我在这里找到了有趣的解决方案。提供这些相对的elta参数，可以得到一个月的最后一天：day=31，days=+1，seconds=-1（这是前一天的最后一秒）：

import datetime
from dateutil.relativedelta import relativedelta

day_of_febuary = datetime.datetime(2022, 2, 21)
last_day_of_febuary = day_of_febuary + relativedelta(day=31, days=+1, seconds=-1)
print(last_day_of_febuary)
# Output: 2022-02-28 23:59:59

这是一个很长（容易理解）的版本，但考虑了闰年。

def last_day_month(year, month):
    leap_year_flag = 0
    end_dates = {
        1: 31,
        2: 28,
        3: 31,
        4: 30,
        5: 31,
        6: 30,
        7: 31,
        8: 31,
        9: 30,
        10: 31,
        11: 30,
        12: 31
    }

    # Checking for regular leap year    
    if year % 4 == 0:
        leap_year_flag = 1
    else:
        leap_year_flag = 0

    # Checking for century leap year    
    if year % 100 == 0:
        if year % 400 == 0:
            leap_year_flag = 1
        else:
            leap_year_flag = 0
    else:
        pass

    # return end date of the year-month
    if leap_year_flag == 1 and month == 2:
        return 29
    elif leap_year_flag == 1 and month != 2:
        return end_dates[month]
    else:
        return end_dates[month]

使用熊猫！

def isMonthEnd(date):
    return date + pd.offsets.MonthEnd(0) == date

isMonthEnd(datetime(1999, 12, 31))
True
isMonthEnd(pd.Timestamp('1999-12-31'))
True
isMonthEnd(pd.Timestamp(1965, 1, 10))
False

对我来说，更简单的方法是使用panda（两行解决方案）：

    from datetime import datetime
    import pandas as pd

    firstday_month = datetime(year, month, 1)
    lastday_month = firstday_month + pd.offsets.MonthEnd(1)

另一种方法是：取一个月的第一天，然后再加一个月，再折一天：

    from datetime import datetime
    import pandas as pd

    firstday_month = datetime(year, month, 1)
    lastday_month = firstday_month + pd.DateOffset(months=1) - pd.DateOffset(days=1)

编辑：查看@Blair Conrad的答案以获得更清洁的解决方案

>>> import datetime
>>> datetime.date(2000, 2, 1) - datetime.timedelta(days=1)
datetime.date(2000, 1, 31)

我喜欢这样

import datetime
import calendar

date=datetime.datetime.now()
month_end_date=datetime.datetime(date.year,date.month,1) + datetime.timedelta(days=calendar.monthrange(date.year,date.month)[1] - 1)