Python 在groupby和melt之后将数据帧转换为嵌套JSON

我有一个数据框，如下所示：

PayeeID     TransactionID     Res_1    Res_2
1001        aa1001234         OK       OK
1001        aa1001235         OK       NOT OK
1002        aa1002567         NOT OK   NOT OK
1002        aa1002568         NOT OK   OK

df_m = df.melt(id_vars=['PayeeID','Trxn_ID'],value_vars = ['Status_I','Status_II'],\
         var_name='Status_Type',value_name='Final')
df_g = df_m.groupby(['PayeeID','Trxn_ID']).\
       apply(lambda x : dict(zip(x['Status_Type'],x['Final']))).reset_index().\
       rename(columns={0:'Final_Status'})
j = df_g.to_json(orient='records')

[{"PayeeID":1001,"Trxn_ID":"aa1001234","Final_Status":{"Status_I":"OK","Status_II":"NOT OK"}}, 
{"PayeeID":1001,"Trxn_ID":"aa1001338","Final_Status":{"Status_I":"NOT OK","Status_II":"OK"}}, 
{"PayeeID":1002,"Trxn_ID":"aa1002455","Final_Status":{"Status_I":"NOT OK","Status_II":"NOT OK"}}, 
{"PayeeID":1002,"Trxn_ID":"aa1002766","Final_Status":{"Status_I":"OK","Status_II":"OK"}}] 

现在我想将其转换为嵌套的JSON字符串，如下所示：

[{"PayeeID":1001,{"Trxn_ID":"aa1001234","Final_Status":{"Status_I":"OK","Status_II":"NOT 
OK"},"Trxn_ID":"aa1001338","Final_Status":{"Status_I":"NOT OK","Status_II":"OK"}}},{"PayeeID":1002, 
{"Trxn_ID":"aa1002455","Final_Status":{"Status_I":"NOT OK","Status_II":"NOT 
OK"},"Trxn_ID":"aa1002766","Final_Status":{"Status_I":"OK","Status_II":"OK"}}}]

即，对于每个PayeId，都应该有嵌套记录

我的做法如下：

PayeeID     TransactionID     Res_1    Res_2
1001        aa1001234         OK       OK
1001        aa1001235         OK       NOT OK
1002        aa1002567         NOT OK   NOT OK
1002        aa1002568         NOT OK   OK

df_m = df.melt(id_vars=['PayeeID','Trxn_ID'],value_vars = ['Status_I','Status_II'],\
         var_name='Status_Type',value_name='Final')
df_g = df_m.groupby(['PayeeID','Trxn_ID']).\
       apply(lambda x : dict(zip(x['Status_Type'],x['Final']))).reset_index().\
       rename(columns={0:'Final_Status'})
j = df_g.to_json(orient='records')

[{"PayeeID":1001,"Trxn_ID":"aa1001234","Final_Status":{"Status_I":"OK","Status_II":"NOT OK"}}, 
{"PayeeID":1001,"Trxn_ID":"aa1001338","Final_Status":{"Status_I":"NOT OK","Status_II":"OK"}}, 
{"PayeeID":1002,"Trxn_ID":"aa1002455","Final_Status":{"Status_I":"NOT OK","Status_II":"NOT OK"}}, 
{"PayeeID":1002,"Trxn_ID":"aa1002766","Final_Status":{"Status_I":"OK","Status_II":"OK"}}] 

但当我看到j时，我得到如下结果：

PayeeID     TransactionID     Res_1    Res_2
1001        aa1001234         OK       OK
1001        aa1001235         OK       NOT OK
1002        aa1002567         NOT OK   NOT OK
1002        aa1002568         NOT OK   OK

df_m = df.melt(id_vars=['PayeeID','Trxn_ID'],value_vars = ['Status_I','Status_II'],\
         var_name='Status_Type',value_name='Final')
df_g = df_m.groupby(['PayeeID','Trxn_ID']).\
       apply(lambda x : dict(zip(x['Status_Type'],x['Final']))).reset_index().\
       rename(columns={0:'Final_Status'})
j = df_g.to_json(orient='records')

[{"PayeeID":1001,"Trxn_ID":"aa1001234","Final_Status":{"Status_I":"OK","Status_II":"NOT OK"}}, 
{"PayeeID":1001,"Trxn_ID":"aa1001338","Final_Status":{"Status_I":"NOT OK","Status_II":"OK"}}, 
{"PayeeID":1002,"Trxn_ID":"aa1002455","Final_Status":{"Status_I":"NOT OK","Status_II":"NOT OK"}}, 
{"PayeeID":1002,"Trxn_ID":"aa1002766","Final_Status":{"Status_I":"OK","Status_II":"OK"}}] 

---

这里我遗漏了什么。

熊猫不知道您想要的数据格式。您需要首先在数据帧中创建它，然后输出到JSON。下面为每个收款人提供一个条目

df = pd.DataFrame([
    [1001,        "aa1001234",         "OK",       "OK",],
    [1001,        "aa1001235",         "OK",       "NOT OK",],
    [1002,        "aa1002567",         "NOT OK",   "NOT OK",],
    [1002,        "aa1002568",         "NOT OK",   "OK"]], columns=["PayeeID", "TransactionID", "Res_1", "Res_2"])

dfg = df.groupby("PayeeID")["TransactionID",     "Res_1",    "Res_2"].aggregate(lambda x: tuple(x))

dfg.reset_index().to_json(orient='records')


[{
    "PayeeID": 1001,
    "TransactionID": ["aa1001234", "aa1001235"],
    "Res_1": ["OK", "OK"],
    "Res_2": ["OK", "NOT OK"]
}, {
    "PayeeID": 1002,
    "TransactionID": ["aa1002567", "aa1002568"],
    "Res_1": ["NOT OK", "NOT OK"],
    "Res_2": ["NOT OK", "OK"]
}]

可能更好的结构可以是：

df['tx'] = df.apply(lambda x: {x['TransactionID']: {'Res_1':x['Res_1'], 'Res_2':x['Res_2']}}, axis=1)
dfg = df.groupby("PayeeID")["tx"].aggregate(lambda x: tuple(x))
dfg.reset_index().to_json(orient="records")

[{
    "PayeeID": 1001,
    "tx": [{
        "aa1001234": {
            "Res_1": "OK",
            "Res_2": "OK"
        }
    }, {
        "aa1001235": {
            "Res_1": "OK",
            "Res_2": "NOT OK"
        }
    }]
}, {
    "PayeeID": 1002,
    "tx": [{
        "aa1002567": {
            "Res_1": "NOT OK",
            "Res_2": "NOT OK"
        }
    }, {
        "aa1002568": {
            "Res_1": "NOT OK",
            "Res_2": "OK"
        }
    }]
}]

---

您正在查找的JSON无效-如何在受款人ID之后有一个对象？该对象没有键。实际上，我正在尝试为一个

payeid

获取一个条目。您需要首先对值进行分组。看看我的答案。