Python-按时间分组的数据帧

Question

有一个数据帧，其中包含以下记录的SAMPLE(非原始数据(：

import pandas as pd
df = pd.DataFrame(dikt, columns=['id', 'price', 'day'])
df:
+-------+-----+-------+-----+
| index | id  | price | day |
+-------+-----+-------+-----+
| 0     | 34  | 12    | 3   |
+-------+-----+-------+-----+
| 1     | 34  | 6     | 5   |
+-------+-----+-------+-----+
| 2     | 56  | 23    | 8   |
+-------+-----+-------+-----+
| 3     | 56  | 21    | 9   |
+-------+-----+-------+-----+
| 4     | 56  | 67    | 22  |
+-------+-----+-------+-----+
| ...   | ... | ...   |     |
+-------+-----+-------+-----+

我想在一周内将价格分组如下：

+-------+-----+---------------------+
| index | id  | price               |
+-------+-----+---------------------+
| 0     | 34  | [12, 6]             |
+-------+-----+---------------------+
| 1     | 56  | [23, 21], [67]      |
+-------+-----+---------------------+
| ...   | ... | ...                 |
+-------+-----+---------------------+

在上表中，价格按日期分组。例如，第12天和第6天可能在第一周的第3天和第5天。所以他们在一起，等等

Answer 1

将一天除以7，添加一列作为周数，并将其分组到该单元中。哪些分组的数据帧将在没有周数的分组中组合。

df['weeknum'] = df['day'] // 7
df2 = df.groupby(['id','weeknum'])['price'].agg(list).to_frame()
df2['price'] = df2['price'].astype(str)
df2.groupby('id')['price'].agg(','.join).to_frame()
price
id  
34  [12, 6]
56  [23, 21],[67]