Ale*_*lex 153 python time datediff
必须有一种更简单的方法来做到这一点.我有想要每隔一段时间刷新的对象,所以我想记录它们的创建时间,检查当前的时间戳,并根据需要进行刷新.
datetime.datetime已被证明是困难的,我不想深入了解ctime库.这种事情有什么比较容易的吗?
Ric*_*ell 445
如果要计算两个已知日期之间的差异,请使用total_seconds
如下所示:
import datetime as dt
a = dt.datetime(2013,12,30,23,59,59)
b = dt.datetime(2013,12,31,23,59,59)
(b-a).total_seconds()
Run Code Online (Sandbox Code Playgroud)
86400.0
#note that seconds doesn't give you what you want:
(b-a).seconds
Run Code Online (Sandbox Code Playgroud)
0
mat*_*eek 31
import time
current = time.time()
...job...
end = time.time()
diff = end - current
Run Code Online (Sandbox Code Playgroud)
那会对你有用吗?
Jar*_*die 14
>>> from datetime import datetime
>>> a = datetime.now()
# wait a bit
>>> b = datetime.now()
>>> d = b - a # yields a timedelta object
>>> d.seconds
7
Run Code Online (Sandbox Code Playgroud)
(7将是你等待一点时间的任何时间)
我发现datetime.datetime非常有用,所以如果你遇到过一个复杂或尴尬的场景,请告诉我们.
编辑:感谢@WoLpH指出,并不总是需要频繁刷新,以便日期时间紧密相连.通过计算增量中的天数,您可以处理更长的时间戳差异:
>>> a = datetime(2010, 12, 5)
>>> b = datetime(2010, 12, 7)
>>> d = b - a
>>> d.seconds
0
>>> d.days
2
>>> d.seconds + d.days * 86400
172800
Run Code Online (Sandbox Code Playgroud)
Pra*_*aur 12
我们有Python 2.7的函数total_seconds()请参阅下面的python 2.6代码
import datetime
import time
def diffdates(d1, d2):
#Date format: %Y-%m-%d %H:%M:%S
return (time.mktime(time.strptime(d2,"%Y-%m-%d %H:%M:%S")) -
time.mktime(time.strptime(d1, "%Y-%m-%d %H:%M:%S")))
d1 = datetime.now()
d2 = datetime.now() + timedelta(days=1)
diff = diffdates(d1, d2)
Run Code Online (Sandbox Code Playgroud)