两个日期之间的天数？

发布于 2024-12-17 17:59:10 字数 120 浏览 0 评论 0原文

查看两个日期之间过去了多少整天的最短方法是什么？这就是我现在正在做的事情。

math.floor((b - a).total_seconds()/float(86400))

原文

What's the shortest way to see how many full days have passed between two dates?
Here's what I'm doing now.

math.floor((b - a).total_seconds()/float(86400))

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无所的.畏惧 2024-12-24 17:59:10

假设您实际上有两个日期对象，您可以从另一个中减去一个并查询结果 timedelta 对象表示天数：

>>> from datetime import date
>>> a = date(2011,11,24)
>>> b = date(2011,11,17)
>>> a-b
datetime.timedelta(7)
>>> (a-b).days
7

它也适用于日期时间 - 我认为它会向下舍入到最近的一天：

>>> from datetime import datetime
>>> a = datetime(2011,11,24,0,0,0)
>>> b = datetime(2011,11,17,23,59,59)
>>> a-b
datetime.timedelta(6, 1)
>>> (a-b).days
6

Assuming you’ve literally got two date objects, you can subtract one from the other and query the resulting timedelta object for the number of days:

>>> from datetime import date
>>> a = date(2011,11,24)
>>> b = date(2011,11,17)
>>> a-b
datetime.timedelta(7)
>>> (a-b).days
7

And it works with datetimes too — I think it rounds down to the nearest day:

>>> from datetime import datetime
>>> a = datetime(2011,11,24,0,0,0)
>>> b = datetime(2011,11,17,23,59,59)
>>> a-b
datetime.timedelta(6, 1)
>>> (a-b).days
6

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没︽人懂的悲伤 2024-12-24 17:59:10

您是指完整的日历日，还是 24 小时组？

仅 24 小时，假设您使用 Python 的日期时间，那么 timedelta 对象已经有一个days 属性：

days = (a - b).days

对于日历日，您需要将 a 向下舍入到最近的一天，将 b 向上舍入到最近的一天，去掉两边的部分日期：

roundedA = a.replace(hour = 0, minute = 0, second = 0, microsecond = 0)
roundedB = b.replace(hour = 0, minute = 0, second = 0, microsecond = 0)
days = (roundedA - roundedB).days

Do you mean full calendar days, or groups of 24 hours?

For simply 24 hours, assuming you're using Python's datetime, then the timedelta object already has a days property:

days = (a - b).days

For calendar days, you'll need to round a down to the nearest day, and b up to the nearest day, getting rid of the partial day on either side:

roundedA = a.replace(hour = 0, minute = 0, second = 0, microsecond = 0)
roundedB = b.replace(hour = 0, minute = 0, second = 0, microsecond = 0)
days = (roundedA - roundedB).days

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杀手六號 2024-12-24 17:59:10

尝试：

(b-a).days

我尝试使用 datetime.date 类型的 b 和 a。

Try:

(b-a).days

I tried with b and a of type datetime.date.

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醉态萌生 2024-12-24 17:59:10

参考我对其他答案的评论。这就是我根据 24 小时和日历日计算天数差异的方法。 days 属性适用于 24 小时，并且该函数最适合日历检查。

from datetime import timedelta, datetime

def cal_days_diff(a,b):

    A = a.replace(hour = 0, minute = 0, second = 0, microsecond = 0)
    B = b.replace(hour = 0, minute = 0, second = 0, microsecond = 0)
    return (A - B).days

if __name__ == '__main__':

    x = datetime(2013, 06, 18, 16, 00)
    y = datetime(2013, 06, 19, 2, 00)

    print (y - x).days          # 0
    print cal_days_diff(y, x)   # 1 

    z = datetime(2013, 06, 20, 2, 00)

    print (z - x).days          # 1
    print cal_days_diff(z, x)   # 2

Referencing my comments on other answers. This is how I would work out the difference in days based on 24 hours and calender days. the days attribute works well for 24 hours and the function works best for calendar checks.

from datetime import timedelta, datetime

def cal_days_diff(a,b):

    A = a.replace(hour = 0, minute = 0, second = 0, microsecond = 0)
    B = b.replace(hour = 0, minute = 0, second = 0, microsecond = 0)
    return (A - B).days

if __name__ == '__main__':

    x = datetime(2013, 06, 18, 16, 00)
    y = datetime(2013, 06, 19, 2, 00)

    print (y - x).days          # 0
    print cal_days_diff(y, x)   # 1 

    z = datetime(2013, 06, 20, 2, 00)

    print (z - x).days          # 1
    print cal_days_diff(z, x)   # 2

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