Python 将数据帧转换为嵌套JSON
我不熟悉蟒蛇和熊猫。我正在尝试将Pandas数据帧转换为嵌套的JSON。函数.to_json()没有为我的目标提供足够的灵活性 以下是数据帧的一些数据点(csv格式,逗号分隔): 有很多重复信息,我希望有一个JSON,如下所示:Python 将数据帧转换为嵌套JSON,python,json,pandas,dataframe,Python,Json,Pandas,Dataframe,我不熟悉蟒蛇和熊猫。我正在尝试将Pandas数据帧转换为嵌套的JSON。函数.to_json()没有为我的目标提供足够的灵活性 以下是数据帧的一些数据点(csv格式,逗号分隔): 有很多重复信息,我希望有一个JSON,如下所示: [ { "ID": 1, "Location": "BREST", "Latitude": 48.383, "Longitude": -4.495, "Country": "FRA", "Tide-Data": {
[
{
"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
编辑:
用于复制数据帧的代码:
# 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)
更新:
j = (df.groupby(['ID','Location','Country','Latitude','Longitude'], as_index=False)
.apply(lambda x: x[['timestamp','tide']].to_dict('r'))
.reset_index()
.rename(columns={0:'Tide-Data'})
.to_json(orient='records'))
结果(格式化):
旧答案:
j = (df.groupby(['ID','Location','Country','Latitude','Longitude'], as_index=False)
.apply(lambda x: x[['timestamp','tide']].to_dict('r'))
.reset_index()
.rename(columns={0:'Tide-Data'})
.to_json(orient='records'))
您可以使用groupby()
、apply()
和to_json()
方法来执行此操作:
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如果您不关心标识,可以直接写入JSON文件:
(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'))
有很多选择。看看这些选项是否能满足您的需求。尤其是查看
orient
选项。我不知道如何实现。它一次又一次地重复所有相同的信息,但我希望将timestamp和tide列嵌套。如果要嵌套timestamp
和tide
,最好在调用到_json
之前进行嵌套。抱歉,一开始我误解了这个问题。但我的问题是:如何将它们组合在一起?@Felix,很高兴我能提供帮助:)我刚刚意识到我需要这种格式的数据:“潮汐数据”:{“时间戳”:“1848-07-01”,“潮汐”:“6957.0”}。我必须在你的函数中更改什么?@Felix,你能在你的问题中更新你想要的JSON吗,这样我就可以看到多个(分组)条目的样子了?我认为你的更新是正确的JSON格式。我明天一开始画折线图就告诉你。非常感谢您的更新@MaxU作为新版本的pandas(例如1.2.1),它不再适用于我。我得到这个错误:ValueError:1列传递,传递的数据有n列(在我的例子中,n是5)。是什么改变了熊猫,使之成为现实?
(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'))