我的資料來自這樣的資料庫:
| 帳戶 | 姓名 | 地址1 | 狀態 | 壓縮 | 貸款型別 | 到期 |
|---|---|---|---|---|---|---|
| 100 | 山姆 | 5號街 | 紐約 | NY001 | 乙 | 2019年 |
| 100 | 山姆 | 5號街 | 紐約 | NY001 | 噸 | 2020年 |
| 100 | 山姆 | 10號街 | 新澤西州 | NJ001 | 乙 | 2019年 |
| 100 | 山姆 | 10號街 | 新澤西州 | NJ001 | 噸 | 2020年 |
| 101 | 約翰 | 1號街 | 認證機構 | CA001 | 乙 | 2019年 |
| 101 | 約翰 | 1號街 | 認證機構 | CA001 | 噸 | 2020年 |
我需要使用 python pandas 將上述資料轉換為以下 json 格式。我正在嘗試 df.to_json(orient = 'index') 但它沒有創建如下嵌套的甲酸鹽。有什么建議 ?
{
results: [
{
account:100,
Name: Sam,
LoanDetails : [
{
Address1: Street 5,
State : NY,
ZIP: NY0001,
LoanList : [
{
Loantype: E,
expiry: 2012
}
{
Loantype: T,
expiry: 2020
}
]
}
{
Address1: Street 10,
State: NJ,
ZIP: Nj0001,
LoanList: [
{
Loantype: E,
expiry: 2019
}
{
Loantype: T,
expiry: 2020
}
]
}
}
{
account:100,
Name: John,
LoanDetails :
{
Address1: Street 1,
State : CA,
ZIP: CA0001,
LoanList : [
{
Loantype: E,
expiry: 2012
}
{
Loantype: T,
expiry: 2020
}
]
}
}
]
}
uj5u.com熱心網友回復:
我在下面嘗試過,它奏效了:
import pandas as pd
import json
df = pd.DataFrame({'account':['100','100','100','100','101'],
'name':['sam','sam','sam','sam','john'],
'address1':['street 5','street 5','street 10','street 10','street 1'],
'state':['ny','ny','nj','nj','ca'],
'zip':['ny0001','ny0001','nj0001','nj0001','CA001'],
'loantype':['e','t','e','t','e'],
'expiry':[2019,2020,2019,2020,2019]
})
k = df.groupby(['account','name','address1','state']).apply(lambda x:x[['loantype','expiry']].to_dict('records')).reset_index().rename(columns={0:'Loanlist'})#.to_json(orient = 'records')
j = k.groupby(['account','name',]).apply(lambda x:x[['address1','state','Loanlist']].to_dict('records')).reset_index().rename(columns={0:'Loandetails'}).to_json(orient = 'records')
print(j)
uj5u.com熱心網友回復:
這只是有助于獲得您的輸出。可能有替代方法。
a = {}
for i in df.index:
a[i]={}
a[i]["LoanDetails"] = {}
a[i]["LoanList"] = {}
for col in df:
if col in (["Account","Name"]):
a[i][col] = df[col].iloc[i]
if col in (["Address1","State","Zip"]):
a[i]["LoanDetails"][col] = df[col].iloc[i]
if col in (["Loantype","expiry"]):
a[i]["LoanList"][col] = df[col].iloc[i]
b ={}
b["Result"] =[]
for i,v in a.items():
b["Result"].append(v)
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