Python 月份开始日期和结束日期之间的两个日期

Question

我需要在两个日期之间设置每个月的开始日期和结束日期。目前，我每次需要时都会手动创建列表。

start = 2019-10-01
end = today

预期输出:

start_times = ['10/1/2019 05:00:00 AM','11/1/2019 05:00:00 AM','12/1/2019 05:00:00 AM','1/1/2020 05:00:00 AM','2/1/2020 05:00:00 AM','3/1/2020 05:00:00 AM'+'4/1/2020 05:00:00 AM','5/1/2020 05:00:00 AM','6/1/2020 05:00:00 AM','7/1/2020 05:00:00 AM','8/1/2020 05:00:00 AM']
end_times = ['10/31/2019 08:00:00 PM','11/30/2019 08:00:00 PM','12/31/2019 08:00:00 PM','1/31/2020 08:00:00 PM','2/29/2020 08:00:00 PM','3/31/2020 08:00:00 PM','4/30/2020 08:00:00 PM','5/31/2020 08:00:00 PM','6/30/2020 08:00:00 PM','7/31/2020 08:00:00 PM','8/31/2020 08:00:00 PM']

关于如何继续这里的任何想法。

Answer 1

因为您需要日期范围和特定格式，所以我建议使用 numpy 进行日期范围创建和计算，然后使用 python datetime 进行格式化，如下所示:

import numpy as np
import datetime as dt

def dateRange(start='2019-10-01', end='2020-08-28'):
    
    start_date = np.datetime64(start[:-3])  
    #Add one month to end date so last month is included
    end_date = np.datetime64(end[:-3]) + np.timedelta64(1, 'M')
    #Create Date Range
    date_range = np.arange(start_date, end_date, dtype='datetime64[M]')
    #Add 5 AM to start_dates
    start_dates = date_range + np.timedelta64(5, 'h')
    #Add 1 Month and subrtract 4 hours to 8 PM on last day of month
    end_dates = date_range + np.timedelta64(1, 'M') - np.timedelta64(4, 'h')
    #Apply Formatting   
    final_start_dates = [parse_date(np.datetime_as_string(d, unit='s')) for d in start_dates]
    final_end_dates = [parse_date(np.datetime_as_string(d, unit='s')) for d in end_dates]
    
    return final_start_dates, final_end_dates

def parse_date(date_string):
    #parse numpy date string
    date_time = date_string.split('T')
    date_parts = date_time[0].split('-')
    time_parts = date_time[1][:-1].split(':')
    #convert string values to ints
    datetime_ints = [int(d) for d in date_parts] + [int(time_parts[0])]
    
    #apply formatting
    return dt.datetime(*datetime_ints).strftime('%m/%d/%Y %I:%M:%S %p')

输出:

>>> start,end=dateRange()
>>> start
['10/01/2019 05:00:00 AM', '11/01/2019 05:00:00 AM', '12/01/2019 05:00:00 AM', '01/01/2020 05:00:00 AM', '02/01/2020 05:00:00 AM', '03/01/2020 05:00:00 AM', '04/01/2020 05:00:00 AM', '05/01/2020 05:00:00 AM', '06/01/2020 05:00:00 AM', '07/01/2020 05:00:00 AM', '08/01/2020 05:00:00 AM']
>>> end
['10/31/2019 08:00:00 PM', '11/30/2019 08:00:00 PM', '12/31/2019 08:00:00 PM', '01/31/2020 08:00:00 PM', '02/29/2020 08:00:00 PM', '03/31/2020 08:00:00 PM', '04/30/2020 08:00:00 PM', '05/31/2020 08:00:00 PM', '06/30/2020 08:00:00 PM', '07/31/2020 08:00:00 PM', '08/31/2020 08:00:00 PM']

我认为这就是您在每天制作的问题中所寻找的。实际上更容易。

def all_dates(start='2019-10-01', end='2019-10-15'):
    
    start_date = np.datetime64(start)   
    #Add one day to end date so last say is included
    end_date = np.datetime64(end) + np.timedelta64(1, 'D')
    #Create Date Range
    date_range = np.arange(start_date, end_date, dtype='datetime64[D]')
    
    final_start_dates = [parse_date(np.datetime_as_string(d + np.timedelta64(5, 'h'), unit='s')) for d in date_range]
    
    final_end_dates = [parse_date(np.datetime_as_string(d + np.timedelta64(20, 'h'), unit='s')) for d in date_range]
    
    return final_start_dates, final_end_dates

输出:

>>> start, end = all_dates()
>>> start
['10/1/2019 05:00:00 AM', '10/2/2019 05:00:00 AM', '10/3/2019 05:00:00 AM', '10/4/2019 05:00:00 AM', '10/5/2019 05:00:00 AM', '10/6/2019 05:00:00 AM', '10/7/2019 05:00:00 AM', '10/8/2019 05:00:00 AM', '10/9/2019 05:00:00 AM', '10/10/2019 05:00:00 AM', '10/11/2019 05:00:00 AM', '10/12/2019 05:00:00 AM', '10/13/2019 05:00:00 AM', '10/14/2019 05:00:00 AM', '10/15/2019 05:00:00 AM']
>>> end
['10/1/2019 08:00:00 PM', '10/2/2019 08:00:00 PM', '10/3/2019 08:00:00 PM', '10/4/2019 08:00:00 PM', '10/5/2019 08:00:00 PM', '10/6/2019 08:00:00 PM', '10/7/2019 08:00:00 PM', '10/8/2019 08:00:00 PM', '10/9/2019 08:00:00 PM', '10/10/2019 08:00:00 PM', '10/11/2019 08:00:00 PM', '10/12/2019 08:00:00 PM', '10/13/2019 08:00:00 PM', '10/14/2019 08:00:00 PM', '10/15/2019 08:00:00 PM']