我使用replace()方法并编写了这个递归函数。dt是一个日期时间。datetime对象:

def month_timedelta(dt, m):
    y = m // 12
    dm = m % 12
    if y == 0:
        if dt.month + m <= 12:
            return dt.replace(month = dt.month + m)
        else:
            dy = (dt.month + m) // 12
            ndt = dt.replace(year=dt.year + dy)
            return ndt.replace(month=(ndt.month + m) % 12)
    else:
        return month_timedelta(dt.replace(year=dt.year + y),dm)

按月开始计算:

from datetime import timedelta
from dateutil.relativedelta import relativedelta

end_date = start_date + relativedelta(months=delta_period) + timedelta(days=-delta_period)

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)

下面是一个示例，它允许用户决定如何返回一个日期，其中一天大于一个月中的天数。

def add_months(date, months, endOfMonthBehaviour='RoundUp'):
    assert endOfMonthBehaviour in ['RoundDown', 'RoundIn', 'RoundOut', 'RoundUp'], \
        'Unknown end of month behaviour'
    year = date.year + (date.month + months - 1) / 12
    month = (date.month + months - 1) % 12 + 1
    day = date.day
    last = monthrange(year, month)[1]
    if day > last:
        if endOfMonthBehaviour == 'RoundDown' or \
            endOfMonthBehaviour == 'RoundOut' and months < 0 or \
            endOfMonthBehaviour == 'RoundIn' and months > 0:
            day = last
        elif endOfMonthBehaviour == 'RoundUp' or \
            endOfMonthBehaviour == 'RoundOut' and months > 0 or \
            endOfMonthBehaviour == 'RoundIn' and months < 0:
            # we don't need to worry about incrementing the year
            # because there will never be a day in December > 31
            month += 1
            day = 1
    return datetime.date(year, month, day)


>>> from calendar import monthrange
>>> import datetime
>>> add_months(datetime.datetime(2016, 1, 31), 1)
datetime.date(2016, 3, 1)
>>> add_months(datetime.datetime(2016, 1, 31), -2)
datetime.date(2015, 12, 1)
>>> add_months(datetime.datetime(2016, 1, 31), -2, 'RoundDown')
datetime.date(2015, 11, 30)

    def addDay(date, number):
        for i in range(number)
            #try to add a day
            try:
                date = date.replace(day = date.day + 1)
            #in case it's impossible ex:january 32nd add a month and restart at day 1
            except:
                #add month part
                try:
                    date = date.replace(month = date.month +1, day = 1)
                except:
                    date = date.replace(year = date.year +1, month = 1, day = 1)

写给还在读这篇文章的人。我认为这段代码更加清晰，特别是与使用模(%)的代码相比。

很抱歉有语法错误，英语不是我的主要语言

import time

def add_month(start_time, months):  

        ret = time.strptime(start_time, '%Y-%m-%d')
        t = list(ret)

        t[1] += months

        if t[1] > 12:
            t[0] += 1 + int(months / 12)

            t[1] %= 12

        return int(time.mktime(tuple(t)))