格式化熊猫数据系列，该数据系列由两列分组，并在第三列重新采样，并以字典的平均值为准

Question

我有一个这样的数据框：

°  item_name   item_category   scraping_date        price
0   Michel1     Category1       2018-04-14           21.0
1   Michel1     Category1       2018-04-16           42.1
2   Michel1     Category1       2018-04-17           84.0
3   Michel1     Category1       2018-04-19           126.2
4   Michel1     Category1       2018-04-20           168.3
5   Michel1     Category2       2018-04-23           21.2
6   Michel1     Category2       2018-05-08           42.0
7   Michel1     Category2       2018-03-26           84.1
8   Michel1     Category2       2018-03-31           126.2
9   Michel1     Category2       2018-04-01           168.3
10  Michel2     Category1       2018-04-04           21.0
11  Michel2     Category1       2018-04-05           42.1
12  Michel2     Category1       2018-04-09           84.2
13  Michel2     Category1       2018-04-11           126.3
14  Michel2     Category1       2018-04-12           168.4
15  Michel2     Category2       2018-04-13           21.0
16  Michel2     Category2       2018-05-03           42.1
17  Michel2     Category2       2018-04-25           84.2
18  Michel2     Category2       2018-04-28           126.3
19  Michel2     Category2       2018-04-29           168.4

我想按商品名称和类别分组，按周重新采样，并获得每周的平均价格。最后，我想在这样的字典中输出日期：

[
  {
    "item_name": "Michel1",
    "item_category": "Category1", 
    "prices": [
                {"week": "1", "average": "84.2"},
                {"week": "2", "average": "84.2"}
              ]
  },
  {
    "item_name": "Michel1",
    "item_category": "Category2", 
    "prices": [
                {"week": "1", "average": "84.2"},
                {"week": "2", "average": "84.2"}
              ]
  },....
]

我附带了一些要分组的东西，并得到了平均值，但我无法将其转化为字典：

df["price"] = df["price"].astype(float)
df["scraping_date"] = pd.to_datetime(df["scraping_date"])
df.set_index("scraping_date").groupby(["item_name","item_category"])["price"].resample("W").mean()

如果我执行.to_dict()，我得到了，这几乎不是我想要的：

{('Michel1', 'Category1', Timestamp('2017-12-03 00:00:00')): 20.0,
 ('Michel1', 'Category1', Timestamp('2017-12-10 00:00:00')): 20.0,
 ('Michel1', 'Category2', Timestamp('2017-12-17 00:00:00')): 20.0,
 ('Michel1', 'Category2', Timestamp('2017-12-24 00:00:00')): 20.0,
 ('Michel2', 'Category1', Timestamp('2017-12-31 00:00:00')): 20.0,
 ('Michel2', 'Category1', Timestamp('2018-01-07 00:00:00')): 20.0,
}

Answer 1

我不能保证速度，与apply一起使用分组

df['Week']=pd.to_datetime(df.scraping_date).dt.week
df.groupby(['item_name','item_category']).apply(lambda x : x.groupby(['Week']).price.mean().to_frame('average')
.reset_index().to_dict('r')).to_frame('price').reset_index().to_dict('r')
Out[51]: 
[{'item_category': 'Category1',
  'item_name': 'Michel1',
  'price': [{'Week': 15.0, 'average': 21.0},
   {'Week': 16.0, 'average': 105.15}]},
 {'item_category': 'Category2',
  'item_name': 'Michel1',
  'price': [{'Week': 13.0, 'average': 126.2},
   {'Week': 17.0, 'average': 21.2},
   {'Week': 19.0, 'average': 42.0}]},
 {'item_category': 'Category1',
  'item_name': 'Michel2',
  'price': [{'Week': 14.0, 'average': 31.55},
   {'Week': 15.0, 'average': 126.3}]},
 {'item_category': 'Category2',
  'item_name': 'Michel2',
  'price': [{'Week': 15.0, 'average': 21.0},
   {'Week': 17.0, 'average': 126.3},
   {'Week': 18.0, 'average': 42.1}]}]