使用Python标准库，即没有dateutil或其他，并解决“2月31日”问题:

import datetime
import calendar

def add_months(date, months):
    months_count = date.month + months

    # Calculate the year
    year = date.year + int(months_count / 12)

    # Calculate the month
    month = (months_count % 12)
    if month == 0:
        month = 12

    # Calculate the day
    day = date.day
    last_day_of_month = calendar.monthrange(year, month)[1]
    if day > last_day_of_month:
        day = last_day_of_month

    new_date = datetime.date(year, month, day)
    return new_date

测试:

>>>date = datetime.date(2018, 11, 30)

>>>print(date, add_months(date, 3))
(datetime.date(2018, 11, 30), datetime.date(2019, 2, 28))

>>>print(date, add_months(date, 14))
(datetime.date(2018, 12, 31), datetime.date(2020, 2, 29))

我认为这样做会比手动添加天数更安全:

import datetime
today = datetime.date.today()

def addMonths(dt, months = 0):
    new_month = months + dt.month
    year_inc = 0
    if new_month>12:
        year_inc +=1
        new_month -=12
    return dt.replace(month = new_month, year = dt.year+year_inc)

newdate = addMonths(today, 6)

当我需要添加几个月或几年的时间，并且不想导入更多的库时，我就会这样做。

import datetime
__author__ = 'Daniel Margarido'


# Check if the int given year is a leap year
# return true if leap year or false otherwise
def is_leap_year(year):
    if (year % 4) == 0:
        if (year % 100) == 0:
            if (year % 400) == 0:
                return True
            else:
                return False
        else:
            return True
    else:
        return False


THIRTY_DAYS_MONTHS = [4, 6, 9, 11]
THIRTYONE_DAYS_MONTHS = [1, 3, 5, 7, 8, 10, 12]

# Inputs -> month, year Booth integers
# Return the number of days of the given month
def get_month_days(month, year):
    if month in THIRTY_DAYS_MONTHS:   # April, June, September, November
        return 30
    elif month in THIRTYONE_DAYS_MONTHS:   # January, March, May, July, August, October, December
        return 31
    else:   # February
        if is_leap_year(year):
            return 29
        else:
            return 28

# Checks the month of the given date
# Selects the number of days it needs to add one month
# return the date with one month added
def add_month(date):
    current_month_days = get_month_days(date.month, date.year)
    next_month_days = get_month_days(date.month + 1, date.year)

    delta = datetime.timedelta(days=current_month_days)
    if date.day > next_month_days:
        delta = delta - datetime.timedelta(days=(date.day - next_month_days) - 1)

    return date + delta


def add_year(date):
    if is_leap_year(date.year):
        delta = datetime.timedelta(days=366)
    else:
        delta = datetime.timedelta(days=365)

    return date + delta


# Validates if the expected_value is equal to the given value
def test_equal(expected_value, value):
    if expected_value == value:
        print "Test Passed"
        return True

    print "Test Failed : " + str(expected_value) + " is not equal to " str(value)
    return False

# Test leap year
print "---------- Test leap year ----------"
test_equal(True, is_leap_year(2012))
test_equal(True, is_leap_year(2000))
test_equal(False, is_leap_year(1900))
test_equal(False, is_leap_year(2002))
test_equal(False, is_leap_year(2100))
test_equal(True, is_leap_year(2400))
test_equal(True, is_leap_year(2016))

# Test add month
print "---------- Test add month ----------"
test_equal(datetime.date(2016, 2, 1), add_month(datetime.date(2016, 1, 1)))
test_equal(datetime.date(2016, 6, 16), add_month(datetime.date(2016, 5, 16)))
test_equal(datetime.date(2016, 3, 15), add_month(datetime.date(2016, 2, 15)))
test_equal(datetime.date(2017, 1, 12), add_month(datetime.date(2016, 12, 12)))
test_equal(datetime.date(2016, 3, 1), add_month(datetime.date(2016, 1, 31)))
test_equal(datetime.date(2015, 3, 1), add_month(datetime.date(2015, 1, 31)))
test_equal(datetime.date(2016, 3, 1), add_month(datetime.date(2016, 1, 30)))
test_equal(datetime.date(2016, 4, 30), add_month(datetime.date(2016, 3, 30)))
test_equal(datetime.date(2016, 5, 1), add_month(datetime.date(2016, 3, 31)))

# Test add year
print "---------- Test add year ----------"
test_equal(datetime.date(2016, 2, 2), add_year(datetime.date(2015, 2, 2)))
test_equal(datetime.date(2001, 2, 2), add_year(datetime.date(2000, 2, 2)))
test_equal(datetime.date(2100, 2, 2), add_year(datetime.date(2099, 2, 2)))
test_equal(datetime.date(2101, 2, 2), add_year(datetime.date(2100, 2, 2)))
test_equal(datetime.date(2401, 2, 2), add_year(datetime.date(2400, 2, 2)))

只需创建一个datetime.date()对象，调用add_month(date)来添加一个月，调用add_year(date)来添加一个年。

在这个函数中，n可以是正数，也可以是负数。

def addmonth(d, n):
    n += 1
    dd = datetime.date(d.year + n/12, d.month + n%12, 1)-datetime.timedelta(1)
    return datetime.date(dd.year, dd.month, min(d.day, dd.day))

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)

修改了AddMonths()在Zope中使用和处理无效的天数:

def AddMonths(d,x):
    days_of_month = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    newmonth = ((( d.month() - 1) + x ) % 12 ) + 1
    newyear  = d.year() + ((( d.month() - 1) + x ) // 12 ) 
    if d.day() > days_of_month[newmonth-1]:
      newday = days_of_month[newmonth-1]
    else:
      newday = d.day() 
    return DateTime( newyear, newmonth, newday)