我插话晚了，但是

查看肯雷茨玛雅模块，

https://github.com/kennethreitz/maya

像这样的方法可能会对你有帮助，只需将小时=1改为天=1或年=1

>>> from maya import MayaInterval

# Create an event that is one hour long, starting now.
>>> event_start = maya.now()
>>> event_end = event_start.add(hours=1)

>>> event = MayaInterval(start=event_start, end=event_end)

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

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)))

这是我想到的。它移动了正确的月份和年份，但忽略了天数(这是我的情况所需要的)。

import datetime

month_dt = 4
today = datetime.date.today()
y,m = today.year, today.month
m += month_dt-1
year_dt = m//12
new_month = m%12
new_date = datetime.date(y+year_dt, new_month+1, 1)

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)

我找不到这个问题的确切解决方案，所以我将发布我的解决方案，以防使用标准日历和datetime库可能有任何帮助。这适用于添加和减去月份，并考虑月末滚动和最后一个月比第一个月天数少的情况。如果你正在寻找更复杂的操作，我还有一个更通用的解决方案，它添加了定期间隔(天，月，年，季度，学期等)，如:“1m”，“-9m”，“-1.5y”，“-3q”，“1s”等。

from datetime import datetime
from calendar import monthrange
def date_bump_months(start_date, months):
    """
    bumps months back and forth. 
    --> if initial date is end-of-month, i will move to corresponding month-end
    --> ir inital date.day is greater than end of month of final date, it casts it to momth-end
    """
    signbit = -1 if months < 0 else 1
    d_year, d_month = divmod(abs(months),12)    
    end_year = start_date.year + d_year*signbit 
    end_month = 0
    if signbit ==-1:            
        if d_month < start_date.month:
            end_month = start_date.month - d_month
        else:
            end_year -=1
            end_month = 12 - (d_month - start_date.month)
    else:
        end_month +=start_date.month
        if end_month  > 12:
            end_year +=1
            end_month -=12
    # check if we are running end-of-month dates
    eom_run = monthrange(start_date.year, start_date.month)[1]==start_date.day
    eom_month = monthrange((end_year), (end_month))[1]
    if eom_run:
        end_day = eom_month 
    else:
        end_day = min(start_date.day, eom_month )    
    return date(end_year, end_month, end_day)