python - 如何将周中的几天分成 1 周并对其执行功能

Question

所以我有这样的股市数据

    Date           Open        High        Low        Close     Weekday

0   2007-09-17  4518.450195 4549.049805 4482.850098 4494.649902 Monday
1   2007-09-18  4494.100098 4551.799805 4481.549805 4546.200195 Tuesday
2   2007-09-19  4550.25     4739.0      4550.25     4732.350098 Wednesday
3   2007-09-20  4734.850098 4760.850098 4721.149902 4747.549805 Thursday
4   2007-09-21  4752.950195 4855.700195 4733.700195 4837.549805 Friday

5   2007-09-24  4837.149902 4941.149902 4837.149902 4932.200195 Monday
6   2007-09-25  4939.100098 4953.899902 4878.149902 4938.850098 Tuesday
7   2007-09-26  4937.600098 4980.850098 4930.350098 4940.5      Wednesday
8   2007-09-27  4942.700195 5016.399902 4942.700195 5000.549805 Thursday
9   2007-09-28  4996.450195 5055.799805 4996.450195 5021.350098 Friday

10  2007-10-01  5021.5      5089.299805 5001.350098 5068.950195 Monday
11  2007-10-03  5069.0      5261.350098 5034.149902 5210.799805 Wednesday
12  2007-10-04  5211.649902 5233.100098 5126.049805 5208.649902 Thursday
13  2007-10-05  5208.149902 5248.549805 5164.5      5185.850098 Friday

(为了更好的视觉表现，我把周五之后的空白放在了一起)
我已在 Weekday 中将日期字符串转换为星期几柱子。我想要做的是我想要找到的每一周，哪一天的开盘价最低，并增加到那一天的变量。所以在示例数据中它应该是
第 1 周(指数 0-4)——周二最低 Open价钱
第 2 周(指数 5-9)——周一最低 Open价钱
第 3 周(指数 10-13)——周一最低 Open价钱
所以它应该打印 Monday -2, Tuesday -1, Wednesday - 0, Thursday -0, Friday -0我怎样才能做到这一点？

Answer 1

好像你想groupby每 5 个 obs -

df.groupby(df.index // 5)['Open'].min()

0    4494.100098
1    4837.149902
2    5021.500000
Name: Open, dtype: float64

获取每周最低开盘价的星期几 -

df.loc[df.groupby(df.index // 5)['Open'].idxmin(),'Weekday']

1     Tuesday
5      Monday
10     Monday
Name: Weekday, dtype: object

获取月数 -

df.loc[df.groupby(df.index // 5)['Open'].idxmin(),'Weekday'].value_counts()

Monday     2
Tuesday    1
Name: Weekday, dtype: int64

更新
要处理缺失的天数，请利用 Date要分组的列 -

df.loc[df.groupby(df['Date'].dt.week)['Open'].idxmin(),'Weekday'].value_counts()

Monday     2
Tuesday    1
Name: Weekday, dtype: int64