我有一个由csv文件制成的Pandas数据框,我想将其转换为自定义嵌套JSON
df = pd.read_csv("/content/analog.csv", sep=",")
数据帧就像:
datetime key1 key2 key3 col1 col2 col3 ... col100
2020-08-27T02:28:00Z WECP001 27001 con0000 1001 1002 1003 ... 1005
2020-08-27T02:28:01Z WECP001 27001 con0000 11 22 33 ... 99
我想将此df转换为嵌套的json,例如:
[
{
"datetime":"2020-08-27T02:28:00Z"
"key1:"WECP001"
"key2:"27001"
"key3":"con0000"
"DATA":{
"col1":1001,
"col2":1002,
"col3":1003,
...
"col100":1005
}
},
{
"datetime":"2020-08-27T02:28:01Z"
"key1:"WECP001"
"key2:"27001"
"key3":"con0000"
"DATA":{
"col1":11,
"col2":22,
"col3":33,
...
"col100":99
}
}
]
这是我讨厌的代码,但是我得到错误的结果
cols = df.columns.difference(['datetime','key1','key2','key3'])
j = (df.groupby(['datetime','key1','key2','key3'])[cols]
.apply(lambda x: x.to_dict('r'))
.reset_index(name='DATA')
.to_json(orient='records'))
print (j)
输出就像:
{
"datetime":"2020-08-27T02:28:00Z"
"key1:"WECP001"
"key2:"27001"
"key3":"con0000"
"DATA":[{
"col1":1001,
"col2":1002,
"col3":1003,
...
"col100
}]
我的代码的问题是“数据”部分成为列表而不是字典 谢谢你的帮助
答案 0 :(得分:1)
尝试一下
df2 = df.iloc[:, :4]
df2['Data'] = df[[f'col{i}' for i in range(1, 4)]].to_dict(orient='records')
print(df2.to_json(orient='records'))
输出:
[
{
"datetime": "2020-08-27T02:28:00Z",
"key1": "WECP001",
"key2": 27001,
"key3": "con0000",
"Data": {
"col1": 1001,
"col2": 1002,
"col3": 1003
}
},
{
"datetime": "2020-08-27T02:28:01Z",
"key1": "WECP001",
"key2": 27001,
"key3": "con0000",
"Data": {
"col1": 11,
"col2": 22,
"col3": 33
}
}
]
答案 1 :(得分:0)
import pandas as pd
df = pd.read_csv("***abc.csv", sep=",")
df2 = df.iloc[:, :4]
df2['DATA'] = df.iloc[:, 5:].to_dict(orient='records')
print(df2.to_json(orient='records', lines=True))
结果:
{
"datetime":"2020-08-27T02:28:00Z"
"key1:"WECP001"
"key2:"27001"
"key3":"con0000"
"DATA":{
"col1":1001,
"col2":1002,
"col3":1003,
...
"col100":
}
}
{....}
{....}