fields = eval(fields[0])
    headers = [ f['label'] for f in fields]
    results = re.findall ('"results": (\[.*\]).*}', fstr, re.M+re.S+re.I)
    results = eval(results[0])
    # print (fields, results, headers)

with open ("output.csv", "w", newline="") as fptr :
    recordWrite = csv.writer(fptr, dialect='excel')
    recordWrite.writerow(headers)
    recordWrite.writerows(results)

注意:查看上面的json,可以使用“eval”而不是“re”来评估整个文件内容。唯一需要的更改是在列表结束和“字段”开始后添加逗号。

可以读取json文件,并可以使用“re”模块对其进行评估,以分隔字段和结果。然后可以将结果写入输出csv文件

建议代码:

import re
import csv
with open("sample.json") as fptr :
    fstr = fptr.read()
    #jsonStr = eval(fstr)
    fields = re.findall ('"fields": (\[.*?\])', fstr, re.M+re.S+re.I)
    fields = eval(fields[0])
    headers = [ f['label'] for f in fields]
    results = re.findall ('"results": (\[.*\]).*}', fstr, re.M+re.S+re.I)
    results = eval(results[0])
    # print (fields, results, headers)

with open ("output.csv", "w", newline="") as fptr :
    recordWrite = csv.writer(fptr, dialect='excel')
    recordWrite.writerow(headers)
    recordWrite.writerows(results)

注意:查看上面的json,可以使用“eval”而不是“re”来评估整个文件内容。唯一需要的更改是在列表结束和“字段”开始后添加逗号。

格式设置与此无关。语法和格式无关。语法是