我有一个更好的办法来解决“2月31日”的问题:

def add_months(start_date, months):
    import calendar

    year = start_date.year + (months / 12)
    month = start_date.month + (months % 12)
    day = start_date.day

    if month > 12:
        month = month % 12
        year = year + 1

    days_next = calendar.monthrange(year, month)[1]
    if day > days_next:
        day = days_next

    return start_date.replace(year, month, day)

我认为它也适用于负数(减去月份)，但我还没有对此进行过多测试。

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

获得x个月之后或之前的下一个日期的一般函数。

from datetime import date

def after_month(given_date, month):
    yyyy = int(((given_date.year * 12 + given_date.month) + month)/12)
    mm = int(((given_date.year * 12 + given_date.month) + month)%12)

    if mm == 0:
        yyyy -= 1
        mm = 12
    return given_date.replace(year=yyyy, month=mm)


if __name__ == "__main__":
    today = date.today()
    print(today)

    for mm in [-12, -1, 0, 1, 2, 12, 20 ]:
        next_date = after_month(today, mm)
        print(next_date)

按月开始计算:

from datetime import timedelta
from dateutil.relativedelta import relativedelta

end_date = start_date + relativedelta(months=delta_period) + timedelta(days=-delta_period)

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)

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)