将 Pandas 数据框转换为嵌套 JSON
问题描述
我是 Python 和 Pandas 的新手.我正在尝试将 Pandas Dataframe 转换为嵌套的 JSON..to_json() 函数没有给我足够的灵活性来实现我的目标.
I am new to Python and Pandas. I am trying to convert a Pandas Dataframe to a nested JSON. The function .to_json() doens't give me enough flexibility for my aim.
以下是数据框的一些数据点(以 csv 格式，逗号分隔):
Here are some data points of the dataframe (in csv, comma separated):
,ID,Location,Country,Latitude,Longitude,timestamp,tide
0,1,BREST,FRA,48.383,-4.495,1807-01-01,6905.0
1,1,BREST,FRA,48.383,-4.495,1807-02-01,6931.0
2,1,BREST,FRA,48.383,-4.495,1807-03-01,6896.0
3,1,BREST,FRA,48.383,-4.495,1807-04-01,6953.0
4,1,BREST,FRA,48.383,-4.495,1807-05-01,7043.0
2508,7,CUXHAVEN 2,DEU,53.867,8.717,1843-01-01,7093.0
2509,7,CUXHAVEN 2,DEU,53.867,8.717,1843-02-01,6688.0
2510,7,CUXHAVEN 2,DEU,53.867,8.717,1843-03-01,6493.0
2511,7,CUXHAVEN 2,DEU,53.867,8.717,1843-04-01,6723.0
2512,7,CUXHAVEN 2,DEU,53.867,8.717,1843-05-01,6533.0
4525,9,MAASSLUIS,NLD,51.918,4.25,1848-02-01,6880.0
4526,9,MAASSLUIS,NLD,51.918,4.25,1848-03-01,6700.0
4527,9,MAASSLUIS,NLD,51.918,4.25,1848-04-01,6775.0
4528,9,MAASSLUIS,NLD,51.918,4.25,1848-05-01,6580.0
4529,9,MAASSLUIS,NLD,51.918,4.25,1848-06-01,6685.0
6540,8,WISMAR 2,DEU,53.898999999999994,11.458,1848-07-01,6957.0
6541,8,WISMAR 2,DEU,53.898999999999994,11.458,1848-08-01,6944.0
6542,8,WISMAR 2,DEU,53.898999999999994,11.458,1848-09-01,7084.0
6543,8,WISMAR 2,DEU,53.898999999999994,11.458,1848-10-01,6898.0
6544,8,WISMAR 2,DEU,53.898999999999994,11.458,1848-11-01,6859.0
8538,10,SAN FRANCISCO,USA,37.806999999999995,-122.465,1854-07-01,6909.0
8539,10,SAN FRANCISCO,USA,37.806999999999995,-122.465,1854-08-01,6940.0
8540,10,SAN FRANCISCO,USA,37.806999999999995,-122.465,1854-09-01,6961.0
8541,10,SAN FRANCISCO,USA,37.806999999999995,-122.465,1854-10-01,6952.0
8542,10,SAN FRANCISCO,USA,37.806999999999995,-122.465,1854-11-01,6952.0
有很多重复的信息，我想要一个这样的JSON:
There is a lot of repetitive information and I would like to have a JSON like this:
[
{
"ID": 1,
"Location": "BREST",
"Latitude": 48.383,
"Longitude": -4.495,
"Country": "FRA",
"Tide-Data": {
"1807-02-01": 6931,
"1807-03-01": 6896,
"1807-04-01": 6953,
"1807-05-01": 7043
}
},
{
"ID": 5,
"Location": "HOLYHEAD",
"Latitude": 53.31399999999999,
"Longitude": -4.62,
"Country": "GBR",
"Tide-Data": {
"1807-02-01": 6931,
"1807-03-01": 6896,
"1807-04-01": 6953,
"1807-05-01": 7043
}
}
]
我怎样才能做到这一点?
How could I achieve this?
