最干净和最Pythonic的方式来获取明天的日期？

Question

什么是最干净，最Pythonic的方式来获取明天的日期？必须有一个更好的办法，而不是一个添加到天，处理日在月底等

Answer 1

datetime.date.today() + datetime.timedelta(days=1)应该做的伎俩

Answer 2

timedelta可以处理增加天，秒，微秒，毫秒，几分钟，几小时或几周。

>>> import datetime 
>>> today = datetime.date.today() 
>>> today 
datetime.date(2009, 10, 1) 
>>> today + datetime.timedelta(days=1) 
datetime.date(2009, 10, 2) 
>>> datetime.date(2009,10,31) + datetime.timedelta(hours=24) 
datetime.date(2009, 11, 1) 

如问一个评论，飞跃天不成问题：

>>> datetime.date(2004, 2, 28) + datetime.timedelta(days=1) 
datetime.date(2004, 2, 29) 
>>> datetime.date(2004, 2, 28) + datetime.timedelta(days=2) 
datetime.date(2004, 3, 1) 
>>> datetime.date(2005, 2, 28) + datetime.timedelta(days=1) 
datetime.date(2005, 3, 1) 

Answer 3

leap seconds寿无处理：

>>> from datetime import datetime, timedelta 
>>> dt = datetime(2008,12,31,23,59,59) 
>>> str(dt) 
'2008-12-31 23:59:59' 
>>> # leap second was added at the end of 2008, 
>>> # adding one second should create a datetime 
>>> # of '2008-12-31 23:59:60' 
>>> str(dt+timedelta(0,1)) 
'2009-01-01 00:00:00' 
>>> str(dt+timedelta(0,2)) 
'2009-01-01 00:00:01' 

织补。

编辑 - @马克：文档说“是”，但代码说“没有那么多”：

>>> time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S") 
(2008, 12, 31, 23, 59, 60, 2, 366, -1) 
>>> time.mktime(time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S")) 
1230789600.0 
>>> time.gmtime(time.mktime(time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S"))) 
(2009, 1, 1, 6, 0, 0, 3, 1, 0) 
>>> time.localtime(time.mktime(time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S"))) 
(2009, 1, 1, 0, 0, 0, 3, 1, 0) 

我认为gmtime的或本地时间将采取由mktime返回，给我的价值返回原始元组，其中60为秒数。而这个测试表明，这些闰秒可慢慢消失......

>>> a = time.mktime(time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S")) 
>>> b = time.mktime(time.strptime("2009-01-01 00:00:00","%Y-%m-%d %H:%M:%S")) 
>>> a,b 
(1230789600.0, 1230789600.0) 
>>> b-a 
0.0 

Answer 4

连基本的time模块可以处理这个问题：

import time 
time.localtime(time.time() + 24*3600)