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)))

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)

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)

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。

这是我想到的。它移动了正确的月份和年份，但忽略了天数(这是我的情况所需要的)。

import datetime

month_dt = 4
today = datetime.date.today()
y,m = today.year, today.month
m += month_dt-1
year_dt = m//12
new_month = m%12
new_date = datetime.date(y+year_dt, new_month+1, 1)

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)