在python中添加年份

如果我想在程序中添加 100 年，为什么它显示错误的日期？

import datetime
stringDate= "January 10, 1920"
dateObject= datetime.datetime.strptime(stringDate, "%B %d, %Y")
endDate= dateObject+datetime.timedelta(days=100*365)
print dateObject.date()
print endDate.date()

一年的秒数不固定 https://stackoverflow.com/a/32658742/4279. 你知道一年有多少天吗？再想一想。 https://www.washingtonpost.com/news/speaking-of-science/wp/2017/02/24/think-you-know-how-many-days-are-in-a-year-think-again/?utm_term=.f05c01f51363

要执行周期（日历）算术，您可以使用dateutil.relativedelta http://dateutil.readthedocs.org/en/latest/relativedelta.html:

#!/usr/bin/env python
from datetime import date
from dateutil.relativedelta import relativedelta # $ pip install python-dateutil

print(date(1920, 1, 10) + relativedelta(years=+100))
# -> 2020-01-10

要明白，为什么d.replace(year=d.year + 100)失败，请考虑：

print(date(2000, 2, 29) + relativedelta(years=+100))
2100-02-28

请注意2100不是闰年，而2000是闰年。

如果您想要添加的唯一单位是年，那么您可以仅使用 stdlib 来实现它：

from calendar import isleap

def add_years(d, years):
    new_year = d.year + years
    try:
        return d.replace(year=new_year)
    except ValueError:
        if (d.month == 2 and d.day == 29 and # leap day
            isleap(d.year) and not isleap(new_year)):
            return d.replace(year=new_year, day=28)
        raise

Example:

from datetime import date

print(add_years(date(1920, 1, 10), 100))
# -> 2020-01-10
print(add_years(date(2000, 2, 29), 100))
# -> 2100-02-28
print(add_years(date(2000, 2, 29), 4))
# -> 2004-02-29