假设你的datetime变量叫做date:

date=datetime.datetime(year=date.year+int((date.month+6)/12),
                       month=(date.month+6)%13 + (1 if (date.month + 
                       months>12) else 0), day=date.day)

我知道这个问题已经有很多答案，但是使用collections.deque和rotate()方法，可以创建一个函数，该函数接受一个datetime对象作为输入，然后输出一个比当前对象晚一个“业务月”的新datetime对象。如果该月的某一天在下个月不存在，则减去1，直到它到达该月的有效日期，然后返回该对象。

import collections
import datetime

def next_month(dt: datetime.datetime):
    month_list = list(range(1, 12 + 1))
    months = collections.deque(month_list)
    while True:
        this_month = list(months)[0]
        if dt.month == this_month:
            break
        months.rotate(-1)
    months.rotate(-1)
    month_plus = list(months)[0]
    for i in range(4):
        try:
            return dt.replace(month=month_plus, day=dt.day - i)
        except ValueError:
            continue

使用itertools.cycle也可以得到相同的结果。

import datetime
import itertools

def next_month(dt: datetime.datetime):
    month_list = list(range(1, 12 + 1))
    month = itertools.cycle(month_list)
    while True:
        if next(month) == dt.month:
            break
    month_plus = next(month)
    for i in range(4):
        try:
            return dt.replace(month=month_plus, day=dt.day - i)
        except ValueError:
            continue

这个解决方案适用于12月，而本页上的大多数答案都不适用。 在使用模量(%)或整数除法(//)之前，您需要首先将月份从基于1的索引(例如Jan = 1)转移到基于0的索引(例如Jan = 0)，否则11月(11)加上1个月得到12，当找到余数(12% 12)时，得到0。

(不要建议“(月% 12)+ 1”或10月+ 1 = 12月!)

def AddMonths(d,x):
    newmonth = ((( d.month - 1) + x ) % 12 ) + 1
    newyear  = int(d.year + ((( d.month - 1) + x ) / 12 ))
    return datetime.date( newyear, newmonth, d.day)

然而……这并不能解释1月31日+一个月的问题。所以我们回到OP，你说增加一个月是什么意思?一种解决方案是回溯，直到找到一个有效的日期，因为大多数人会假设一月的最后一天加一个月等于二月的最后一天。 这也适用于负数的月份。 证明:

>>> import datetime
>>> AddMonths(datetime.datetime(2010,8,25),1)
datetime.date(2010, 9, 25)
>>> AddMonths(datetime.datetime(2010,8,25),4)
datetime.date(2010, 12, 25)
>>> AddMonths(datetime.datetime(2010,8,25),5)
datetime.date(2011, 1, 25)
>>> AddMonths(datetime.datetime(2010,8,25),13)
datetime.date(2011, 9, 25)
>>> AddMonths(datetime.datetime(2010,8,25),24)
datetime.date(2012, 8, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-1)
datetime.date(2010, 7, 25)
>>> AddMonths(datetime.datetime(2010,8,25),0)
datetime.date(2010, 8, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-12)
datetime.date(2009, 8, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-8)
datetime.date(2009, 12, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-7)
datetime.date(2010, 1, 25)>>> 

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。

我知道这是6个月，但是如果你要添加一个月，答案在谷歌中显示为“在python中添加月”:

import calendar

date = datetime.date.today()    //Or your date

datetime.timedelta(days=calendar.monthrange(date.year,date.month)[1])

这将计算当前月份的天数，并将它们添加到当前日期，如果您在日期上迭代，则使用365/12将AD年的1/12会导致短/长月份出现问题。

下面是一个示例，它允许用户决定如何返回一个日期，其中一天大于一个月中的天数。

def add_months(date, months, endOfMonthBehaviour='RoundUp'):
    assert endOfMonthBehaviour in ['RoundDown', 'RoundIn', 'RoundOut', 'RoundUp'], \
        'Unknown end of month behaviour'
    year = date.year + (date.month + months - 1) / 12
    month = (date.month + months - 1) % 12 + 1
    day = date.day
    last = monthrange(year, month)[1]
    if day > last:
        if endOfMonthBehaviour == 'RoundDown' or \
            endOfMonthBehaviour == 'RoundOut' and months < 0 or \
            endOfMonthBehaviour == 'RoundIn' and months > 0:
            day = last
        elif endOfMonthBehaviour == 'RoundUp' or \
            endOfMonthBehaviour == 'RoundOut' and months > 0 or \
            endOfMonthBehaviour == 'RoundIn' and months < 0:
            # we don't need to worry about incrementing the year
            # because there will never be a day in December > 31
            month += 1
            day = 1
    return datetime.date(year, month, day)


>>> from calendar import monthrange
>>> import datetime
>>> add_months(datetime.datetime(2016, 1, 31), 1)
datetime.date(2016, 3, 1)
>>> add_months(datetime.datetime(2016, 1, 31), -2)
datetime.date(2015, 12, 1)
>>> add_months(datetime.datetime(2016, 1, 31), -2, 'RoundDown')
datetime.date(2015, 11, 30)