Pandas groupby：创建带有两列的groupby时，如何以正确的顺序对工作日进行排序？

Question

以下数据框包含一年中每小时的值（kWh）。

cons2016.head()

    Date        Hour    kWh     Month   Weekday
0   2016-01-01  00:00   71.48   January Friday
1   2016-01-01  01:00   65.32   January Friday
2   2016-01-01  02:00   65.38   January Friday
3   2016-01-01  03:00   62.44   January Friday
4   2016-01-01  04:00   57.56   January Friday

我想从此数据框中创建一个Seaborn热图（在工作日中，正确顺序为纵轴，小时为横轴）。所以我分组：

weekdayhour = cons2016.groupby(["Weekday", "Hour"]).mean()
weekdayhour = weekdayhour.reset_index()
weekdayhour.head()

    Weekday Hour    kWh
0   Friday  00:00   61.188113
1   Friday  01:00   57.231698
2   Friday  02:00   55.818679
3   Friday  03:00   55.074151
4   Friday  04:00   55.049811

但是现在工作日按字母顺序排列（也在热图中）：

heat_weekdayhour = weekdayhour.pivot(index="Weekday", columns="Hour", values="kWh")
sns.heatmap(heat_weekdayhour)

从星期一到星期日，我如何正常地获得工作日？我试过像这样添加.reindex：

weekdays = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"]
weekdayhour = cons2016.groupby(["Weekday", "Hour"]).mean().reindex(labels=weekdays)

但这给了我TypeError: Expected tuple, got str

谢谢您的帮助！

Answer 1

使用Categorical

weekdays = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"]
weekdayhour.Weekday = pd.Categorical(weekdayhour.Weekday,categories=weekdays)
weekdayhour = weekdayhour.sort_values('Weekday')
  Weekday   Hour    kWh
0  Friday  00:00  71.48
1  Friday  01:00  65.32
2  Friday  02:00  65.38
3  Friday  03:00  62.44
4  Friday  04:00  57.56

更多信息：

weekdayhour.Weekday
0    Friday
1    Friday
2    Friday
3    Friday
4    Friday
Name: Weekday, dtype: category
Categories (7, object): [Monday < Tuesday < Wednesday < Thursday < Friday < Saturday < Sunday]

Answer 2

import pandas as pd

#You first create your list in the order you want it
days = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"]

#Using Categorical() function to set the order according to how it is arranged above
df["DOTW_Appointment"] = pd.Categorical(df.DOTW_Appointment, categories=days, ordered=True)