我们可能应该使用dateutil。relativedelta

然而，出于学术兴趣，我只想补充一点，在我发现它之前，我打算用这个:

亿: 今天代表。年度+ (K .今日month + 6) / 12 / (K . today . 12 month + 5%) + 1, K .今日day)。 except: 今天代表。年+ (K.today.month+6)//12 (K.today.month+6)%12+1, 1) -时间轴(天= 1)

它看起来很简单，但仍然可以捕捉到所有的问题，如29、30、31

它也适用于- 6 MTHS通过执行-timedelta

别被k弄糊涂了，今天它只是我程序中的一个变量

我找不到这个问题的确切解决方案，所以我将发布我的解决方案，以防使用标准日历和datetime库可能有任何帮助。这适用于添加和减去月份，并考虑月末滚动和最后一个月比第一个月天数少的情况。如果你正在寻找更复杂的操作，我还有一个更通用的解决方案，它添加了定期间隔(天，月，年，季度，学期等)，如:“1m”，“-9m”，“-1.5y”，“-3q”，“1s”等。

from datetime import datetime
from calendar import monthrange
def date_bump_months(start_date, months):
    """
    bumps months back and forth. 
    --> if initial date is end-of-month, i will move to corresponding month-end
    --> ir inital date.day is greater than end of month of final date, it casts it to momth-end
    """
    signbit = -1 if months < 0 else 1
    d_year, d_month = divmod(abs(months),12)    
    end_year = start_date.year + d_year*signbit 
    end_month = 0
    if signbit ==-1:            
        if d_month < start_date.month:
            end_month = start_date.month - d_month
        else:
            end_year -=1
            end_month = 12 - (d_month - start_date.month)
    else:
        end_month +=start_date.month
        if end_month  > 12:
            end_year +=1
            end_month -=12
    # check if we are running end-of-month dates
    eom_run = monthrange(start_date.year, start_date.month)[1]==start_date.day
    eom_month = monthrange((end_year), (end_month))[1]
    if eom_run:
        end_day = eom_month 
    else:
        end_day = min(start_date.day, eom_month )    
    return date(end_year, end_month, end_day)

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)

当我需要添加几个月或几年的时间，并且不想导入更多的库时，我就会这样做。

import datetime
__author__ = 'Daniel Margarido'


# Check if the int given year is a leap year
# return true if leap year or false otherwise
def is_leap_year(year):
    if (year % 4) == 0:
        if (year % 100) == 0:
            if (year % 400) == 0:
                return True
            else:
                return False
        else:
            return True
    else:
        return False


THIRTY_DAYS_MONTHS = [4, 6, 9, 11]
THIRTYONE_DAYS_MONTHS = [1, 3, 5, 7, 8, 10, 12]

# Inputs -> month, year Booth integers
# Return the number of days of the given month
def get_month_days(month, year):
    if month in THIRTY_DAYS_MONTHS:   # April, June, September, November
        return 30
    elif month in THIRTYONE_DAYS_MONTHS:   # January, March, May, July, August, October, December
        return 31
    else:   # February
        if is_leap_year(year):
            return 29
        else:
            return 28

# Checks the month of the given date
# Selects the number of days it needs to add one month
# return the date with one month added
def add_month(date):
    current_month_days = get_month_days(date.month, date.year)
    next_month_days = get_month_days(date.month + 1, date.year)

    delta = datetime.timedelta(days=current_month_days)
    if date.day > next_month_days:
        delta = delta - datetime.timedelta(days=(date.day - next_month_days) - 1)

    return date + delta


def add_year(date):
    if is_leap_year(date.year):
        delta = datetime.timedelta(days=366)
    else:
        delta = datetime.timedelta(days=365)

    return date + delta


# Validates if the expected_value is equal to the given value
def test_equal(expected_value, value):
    if expected_value == value:
        print "Test Passed"
        return True

    print "Test Failed : " + str(expected_value) + " is not equal to " str(value)
    return False

# Test leap year
print "---------- Test leap year ----------"
test_equal(True, is_leap_year(2012))
test_equal(True, is_leap_year(2000))
test_equal(False, is_leap_year(1900))
test_equal(False, is_leap_year(2002))
test_equal(False, is_leap_year(2100))
test_equal(True, is_leap_year(2400))
test_equal(True, is_leap_year(2016))

# Test add month
print "---------- Test add month ----------"
test_equal(datetime.date(2016, 2, 1), add_month(datetime.date(2016, 1, 1)))
test_equal(datetime.date(2016, 6, 16), add_month(datetime.date(2016, 5, 16)))
test_equal(datetime.date(2016, 3, 15), add_month(datetime.date(2016, 2, 15)))
test_equal(datetime.date(2017, 1, 12), add_month(datetime.date(2016, 12, 12)))
test_equal(datetime.date(2016, 3, 1), add_month(datetime.date(2016, 1, 31)))
test_equal(datetime.date(2015, 3, 1), add_month(datetime.date(2015, 1, 31)))
test_equal(datetime.date(2016, 3, 1), add_month(datetime.date(2016, 1, 30)))
test_equal(datetime.date(2016, 4, 30), add_month(datetime.date(2016, 3, 30)))
test_equal(datetime.date(2016, 5, 1), add_month(datetime.date(2016, 3, 31)))

# Test add year
print "---------- Test add year ----------"
test_equal(datetime.date(2016, 2, 2), add_year(datetime.date(2015, 2, 2)))
test_equal(datetime.date(2001, 2, 2), add_year(datetime.date(2000, 2, 2)))
test_equal(datetime.date(2100, 2, 2), add_year(datetime.date(2099, 2, 2)))
test_equal(datetime.date(2101, 2, 2), add_year(datetime.date(2100, 2, 2)))
test_equal(datetime.date(2401, 2, 2), add_year(datetime.date(2400, 2, 2)))

只需创建一个datetime.date()对象，调用add_month(date)来添加一个月，调用add_year(date)来添加一个年。

“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))

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)