Python Pandas:使用子计数从数据帧获取嵌套json

Python Pandas:使用子计数从数据帧获取嵌套json,python,json,pandas,Python,Json,Pandas,这是我的源csv文件 provider,calling_num,called_num,result Telia,123,120,Answer Telia,123,121,No Answer Telia,123,122,Answer Telia,456,124,Busy Telia,456,500,Busy Telia,456,480,Answer Telia,456,478,No Answer Telia,789,320,Answer Telia,789,754,Answer GlobalTe

这是我的源csv文件

provider,calling_num,called_num,result

Telia,123,120,Answer
Telia,123,121,No Answer
Telia,123,122,Answer
Telia,456,124,Busy
Telia,456,500,Busy
Telia,456,480,Answer
Telia,456,478,No Answer
Telia,789,320,Answer
Telia,789,754,Answer
GlobalTelecom,512,245,Answer
GlobalTelecom,512,741,Answer
GlobalTelecom,512,841,No Answer
GlobalTelecom,486,952,Busy
GlobalTelecom,486,762,Busy
GlobalTelecom,700,834,Busy
GlobalTelecom,700,295,Answer
GlobalTelecom,700,423,No Answer
GlobalTelecom,700,384,Answer
TotalNetwork,840,756,Answer
TotalNetwork,840,982,Answer
TotalNetwork,840,725,Busy
TotalNetwork,920,953,No Answer
TotalNetwork,920,742,No Answer
TotalNetwork,920,1200,Answer
TotalNetwork,920,175,Busy
TotalNetwork,730,886,Busy
TotalNetwork,730,955,Answer
TotalNetwork,730,772,No Answer
TotalNetwork,730,955,No Answer
TotalNetwork,730,355,Busy
我完成了分组:

df = pd.read_csv("./calls.csv")
df1 = df.groupby(['provider']).size().fillna(0).astype('int64')
df2 = df.groupby(['provider','calling_num']).size().fillna(0).astype('int64')
df3 = df.groupby(['provider','calling_num','result']).size().fillna(0).astype('int64')
我需要获取嵌套的json文件,如下所示:

[     
  {
    name: 'GlobalTelecom',
    children:[
        {
          name:'486',
          children:[ 
             {
               name:'Busy'
               children:[],
               value:'2'
             },
            
          ],
          value:'2'
        },        
        {       
          name:'512',
          children:[ 
             {
               name:'Answer'
               children:[],
               value:'2'
             },
             {
               name:'No Answer'
               children:[],
               value:'1'
             },                         
          ],
          value:'3'
        },        
        {       
          name:'700',
          children:[ 
             {
               name:'Answer'
               children:[],
               value:'2'
             },
             {
               name:'No Answer'
               children:[],
               value:'1'
             },
             {
               name:'Busy'
               children:[],
               value:'1'
             },                                                  
          ],
          value:'4'
        },                
    ],
    value:'9'
  },  
..... and so on for each provider
]
我读了这个好问题

但除了这个问题,我不仅需要得到每个“结果”(最详细)的计数,还需要得到“提供者”的总计数,以及“调用_num”的子计数