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)))

Dateutil包实现了这样的功能。但要知道，这将是天真的，因为其他人已经指出。

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))

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))

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)

我使用replace()方法并编写了这个递归函数。dt是一个日期时间。datetime对象:

def month_timedelta(dt, m):
    y = m // 12
    dm = m % 12
    if y == 0:
        if dt.month + m <= 12:
            return dt.replace(month = dt.month + m)
        else:
            dy = (dt.month + m) // 12
            ndt = dt.replace(year=dt.year + dy)
            return ndt.replace(month=(ndt.month + m) % 12)
    else:
        return month_timedelta(dt.replace(year=dt.year + y),dm)