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))

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)

一个简单的建议是《绿箭侠》

PIP安装箭头

>>> import arrow

>>> arrow.now().date()
datetime.date(2019, 6, 28)
>>> arrow.now().shift(months=6).date()
datetime.date(2019, 12, 28)

还有另一个解决方案，希望有人会喜欢:

def add_months(d, months):
    return d.replace(year=d.year+months//12).replace(month=(d.month+months)%12)

这个解决方案并不适用于所有情况下的第29、30、31天，所以需要更健壮的解决方案(这已经不是那么好了:)):

def add_months(d, months):
    for i in range(4):
        day = d.day - i
        try:
            return d.replace(day=day).replace(year=d.year+int(months)//12).replace(month=(d.month+int(months))%12)
        except:
            pass
    raise Exception("should not happen")

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)