Question

无法使用Python在JSON中生成特定的组和子组

我正在尝试使用Python Pandas生成嵌套的JSON。但是，以某种方式无法弄清楚子分组是如何工作的，或者我可以生成它。不确定如何先打包子组然后再打包。

在python或pandas中是否有任何内置函数，或者任何相关的Python包都可以执行这些操作而无需编写大量代码？

我写的是下面的内容：

j = (df.groupby(['empno', 'work_id'], as_index=False)
         .apply(lambda x: x[['status_id', 'type', 'languageId', 'language',
                             'email', 'game_name', 'experience_level', 'CellNo'
                             ]].to_dict('r'))
         .reset_index()
         .rename(columns={0: 'workPostDetails'})
         .to_json(orient='records'))

    print("JSON::")
    print(j)

样本数据：

empno   work_id     status_id   type    languageId  Language    send_by         recived_by      game_name       experience_level
-----   ----------  ----------  -----   ----------- ---------   ---------       ----------      -------------   -----------------
0017    X123        2101        email   1           All         a@abc.com       b@xyz.com       C++             Expert
0017    X123        2103        phone   1           All         +1 9282828282   +1 9383838383   A++             Intermediate

期望的JSON：

{     
   "empno": "0017",     
   "work_id": "X123",     
   "workPostDetails": {         
                    "workDetails": [             
                                     {                 
                                        "status_id": "2101",                 
                                        "type": "email",                 
                                        "languageId": "1",                 
                                        "language": "All-Read-Write",                 
                                        "send_by": {                     
                                                      "email": "a@abc.com"                 
                                                   },                 
                                        "recived_by": [                     
                                                      {                         
                                                        "email": "b@xyz.com"                     
                                                      }                 
                                                    ],                 
                                        "skillDetails": [                     
                                                          {                         
                                                            "game_name": "EA Sports",                         
                                                            "experience_level": "Expert"                     
                                                          }                
                                                        ]             
                                     },             
                                     {                 
                                        "status_id": "2103",                 
                                        "type": "sms",                 
                                        "languageId": "2",                 
                                        "language": "All-Read",                 
                                        "send_by": {                     
                                                     "CellNo": "+1 9282828282"                 
                                                   },                 
                                        "recived_by": [                     
                                                        {                         
                                                          "CellNo": "+1 9383838383"                     
                                                        }                 
                                                      ],                 
                                        "skillDetails": [                     
                                                          {                         
                                                            "game_name": "Candy Crush",                         
                                                            "experience_level": "Intermediate"                     
                                                          }           
                                                        ]             
                                     }         
                                  ]     
                 } 
}

Answer 1

您可以先通过调整列来准备数据框，然后迭代GroupBy对象。它会给出：

# act on a copy to preserve original data
df2 = df.copy()

# prepare columns
df2[['send_by', 'recived_by']] = df[['send_by', 'recived_by']].apply(
    lambda x: x.apply(lambda y: { 'email': y} if '@' in y else {'CellNo':y}))
df2['skillDetails'] = df2.apply(lambda x: {k: x[k]
                       for k in ('game_name','experience_level')}, axis=1)
df2.drop(columns=['game_name','experience_level'], inplace=True)

# generate the json string
j = pandas.DataFrame(((name[0], name[1],
           val.loc(axis=1)['status_id':].to_dict('r'))
          for name, val in df2.groupby(['empno', 'work_id'])),
         columns = ['empno', 'work_id', 'workPostDetail']).to_json(orient ='records')

Answer 2

这是最终版本-

df2 = df.copy()

# prepare columns
df2[['send_by', 'recived_by']] = df[['send_by', 'recived_by']].apply(
lambda x: x.apply(lambda y: { 'email': y} if '@' in y else {'CellNo':y}))
df2['skillDetails'] = df2.apply(lambda x: {k: x[k]
                   for k in ('game_name','experience_level')}, axis=1)
df2.drop(columns=['game_name','experience_level'], inplace=True)
df2['workDetails'] = df2.apply(lambda x1: {k1: x1[k1]
                   for k1 in 
('status_id','type','languageId','Language','send_by','recived_by','skillDetails')}, axis=1)
df2.drop(columns=['status_id','type','languageId', 'Language', 'send_by', 'recived_by', 'skillDetails'], inplace=True)
# generate the json string

j = p.DataFrame(((name[0], name[1],
       val.loc(axis=1)['workDetails':].to_dict('r'))
      for name, val in df2.groupby(['empno', 'work_id'])),
     columns = ['empno', 'work_id', 'workPostDetail']).to_json(orient ='records')

谢谢大家！

如何使用python生成特定的嵌套JSON

2 个答案: