最乾淨和最Pythonic的方式來獲取明天的日期？

Question

什麼是最乾淨，最Pythonic的方式來獲取明天的日期？必須有一個更好的辦法，而不是一個添加到天，處理日在月底等

Answer 1

datetime.date.today() + datetime.timedelta(days=1)應該做的伎倆

Answer 2

timedelta可以處理增加天，秒，微秒，毫秒，幾分鐘，幾小時或幾周。

>>> 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) 

如問一個評論，飛躍天不成問題：

>>> 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壽無處理：

>>> 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' 

織補。

編輯 - @馬克：文檔說「是」，但代碼說「沒有那麼多」：

>>> 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) 

我認爲gmtime的或本地時間將採取由mktime返回，給我的價值返回原始元組，其中60爲秒數。而這個測試表明，這些閏秒可慢慢消失......

>>> 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

連基本的time模塊可以處理這個問題：

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