Question

我正在开发Web API，我从查询结果中得到了一个列表，如下所示：

[{'ORG': 'Asset Management',
  'SURVEY_DATE': datetime.date(2018, 4, 23),
  'NOS': '1'},
 {'ORG': 'Asset Management',
  'SURVEY_DATE': datetime.date(2018, 5, 8),
  'NOS': '1'},
 {'ORG': 'Chief Advocacy Office',
  'SURVEY_DATE': datetime.date(2018, 10, 31),
  'NOS': '50'},
 {'ORG': 'Chief Advocacy Office',
  'SURVEY_DATE': datetime.date(2019, 2, 13),
  'NOS': '1'},
 {'ORG': 'Chief Information Office',
  'SURVEY_DATE': datetime.date(2018, 1, 22),
  'NOS': '1'},
 {'ORG': 'Chief Information Office',
  'SURVEY_DATE': datetime.date(2018, 2, 2),
  'NOS': '1'}]

我试图先将其转换为数据框并像这样进行编码：

 df1 = df1.groupby('ORG').apply(lambda x: dict(zip(x['SURVEY_DATE'],x['NOS']))).to_dict()

但是有没有一种方法不需要在数据框中转换它？

我想将其格式化为以下字典中的响应数据字典：

{
  "Asset Management": [
    {
      "date": "2019-03-30",
      "numberOfSurveys": 76
    },
    {
      "date": "2019-03-31",
      "numberOfSurveys": 83
    }
  ],
  "Chief Advocacy Office": [
   {
      "date": "2019-03-30",
      "numberOfSurveys": 50
   },
   {
     "date": "2019-03-31",
     "numberOfSurveys": 40
   }
  ],
  "Chief Information Office": [
   {
       "date": "2019-03-30",
      "numberOfSurveys": 50
   },
   {
      "date": "2019-03-31",
      "numberOfSurveys": 40
   }
  ]
}

Answer 1

使用collections.defaultdict()：

import collections

result = collections.defaultdict(list)
for row in original_data:
    result[row['ORG']].append({
        'date': row['SURVEY_DATE'],
        'numberOfSurveys': row['NOS'],
    })

如何在python中将列表转换为具有特定格式的字典？

1 个答案: