如何计算大熊猫一周中每一天的最大值的最常见时间

问题描述

使用python中的yahoo财务软件包，我能够下载相关数据以显示OCHL。我的目标是找出一天中的哪个时间是平均水平最高的股票。

以下是下载数据的代码：

import yfinance as yf
import pandas as pd

df = yf.download(
        tickers = "APPL",period = "60d",interval = "5m",auto_adjust = True,group_by = 'ticker',prepost = True,)

maxTimes = df.groupby([df.index.month,df.index.day,df.index.day_name()])['High'].idxmax()

这给了我类似的东西：

Datetime  Datetime  Datetime 
6         2         Tuesday     2020-06-02 19:45:00-04:00
          3         Wednesday   2020-06-03 15:50:00-04:00
          4         Thursday    2020-06-04 10:30:00-04:00
          5         Friday      2020-06-05 11:30:00-04:00
...
8         3         Monday      2020-08-03 14:40:00-04:00
          4         Tuesday     2020-08-04 18:10:00-04:00
          5         Wednesday   2020-08-05 11:10:00-04:00
          6         Thursday    2020-08-06 16:20:00-04:00
          7         Friday      2020-08-07 15:50:00-04:00
Name: High,dtype: datetime64[ns,America/New_York]

我认为我创建的maxTimes对象应该给我每天发生一天中最高时段的时间，但是我需要的是：

Monday    12:00
Tuesday   13:25
Wednesday 09:35
Thurs     16:10
Fri       12:05

有人能帮助我确定如何使我的数据看起来像这样吗？

解决方法

这应该有效：

import yfinance as yf
import pandas as pd

df = yf.download(
        tickers = "AAPL",period = "60d",interval = "5m",auto_adjust = True,group_by = 'ticker',prepost = True,)

maxTimes = df.groupby([df.index.month,df.index.day,df.index.day_name()])['High'].idxmax()

# Drop date
maxTimes = maxTimes.apply(lambda x: x.time())

# Drop unused sub-indexes
maxTimes = maxTimes.droplevel(level=[0,1])

# To seconds
maxTimes = maxTimes.apply(lambda t: (t.hour * 60 + t.minute) * 60 + t.second)

# Get average
maxTimes =  maxTimes.groupby(maxTimes.index).mean()

# Back to time
maxTimes = pd.to_datetime(maxTimes,unit='s').apply(lambda x: x.time())

print (maxTimes)

'''
Output:

Datetime
Friday       11:59:32.727272
Monday              14:15:00
Thursday            13:21:40
Tuesday             10:35:00
Wednesday           11:53:45
Name: High,dtype: object

'''