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)

我知道这个问题已经有很多答案，但是使用collections.deque和rotate()方法，可以创建一个函数，该函数接受一个datetime对象作为输入，然后输出一个比当前对象晚一个“业务月”的新datetime对象。如果该月的某一天在下个月不存在，则减去1，直到它到达该月的有效日期，然后返回该对象。

import collections
import datetime

def next_month(dt: datetime.datetime):
    month_list = list(range(1, 12 + 1))
    months = collections.deque(month_list)
    while True:
        this_month = list(months)[0]
        if dt.month == this_month:
            break
        months.rotate(-1)
    months.rotate(-1)
    month_plus = list(months)[0]
    for i in range(4):
        try:
            return dt.replace(month=month_plus, day=dt.day - i)
        except ValueError:
            continue

使用itertools.cycle也可以得到相同的结果。

import datetime
import itertools

def next_month(dt: datetime.datetime):
    month_list = list(range(1, 12 + 1))
    month = itertools.cycle(month_list)
    while True:
        if next(month) == dt.month:
            break
    month_plus = next(month)
    for i in range(4):
        try:
            return dt.replace(month=month_plus, day=dt.day - i)
        except ValueError:
            continue

我发现这个解决方法很好。(使用python-dateutil扩展名)

from datetime import date
from dateutil.relativedelta import relativedelta

six_months = date.today() + relativedelta(months=+6)

这种方法的优势在于，它可以处理28天、30天、31天的问题。这在处理业务规则和场景(比如发票生成等)时非常有用。

$ date(2010,12,31)+relativedelta(months=+1)
  datetime.date(2011, 1, 31)

$ date(2010,12,31)+relativedelta(months=+2)
  datetime.date(2011, 2, 28)

Python的datetime没有直接的方法来做到这一点。

在python-dateutil中查看relativedelta类型。它允许您指定以月为单位的时间增量。

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))

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)