如果您不介意使用Pandas，那么使用Timestamp.days_in_month可能是最简单的：

import pandas as pd

> pd.Timestamp(year=2020, month=2, day=1).days_in_month

29

我认为这比其他一些答案更具可读性：

from datetime import timedelta as td
from datetime import datetime as dt
today = dt.now()
a_day_next_month = dt(today.year, today.month, 27) + td(days=5)
first_day_next_month =  dt(a_day_next_month.year, a_day_next_month.month, 1)
last_day_this_month = first_day_next_month - td(days=1)

最简单的方法（不必导入日历）是获取下个月的第一天，然后从中减去一天。

import datetime as dt
from dateutil.relativedelta import relativedelta

thisDate = dt.datetime(2017, 11, 17)

last_day_of_the_month = dt.datetime(thisDate.year, (thisDate + relativedelta(months=1)).month, 1) - dt.timedelta(days=1)
print last_day_of_the_month

输出：

datetime.datetime(2017, 11, 30, 0, 0)

PS：与导入日历方法相比，此代码运行速度更快；见下文：

import datetime as dt
import calendar
from dateutil.relativedelta import relativedelta

someDates = [dt.datetime.today() - dt.timedelta(days=x) for x in range(0, 10000)]

start1 = dt.datetime.now()
for thisDate in someDates:
    lastDay = dt.datetime(thisDate.year, (thisDate + relativedelta(months=1)).month, 1) - dt.timedelta(days=1)

print ('Time Spent= ', dt.datetime.now() - start1)


start2 = dt.datetime.now()
for thisDate in someDates:
    lastDay = dt.datetime(thisDate.year, 
                          thisDate.month, 
                          calendar.monthrange(thisDate.year, thisDate.month)[1])

print ('Time Spent= ', dt.datetime.now() - start2)

输出：

Time Spent=  0:00:00.097814
Time Spent=  0:00:00.109791

此代码假设您希望获得当月最后一天的日期（即，不只是DD部分，而是整个YYYYMMDD日期）

如果传入日期范围，则可以使用以下命令：

def last_day_of_month(any_days):
    res = []
    for any_day in any_days:
        nday = any_day.days_in_month -any_day.day
        res.append(any_day + timedelta(days=nday))
    return res

另一个解决方案是这样做：

from datetime import datetime

def last_day_of_month(year, month):
    """ Work out the last day of the month """
    last_days = [31, 30, 29, 28, 27]
    for i in last_days:
        try:
            end = datetime(year, month, i)
        except ValueError:
            continue
        else:
            return end.date()
    return None

使用如下函数：

>>> 
>>> last_day_of_month(2008, 2)
datetime.date(2008, 2, 29)
>>> last_day_of_month(2009, 2)
datetime.date(2009, 2, 28)
>>> last_day_of_month(2008, 11)
datetime.date(2008, 11, 30)
>>> last_day_of_month(2008, 12)
datetime.date(2008, 12, 31)

calendar.monthrange提供以下信息：

日历.月范围（年，月）返回指定年份和月份的月份第一天的工作日和月份天数。

>>> import calendar
>>> calendar.monthrange(2002, 1)
(1, 31)
>>> calendar.monthrange(2008, 2)  # leap years are handled correctly
(4, 29)
>>> calendar.monthrange(2100, 2)  # years divisible by 100 but not 400 aren't leap years
(0, 28)

so:

calendar.monthrange(year, month)[1]

似乎是最简单的方法。