我经常需要一个月的最后一天来保持上个月的最后一天。为了解决这个问题，我在计算前加一天，然后在返回前再减去它。

from datetime import date, timedelta

# it's a lot faster with a constant day
DAY = timedelta(1)

def add_month(a_date, months):
    "Add months to date and retain last day in month."
    next_day = a_date + DAY
    # calculate new year and month
    m_sum = next_day.month + months - 1
    y = next_day.year + m_sum // 12
    m = m_sum % 12 + 1
    try:
        return date(y, m, next_day.day) - DAY
    except ValueError:
        # on fail return last day in month
        # can't fail on december so I don't bother changing the year
        return date(y, m + 1, 1) - DAY

import datetime


'''
Created on 2011-03-09

@author: tonydiep
'''

def add_business_months(start_date, months_to_add):
    """
    Add months in the way business people think of months. 
    Jan 31, 2011 + 1 month = Feb 28, 2011 to business people
    Method: Add the number of months, roll back the date until it becomes a valid date
    """
    # determine year
    years_change = months_to_add / 12

    # determine if there is carryover from adding months
    if (start_date.month + (months_to_add % 12) > 12 ):
        years_change = years_change + 1

    new_year = start_date.year + years_change

    # determine month
    work = months_to_add % 12
    if 0 == work:
        new_month = start_date.month
    else:
        new_month = (start_date.month + (work % 12)) % 12

    if 0 == new_month:
        new_month = 12 

    # determine day of the month
    new_day = start_date.day
    if(new_day in [31, 30, 29, 28]):
        #user means end of the month
        new_day = 31


    new_date = None
    while (None == new_date and 27 < new_day):
        try:
            new_date = start_date.replace(year=new_year, month=new_month, day=new_day)
        except:
            new_day = new_day - 1   #wind down until we get to a valid date

    return new_date


if __name__ == '__main__':
    #tests
    dates = [datetime.date(2011, 1, 31),
             datetime.date(2011, 2, 28),
             datetime.date(2011, 3, 28),
             datetime.date(2011, 4, 28),
             datetime.date(2011, 5, 28),
             datetime.date(2011, 6, 28),
             datetime.date(2011, 7, 28),
             datetime.date(2011, 8, 28),
             datetime.date(2011, 9, 28),
             datetime.date(2011, 10, 28),
             datetime.date(2011, 11, 28),
             datetime.date(2011, 12, 28),
             ]
    months = range(1, 24)
    for start_date in dates:
        for m in months:
            end_date = add_business_months(start_date, m)
            print("%s\t%s\t%s" %(start_date, end_date, m))

这个解决方案适用于12月，而本页上的大多数答案都不适用。 在使用模量(%)或整数除法(//)之前，您需要首先将月份从基于1的索引(例如Jan = 1)转移到基于0的索引(例如Jan = 0)，否则11月(11)加上1个月得到12，当找到余数(12% 12)时，得到0。

(不要建议“(月% 12)+ 1”或10月+ 1 = 12月!)

def AddMonths(d,x):
    newmonth = ((( d.month - 1) + x ) % 12 ) + 1
    newyear  = int(d.year + ((( d.month - 1) + x ) / 12 ))
    return datetime.date( newyear, newmonth, d.day)

然而……这并不能解释1月31日+一个月的问题。所以我们回到OP，你说增加一个月是什么意思?一种解决方案是回溯，直到找到一个有效的日期，因为大多数人会假设一月的最后一天加一个月等于二月的最后一天。 这也适用于负数的月份。 证明:

>>> import datetime
>>> AddMonths(datetime.datetime(2010,8,25),1)
datetime.date(2010, 9, 25)
>>> AddMonths(datetime.datetime(2010,8,25),4)
datetime.date(2010, 12, 25)
>>> AddMonths(datetime.datetime(2010,8,25),5)
datetime.date(2011, 1, 25)
>>> AddMonths(datetime.datetime(2010,8,25),13)
datetime.date(2011, 9, 25)
>>> AddMonths(datetime.datetime(2010,8,25),24)
datetime.date(2012, 8, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-1)
datetime.date(2010, 7, 25)
>>> AddMonths(datetime.datetime(2010,8,25),0)
datetime.date(2010, 8, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-12)
datetime.date(2009, 8, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-8)
datetime.date(2009, 12, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-7)
datetime.date(2010, 1, 25)>>> 

使用python datetime模块为datetime.today()添加6个月的时间增量。

http://docs.python.org/library/datetime.html

你当然要解决Johannes weß提出的问题——你说的6个月是什么意思?

这里有一个dateutil的例子。relativedelta，我发现它对于迭代过去一年很有用，每次跳过一个月到现在的日期:

>>> import datetime
>>> from dateutil.relativedelta import relativedelta
>>> today = datetime.datetime.today()
>>> month_count = 0
>>> while month_count < 12:
...  day = today - relativedelta(months=month_count)
...  print day
...  month_count += 1
... 
2010-07-07 10:51:45.187968
2010-06-07 10:51:45.187968
2010-05-07 10:51:45.187968
2010-04-07 10:51:45.187968
2010-03-07 10:51:45.187968
2010-02-07 10:51:45.187968
2010-01-07 10:51:45.187968
2009-12-07 10:51:45.187968
2009-11-07 10:51:45.187968
2009-10-07 10:51:45.187968
2009-09-07 10:51:45.187968
2009-08-07 10:51:45.187968

和其他答案一样，你必须弄清楚你说的“6个月后”到底是什么意思。如果你的意思是“六年后某个月的今天”，那么这个是:

datetime.datetime.now() + relativedelta(months=6)

这个怎么样?没有使用其他库(dateutil)或timedelta? 基于vartec的回答，我这样做了，我相信它是有效的:

import datetime

today = datetime.date.today()
six_months_from_today = datetime.date(today.year + (today.month + 6)/12, (today.month + 6) % 12, today.day)

我尝试使用timedelta，但因为它是计算天数的，365/2或6*356/12并不总是转换为6个月，而是182天。如。

day = datetime.date(2015, 3, 10)
print day
>>> 2015-03-10

print (day + datetime.timedelta(6*365/12))
>>> 2015-09-08

我相信我们通常会假设某一天的6个月将在6个月后的同一天登陆(即2015-03-10—> 2015-09-10，而不是2015-09-08)

我希望这对你有帮助。