嗯，这取决于你说的6个月后的日期。

使用自然月份: Inc = 6 年=年+(月+ inc - 1) // 12 月份=(月份+ inc - 1) % 12 + 1 用银行家的定义，6*30: 日期+=日期时间。Timedelta (6 * 30)

    def addDay(date, number):
        for i in range(number)
            #try to add a day
            try:
                date = date.replace(day = date.day + 1)
            #in case it's impossible ex:january 32nd add a month and restart at day 1
            except:
                #add month part
                try:
                    date = date.replace(month = date.month +1, day = 1)
                except:
                    date = date.replace(year = date.year +1, month = 1, day = 1)

写给还在读这篇文章的人。我认为这段代码更加清晰，特别是与使用模(%)的代码相比。

很抱歉有语法错误，英语不是我的主要语言

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)

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)

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

“Python -dateutil”(外部扩展)是一个很好的解决方案，但你可以使用内置的Python模块(datetime和datetime)来实现它。

我做了一个简短的代码，来解决它(处理年，月和日)

(运行:Python 3.8.2)

from datetime import datetime
from calendar import monthrange

# Time to increase (in months)
inc = 12

# Returns mod of the division for 12 (months)
month = ((datetime.now().month + inc) % 12) or 1

# Increase the division by 12 (months), if necessary (+ 12 months increase)
year = datetime.now().year + int((month + inc) / 12)

# (IF YOU DON'T NEED DAYS,CAN REMOVE THE BELOW CODE)
# Returns the same day in new month, or the maximum day of new month
day = min(datetime.now().day,monthrange(year, month)[1])

print("Year: {}, Month: {}, Day: {}".format(year, month, day))