使用下面给出的函数，你可以得到x个月之后/之前的日期。

from datetime import date

def next_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 = next_month(today, mm)
        print(next_date)

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)

我经常需要一个月的最后一天来保持上个月的最后一天。为了解决这个问题，我在计算前加一天，然后在返回前再减去它。

from datetime import date, timedelta

# it's a lot faster with a constant day
DAY = timedelta(1)

def add_month(a_date, months):
    "Add months to date and retain last day in month."
    next_day = a_date + DAY
    # calculate new year and month
    m_sum = next_day.month + months - 1
    y = next_day.year + m_sum // 12
    m = m_sum % 12 + 1
    try:
        return date(y, m, next_day.day) - DAY
    except ValueError:
        # on fail return last day in month
        # can't fail on december so I don't bother changing the year
        return date(y, m + 1, 1) - DAY

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。

一个简单的建议是《绿箭侠》

PIP安装箭头

>>> import arrow

>>> arrow.now().date()
datetime.date(2019, 6, 28)
>>> arrow.now().shift(months=6).date()
datetime.date(2019, 12, 28)

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))