将timedelta转换为总秒数

Question

我有时差

time1 = datetime.datetime.fromtimestamp(time.mktime(time.gmtime()))
...
time2 = datetime.datetime.fromtimestamp(time.mktime(time.gmtime()))
diff = time2 - time1

现在,我如何找到通过的总秒数？diff.seconds不计算天数.我可以:

diff.seconds + diff.days * 24 * 3600

有没有内置的方法？

Answer 1

使用timedelta.total_seconds().

>>> import datetime
>>> datetime.timedelta(seconds=24*60*60).total_seconds()
86400.0

Answer 2

你的datetime.datetime.fromtimestamp(time.mktime(time.gmtime()))表达方式有这样或那样的问题 .

(1)如果你需要的只是两个瞬间的差异,time.time()那么这个工作非常简单.

(2)如果你将这些时间戳用于其他目的,你需要考虑你在做什么,因为结果有很大的气味:

gmtime()返回UTC中的时间元组,但mktime()期望在当地时间的时间元组.

我在澳大利亚墨尔本,标准TZ是UTC + 10,但夏令时仍然有效,直到明天早上,所以它是UTC + 11.当我执行以下操作时,它是2011-04-02T20:31当地时间...... UTC是2011-04-02T09:31

>>> import time, datetime
>>> t1 = time.gmtime()
>>> t2 = time.mktime(t1)
>>> t3 = datetime.datetime.fromtimestamp(t2)
>>> print t0
1301735358.78
>>> print t1
time.struct_time(tm_year=2011, tm_mon=4, tm_mday=2, tm_hour=9, tm_min=31, tm_sec=3, tm_wday=5, tm_yday=92, tm_isdst=0) ### this is UTC
>>> print t2
1301700663.0
>>> print t3
2011-04-02 10:31:03 ### this is UTC+1
>>> tt = time.time(); print tt
1301736663.88
>>> print datetime.datetime.now()
2011-04-02 20:31:03.882000 ### UTC+11, my local time
>>> print datetime.datetime(1970,1,1) + datetime.timedelta(seconds=tt)
2011-04-02 09:31:03.880000 ### UTC
>>> print time.localtime()
time.struct_time(tm_year=2011, tm_mon=4, tm_mday=2, tm_hour=20, tm_min=31, tm_sec=3, tm_wday=5, tm_yday=92, tm_isdst=1) ### UTC+11, my local time

您会注意到t3,您的表达式的结果是UTC + 1,它似乎是UTC +(我的本地DST差异)......不是很有意义.datetime.datetime.utcnow()当DST开启/关闭时,你应该考虑使用哪个不会跳过一个小时,并且可能比你提供更精确的time.time()