我的实现基于taleinat的答案:

import datetime
import calendar

def add_months(orig_date, month_count = 1):
    while month_count > 12:
        month_count -= 12
        orig_date = add_months(orig_date, 12)
    new_year = orig_date.year
    new_month = orig_date.month + month_count
    # note: in datetime.date, months go from 1 to 12
    if new_month > 12:
        new_year += 1
        new_month -= 12

    last_day_of_month = calendar.monthrange(new_year, new_month)[1]
    new_day = min(orig_date.day, last_day_of_month)

    return orig_date.replace(year=new_year, month=new_month, day=new_day)

有了这个功能，你可以添加任意多的月份。

from datetime import date
dt = date(2021, 1, 31)

print(add_months(dt, 49))

返回 2025-02-28

修改了Johannes Wei在1new_month = 121情况下的答案。这对我来说非常合适。月份可以是正的，也可以是负的。

def addMonth(d,months=1):
    year, month, day = d.timetuple()[:3]
    new_month = month + months
    return datetime.date(year + ((new_month-1) / 12), (new_month-1) % 12 +1, day)

我使用这个函数来更改年和月，但保留日:

def replace_month_year(date1, year2, month2):
    try:
        date2 = date1.replace(month = month2, year = year2)
    except:
        date2 = datetime.date(year2, month2 + 1, 1) - datetime.timedelta(days=1)
    return date2

你应该这样写:

new_year = my_date.year + (my_date.month + 6) / 12
new_month = (my_date.month + 6) % 12
new_date = replace_month_year(my_date, new_year, new_month)

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)

这个怎么样?没有使用其他库(dateutil)或timedelta? 基于vartec的回答，我这样做了，我相信它是有效的:

import datetime

today = datetime.date.today()
six_months_from_today = datetime.date(today.year + (today.month + 6)/12, (today.month + 6) % 12, today.day)

我尝试使用timedelta，但因为它是计算天数的，365/2或6*356/12并不总是转换为6个月，而是182天。如。

day = datetime.date(2015, 3, 10)
print day
>>> 2015-03-10

print (day + datetime.timedelta(6*365/12))
>>> 2015-09-08

我相信我们通常会假设某一天的6个月将在6个月后的同一天登陆(即2015-03-10—> 2015-09-10，而不是2015-09-08)

我希望这对你有帮助。

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)