必须有一种更简单的方法来做到这一点。我有想要经常刷新的对象,所以我想记录它们的创建时间,检查当前时间戳,并在必要时刷新。
datetime.datetime 已被证明是困难的,我不想深入 ctime 库。这种事情有什么更容易的吗?
如果要计算两个已知日期之间的差异,请total_seconds
像这样使用:
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()
86400.0
#note that seconds doesn't give you what you want:
(b-a).seconds
0
import time
current = time.time()
...job...
end = time.time()
diff = end - current
那对你有用吗?
>>> from datetime import datetime
>>> a = datetime.now()
# wait a bit
>>> b = datetime.now()
>>> d = b - a # yields a timedelta object
>>> d.seconds
7
(7 将是您在上面等待的时间)
我发现 datetime.datetime 相当有用,所以如果您遇到复杂或尴尬的情况,请告诉我们。
编辑:感谢@WoLpH 指出,人们不一定总是希望如此频繁地刷新,以至于日期时间会很接近。通过考虑增量中的天数,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
我们有 Python 2.7 的函数 total_seconds() 请参见下面的 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)
这是为我工作的那个。
from datetime import datetime
date_format = "%H:%M:%S"
# You could also pass datetime.time object in this part and convert it to string.
time_start = str('09:00:00')
time_end = str('18:00:00')
# Then get the difference here.
diff = datetime.strptime(time_end, date_format) - datetime.strptime(time_start, date_format)
# Get the time in hours i.e. 9.60, 8.5
result = diff.seconds / 3600;
希望这可以帮助!
另一种方法是使用时间戳值:
end_time.timestamp() - start_time.timestamp()
通过阅读源代码,我得出一个结论:无法通过以下方式获得时差.seconds
:
@property
def seconds(self):
"""seconds"""
return self._seconds
# in the `__new__`, you can find the `seconds` is modulo by the total number of seconds in a day
def __new__(cls, days=0, seconds=0, microseconds=0,
milliseconds=0, minutes=0, hours=0, weeks=0):
seconds += minutes*60 + hours*3600
# ...
if isinstance(microseconds, float):
microseconds = round(microseconds + usdouble)
seconds, microseconds = divmod(microseconds, 1000000)
# !
days, seconds = divmod(seconds, 24*3600)
d += days
s += seconds
else:
microseconds = int(microseconds)
seconds, microseconds = divmod(microseconds, 1000000)
# !
days, seconds = divmod(seconds, 24*3600)
d += days
s += seconds
microseconds = round(microseconds + usdouble)
# ...
total_seconds 可以得到两个时间之间的准确差异
def total_seconds(self):
"""Total seconds in the duration."""
return ((self.days * 86400 + self.seconds) * 10**6 +
self.microseconds) / 10**6
综上所述:
from datetime import datetime
dt1 = datetime.now()
dt2 = datetime.now()
print((dt2 - dt1).total_seconds())