在Python中添加年份 [英] Adding years in python

问题描述

如果我想在程序中添加100年，为什么显示错误的日期？

If I want to add 100 years in my program, why is it showing the wrong date?

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()

推荐答案

一年中的秒数不固定。 想知道一年中有多少天吗？再想一想。

要执行周期（日历）算法，可以使用 dateutil.relativedelta ：

To perform period (calendar) arithmetic, you could use dateutil.relativedelta:

#!/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

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

To understand, why d.replace(year=d.year + 100) fails, consider:

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

2100 不是a年，而 2000 是a年。

Notice that 2100 is not a leap year while 2000 is a leap year.

如果要添加的唯一单位是year，则可以仅使用stdlib来实现：

If the only units you want to add is year then you could implement it using only stdlib:

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

示例：

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

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