將熊貓數據框轉換爲嵌套JSON

我是Python和Pandas的新手。我試圖將熊貓數據框轉換爲嵌套的JSON。函數.to_json（）不會爲我的目標提供足夠的靈活性。將熊貓數據框轉換爲嵌套JSON

下面是數據幀的一些數據點（以CSV，逗號分隔）：

,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

有大量的重複信息，我想有這樣的JSON：

[ 
{ 
    "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 
    } 
} 
]

我怎麼能做到這一點？

來源

2016-11-07 Felix

['pandas.DataFrame.to_json'（HTTP： //pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.to_json.html）有很多選項。看看你是否可以得到你想要的選項。 –

特別檢查'orient'選項。 –

我不明白。它一次又一次地重複所有相同的信息，但我希望列時間戳和潮汐被嵌套。 – Felix

UPDATE：

In [102]: 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')) 
    ...:

結果（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" 
     } 
    ] 
    } 
]

OLD答案：

可以使用groupby()，apply()和to_json()方法做到這一點：

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'))

輸出：

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如果你不在乎idents，你可以直接寫信給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'))

來源

2016-11-08 15:06:54 MaxU

哇！你太棒了。這很好用！ – Felix

@Felix，很高興我能幫到:) – MaxU

我剛剛意識到我需要這種格式的數據：「潮汐數據」：{「timestamp」：「1848-07-01」，「tide」：「6957.0」} 。我必須改變你的功能？ – Felix

將熊貓數據框轉換爲嵌套JSON

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