重现数据框的代码:
# input json
json_str = '[{"ID":1,"Location":"BREST","Country":"FRA","Latitude":48.383,"Longitude":-4.495,"timestamp":"1807-01-01","tide":6905},{"ID":1,"Location":"BREST","Country":"FRA","Latitude":48.383,"Longitude":-4.495,"timestamp":"1807-02-01","tide":6931},{"ID":1,"Location":"BREST","Country":"DEU","Latitude":48.383,"Longitude":-4.495,"timestamp":"1807-03-01","tide":6896},{"ID":7,"Location":"CUXHAVEN 2","Country":"DEU","Latitude":53.867,"Longitude":-8.717,"timestamp":"1843-01-01","tide":7093},{"ID":7,"Location":"CUXHAVEN 2","Country":"DEU","Latitude":53.867,"Longitude":-8.717,"timestamp":"1843-02-01","tide":6688},{"ID":7,"Location":"CUXHAVEN 2","Country":"DEU","Latitude":53.867,"Longitude":-8.717,"timestamp":"1843-03-01","tide":6493}]'
# load json object
data_list = json.loads(json_str)
# create dataframe
df = json_normalize(data_list, None, None)
解决方案
更新:
j = (df.groupby(['ID','Location','Country','Latitude','Longitude'])
.apply(lambda x: x[['timestamp','tide']].to_dict('records'))
.reset_index()
.rename(columns={0:'Tide-Data'})
.to_json(orient='records'))
结果(格式化):
In [103]: print(json.dumps(json.loads(j), indent=2, sort_keys=True))
[
{
"Country": "FRA",
"ID": 1,
"Latitude": 48.383,
"Location": "BREST",
"Longitude": -4.495,
"Tide-Data": [
{
"tide": 6905.0,
"timestamp": "1807-01-01"
},
{
"tide": 6931.0,
"timestamp": "1807-02-01"
},
{
"tide": 6896.0,
"timestamp": "1807-03-01"
},
{
"tide": 6953.0,
"timestamp": "1807-04-01"
},
{
"tide": 7043.0,
"timestamp": "1807-05-01"
}
]
},
{
"Country": "DEU",
"ID": 7,
"Latitude": 53.867,
"Location": "CUXHAVEN 2",
"Longitude": 8.717,
"Tide-Data": [
{
"tide": 7093.0,
"timestamp": "1843-01-01"
},
{
"tide": 6688.0,
"timestamp": "1843-02-01"
},
{
"tide": 6493.0,
"timestamp": "1843-03-01"
},
{
"tide": 6723.0,
"timestamp": "1843-04-01"
},
{
"tide": 6533.0,
"timestamp": "1843-05-01"
}
]
},
{
"Country": "DEU",
"ID": 8,
"Latitude": 53.899,
"Location": "WISMAR 2",
"Longitude": 11.458,
"Tide-Data": [
{
"tide": 6957.0,
"timestamp": "1848-07-01"
},
{
"tide": 6944.0,
"timestamp": "1848-08-01"
},
{
"tide": 7084.0,
"timestamp": "1848-09-01"
},
{
"tide": 6898.0,
"timestamp": "1848-10-01"
},
{
"tide": 6859.0,
"timestamp": "1848-11-01"
}
]
},
{
"Country": "NLD",
"ID": 9,
"Latitude": 51.918,
"Location": "MAASSLUIS",
"Longitude": 4.25,
"Tide-Data": [
{
"tide": 6880.0,
"timestamp": "1848-02-01"
},
{
"tide": 6700.0,
"timestamp": "1848-03-01"
},
{
"tide": 6775.0,
"timestamp": "1848-04-01"
},
{
"tide": 6580.0,
"timestamp": "1848-05-01"
},
{
"tide": 6685.0,
"timestamp": "1848-06-01"
}
]
},
{
"Country": "USA",
"ID": 10,
"Latitude": 37.807,
"Location": "SAN FRANCISCO",
"Longitude": -122.465,
"Tide-Data": [
{
"tide": 6909.0,
"timestamp": "1854-07-01"
},
{
"tide": 6940.0,
"timestamp": "1854-08-01"
},
{
"tide": 6961.0,
"timestamp": "1854-09-01"
},
{
"tide": 6952.0,
"timestamp": "1854-10-01"
},
{
"tide": 6952.0,
"timestamp": "1854-11-01"
}
]
}
]
旧答案:
您可以使用 groupby()、apply() 和 to_json() 方法:
You can do it using groupby(), apply() and to_json() methods:
j = (df.groupby(['ID','Location','Country','Latitude','Longitude'], as_index=False)
.apply(lambda x: dict(zip(x.timestamp,x.tide)))
.reset_index()
.rename(columns={0:'Tide-Data'})
.to_json(orient='records'))
输出:
In [112]: print(json.dumps(json.loads(j), indent=2, sort_keys=True))
[
{
"Country": "FRA",
"ID": 1,
"Latitude": 48.383,
"Location": "BREST",
"Longitude": -4.495,
"Tide-Data": {
"1807-01-01": 6905.0,
"1807-02-01": 6931.0,
"1807-03-01": 6896.0,
"1807-04-01": 6953.0,
"1807-05-01": 7043.0
}
},
{
"Country": "DEU",
"ID": 7,
"Latitude": 53.867,
"Location": "CUXHAVEN 2",
"Longitude": 8.717,
"Tide-Data": {
"1843-01-01": 7093.0,
"1843-02-01": 6688.0,
"1843-03-01": 6493.0,
"1843-04-01": 6723.0,
"1843-05-01": 6533.0
}
},
{
"Country": "DEU",
"ID": 8,
"Latitude": 53.899,
"Location": "WISMAR 2",
"Longitude": 11.458,
"Tide-Data": {
"1848-07-01": 6957.0,
"1848-08-01": 6944.0,
"1848-09-01": 7084.0,
"1848-10-01": 6898.0,
"1848-11-01": 6859.0
}
},
{
"Country": "NLD",
"ID": 9,
"Latitude": 51.918,
"Location": "MAASSLUIS",
"Longitude": 4.25,
"Tide-Data": {
"1848-02-01": 6880.0,
"1848-03-01": 6700.0,
"1848-04-01": 6775.0,
"1848-05-01": 6580.0,
"1848-06-01": 6685.0
}
},
{
"Country": "USA",
"ID": 10,
"Latitude": 37.807,
"Location": "SAN FRANCISCO",
"Longitude": -122.465,
"Tide-Data": {
"1854-07-01": 6909.0,
"1854-08-01": 6940.0,
"1854-09-01": 6961.0,
"1854-10-01": 6952.0,
"1854-11-01": 6952.0
}
}
]
PS 如果你不关心身份，你可以直接写入 JSON 文件:
PS if you don't care of idents you can write directly to JSON file:
(df.groupby(['ID','Location','Country','Latitude','Longitude'], as_index=False)
.apply(lambda x: dict(zip(x.timestamp,x.tide)))
.reset_index()
.rename(columns={0:'Tide-Data'})
.to_json('/path/to/file_name.json', orient='records'))
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