将Pandas数据框转换为嵌套JSON

Question

将Pandas数据框转换为嵌套JSON

48

我正在尝试将Pandas数据框转换为嵌套JSON。函数.to_json()对于我的目的来说不够灵活。

以下是数据框的一些数据点（以csv逗号分隔）：

,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

有很多重复的信息，我想要一个像这样的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
    }
}
]

我该如何实现这个？

复制数据框的代码：

# 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 = pd.json_normalize(data_list, None, None)

- Felix

3

pandas.DataFrame.to_json 提供了许多选项。请查看选项，看看是否可以通过这些选项得到您想要的结果。 - Eli Sadoff

如果你想嵌套“timestamp”和“tide”，最好在调用“to_json”之前这样做。抱歉，我一开始误解了问题。 - Eli Sadoff

@MaxU 当然可以。我已经用真实数据替换了图像。 - Felix

@Felix，你需要确切的这种输出格式吗？ - MaxU - stand with Ukraine

@Felix 你应该使用lodash和Node.js来完成这个任务，而不是pandas。你也可以在这里使用D3.nest。 - Union find

显示剩余7条评论

2个回答

0

groupby.apply 强制对每个组进行数据操作以创建嵌套结构，这非常慢。使用简单的 for 循环方法，使用 itertuples 和列表推导式来创建嵌套结构，并通过 json.dumps 进行序列化，速度更快。如果组比较小，则此方法特别有用，因为对于那些情况，groupby.apply 的速度非常慢。¹

import json
keys = ['ID', 'Location', 'Country', 'Latitude', 'Longitude']
mydict = {}
for row in df.itertuples(index=False):
    mydict.setdefault(row[:5], {})[row.timestamp] = row.tide
mylist = [{**dict(zip(keys, k)), 'Tide-Data': v} for k, v in mydict.items()]
j = json.dumps(mylist)

请注意，MaxU的groupby.apply方法需要稍作修改（传递给apply的lambda函数应有所不同），以产生预期的输出结果。

j = df.groupby(keys).apply(lambda x: x.set_index('timestamp')['tide'].to_dict()).reset_index(name='Tide-Data').to_json(orient='records')

对于给定的输入，两者都会产生以下输出：

[
  {
    "ID": 1,
    "Location": "BREST",
    "Country": "FRA",
    "Latitude": 48.383,
    "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
    }
  },
  {
    "ID": 7,
    "Location": "CUXHAVEN 2",
    "Country": "DEU",
    "Latitude": 53.867,
    "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
    }
  },
  {
    "ID": 9,
    "Location": "MAASSLUIS",
    "Country": "NLD",
    "Latitude": 51.918,
    "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
    }
  },
  {
    "ID": 8,
    "Location": "WISMAR 2",
    "Country": "DEU",
    "Latitude": 53.899,
    "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
    }
  },
  {
    "ID": 10,
    "Location": "SAN FRANCISCO",
    "Country": "USA",
    "Latitude": 37.807,
    "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
    }
  }
]

¹ 基准测试结果：在具有 100k 行的框架上，如果每个组相对较小，则循环方法比 groupby.apply 方法快约 50 倍。

import numpy as np

def jsonify(df, groupers):
    res = {}
    for row in df.itertuples(index=False):
        res.setdefault(row[:5], {})[row.timestamp] = row.tide
    j = json.dumps([dict(zip(groupers, k)) | {'Tide-Data': v} for k, v in res.items()])
    return j

df = pd.DataFrame(np.random.default_rng().choice(10, size=(100000, 7)), columns=['ID','Location','Country','Latitude','Longitude', 'timestamp', 'tide'])
groupers = ['ID','Location','Country','Latitude','Longitude']

%timeit jsonify(df, groupers)
# 502 ms ± 17.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit df.groupby(groupers).apply(lambda x: x.set_index('timestamp')['tide'].to_dict()).reset_index(name='Tide-Data').to_json(orient='records')
# 25 s ± 1.38 s per loop (mean ± std. dev. of 7 runs, 1 loop each)

如果数据组较大，则差异会小很多，但循环实现仍然比 groupby.apply 更快:

df = pd.DataFrame(np.random.default_rng().choice(3, size=(100000, 7)), columns=['ID','Location','Country','Latitude','Longitude', 'timestamp', 'tide'])

%timeit jsonify(df, groupers)
# 155 ms ± 6.45 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit df.groupby(groupers).apply(lambda x: x.set_index('timestamp')['tide'].to_dict()).reset_index(name='Tide-Data').to_json(orient='records')
# 201 ms ± 6.63 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

- cottontail

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- MaxU - stand with Ukraine · Accepted Answer

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"
      }
    ]
  }
]

旧回答:

你可以使用 groupby()、apply() 和 to_json() 方法来完成：

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 如果您不关心标识符，可以直接编写到 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'))