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)

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?

从这个答案中，请参见parsedatetime。下面是代码示例。更多细节:使用许多自然语言-> YYYY-MM-DD转换示例进行单元测试，以及明显的parsedatetime转换挑战/bug。

#!/usr/bin/env python
# -*- coding: utf-8 -*-
import time, calendar
from datetime import date

# from https://github.com/bear/parsedatetime
import parsedatetime as pdt

def print_todays_date():
    todays_day_of_week = calendar.day_name[date.today().weekday()]
    print "today's date = " + todays_day_of_week + ', ' + \
                              time.strftime('%Y-%m-%d')

def convert_date(natural_language_date):
    cal = pdt.Calendar()
    (struct_time_date, success) = cal.parse(natural_language_date)
    if success:
        formal_date = time.strftime('%Y-%m-%d', struct_time_date)
    else:
        formal_date = '(conversion failed)'
    print '{0:12s} -> {1:10s}'.format(natural_language_date, formal_date)

print_todays_date()
convert_date('6 months')

上面的代码从MacOSX机器生成以下内容:

$ ./parsedatetime_simple.py 
today's date = Wednesday, 2015-05-13
6 months     -> 2015-11-13
$ 

“Python -dateutil”(外部扩展)是一个很好的解决方案，但你可以使用内置的Python模块(datetime和datetime)来实现它。

我做了一个简短的代码，来解决它(处理年，月和日)

(运行:Python 3.8.2)

from datetime import datetime
from calendar import monthrange

# Time to increase (in months)
inc = 12

# Returns mod of the division for 12 (months)
month = ((datetime.now().month + inc) % 12) or 1

# Increase the division by 12 (months), if necessary (+ 12 months increase)
year = datetime.now().year + int((month + inc) / 12)

# (IF YOU DON'T NEED DAYS,CAN REMOVE THE BELOW CODE)
# Returns the same day in new month, or the maximum day of new month
day = min(datetime.now().day,monthrange(year, month)[1])

print("Year: {}, Month: {}, Day: {}".format(year, month, day))

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)

另一种解决方案:计算下个月的天数总和，并将结果添加到当前日期。

import calendar
import datetime

def date_from_now(months):
    today = datetime.datetime.today()

    month = today.month
    year = today.year
    sum_days = 0

    for i in range(int(months)):
        month += 1

        if month == 13:
            month = 1
            year += 1

        sum_days += calendar.monthrange(year, month)[1]

    return datetime.date.today() + datetime.timedelta(sum_days)

print(date_from_now(12)) # if to day is 2017-01-01, output: 2019-01-01