Python 从特定时间开始 date_range

Question

我想在 Python 中指定一个日期范围，从当天的前一天开始。但是，我希望该日期范围从 10:00:00.

开始

这是我目前使用的代码

import pandas as pd
import datetime as dt
date = dt.datetime.today() - dt.timedelta(days=1)
date_range = pd.date_range(date, freq='60min', periods=24)

但是，这从 00:00:00 开始。我尝试了几种修改上述代码的方法，使其从 10:00:00 开始，但 none 有效。有人可以帮忙吗？

澄清一下：我使用 Pandas 因为这个日期范围将用作数据框的索引。

Answer 1

尝试

today = pd.datetime.today().date()
today - pd.offsets.Hour(14)

Timestamp('2016-11-22 10:00:00')

然后使用pd.date_range

pd.date_range(today - pd.offsets.Hour(14), periods=24, freq='H')

DatetimeIndex(['2016-11-22 10:00:00', '2016-11-22 11:00:00',
               '2016-11-22 12:00:00', '2016-11-22 13:00:00',
               '2016-11-22 14:00:00', '2016-11-22 15:00:00',
               '2016-11-22 16:00:00', '2016-11-22 17:00:00',
               '2016-11-22 18:00:00', '2016-11-22 19:00:00',
               '2016-11-22 20:00:00', '2016-11-22 21:00:00',
               '2016-11-22 22:00:00', '2016-11-22 23:00:00',
               '2016-11-23 00:00:00', '2016-11-23 01:00:00',
               '2016-11-23 02:00:00', '2016-11-23 03:00:00',
               '2016-11-23 04:00:00', '2016-11-23 05:00:00',
               '2016-11-23 06:00:00', '2016-11-23 07:00:00',
               '2016-11-23 08:00:00', '2016-11-23 09:00:00'],
              dtype='datetime64[ns]', freq='H')

Answer 2

您可以构造另一个日期时间，但再次只使用日、月、年组件和偏移量：

In [87]:
date = dt.datetime.today() - dt.timedelta(days=1)
date = dt.datetime(date.year, date.month, date.day) + dt.timedelta(hours=10)
date_range = pd.date_range(date, freq='60min', periods=24)
date_range

Out[87]:
DatetimeIndex(['2016-11-22 10:00:00', '2016-11-22 11:00:00',
               '2016-11-22 12:00:00', '2016-11-22 13:00:00',
               '2016-11-22 14:00:00', '2016-11-22 15:00:00',
               '2016-11-22 16:00:00', '2016-11-22 17:00:00',
               '2016-11-22 18:00:00', '2016-11-22 19:00:00',
               '2016-11-22 20:00:00', '2016-11-22 21:00:00',
               '2016-11-22 22:00:00', '2016-11-22 23:00:00',
               '2016-11-23 00:00:00', '2016-11-23 01:00:00',
               '2016-11-23 02:00:00', '2016-11-23 03:00:00',
               '2016-11-23 04:00:00', '2016-11-23 05:00:00',
               '2016-11-23 06:00:00', '2016-11-23 07:00:00',
               '2016-11-23 08:00:00', '2016-11-23 09:00:00'],
              dtype='datetime64[ns]', freq='60T')

Answer 3

创建字符串日期然后将其解析为日期时间日期

 from datetime import timedelta
 from datetime import date
 import datetime as datetime
 previous_day=pd.datetime.today().date()-datetime.timedelta(days=1)
 start_datetime=datetime.datetime.strptime(str(previous_day)+' 10:00:00','%Y-%m-%d %H:%M:%S')
 date_range = pd.date_range(start_datetime, freq='60min', periods=24)
 print(date_range)

输出：

 DatetimeIndex(['2021-03-25 10:00:00', '2021-03-25 11:00:00',
                '2021-03-25 12:00:00', '2021-03-25 13:00:00',
                '2021-03-25 14:00:00', '2021-03-25 15:00:00',
                '2021-03-25 16:00:00', '2021-03-25 17:00:00',
                '2021-03-25 18:00:00', '2021-03-25 19:00:00',
                '2021-03-25 20:00:00', '2021-03-25 21:00:00',
                '2021-03-25 22:00:00', '2021-03-25 23:00:00',
                '2021-03-26 00:00:00', '2021-03-26 01:00:00',
                '2021-03-26 02:00:00', '2021-03-26 03:00:00',
                '2021-03-26 04:00:00', '2021-03-26 05:00:00',
                '2021-03-26 06:00:00', '2021-03-26 07:00:00',
                '2021-03-26 08:00:00', '2021-03-26 09:00:00'],
               dtype='datetime64[ns]', freq='60T')

Python 从特定时间开始 date_range

Python start date_range from a specific hour

python

datetime

pandas

dataframe

date-range