如何在python中获得timedelta的百分比

Question

我有两个datetime对象，它们给了我一个timedelta（time01-time02）。在

我想得到一个具体的时间差，它是前一个时间差的6%。有合适的方法吗？在

以下是我尝试的结果：

import datetime

sunrise = datetime.datetime.fromtimestamp(/*a timestamp*/)
sunset = datetime.datetime.fromtimestamp(/*a timestamp*/)
sunTime = datetime.timedelta(hours=sunset.hour-sunrise.hour, minutes=sunset.minute-sunrise.minute, seconds=sunset.second-sunrise.second)

...

print sunrise-datetime.timedelta(sunTime*0.06)

输出：

^{pr2}$

谢谢！在

Answer 1

时间戳以秒为单位，所以这应该可以工作

import datetime

sunrise = datetime.datetime.fromtimestamp(/*sunrise timestamp*/)
offset = int((/* sunset timestamp */ - /* sunrise timestamp */) * .06)

print sunrise-datetime.timedelta(seconds=offset)

Answer 2

这应该是有效的：

sunrise = datetime.datetime.fromtimestamp(/*a timestamp*/)
sunset = datetime.datetime.fromtimestamp(/*a timestamp*/)
sunTime = sunset-sunrise

...

print sunrise-datetime.timedelta(seconds=sunTime.total_seconds()*0.06)

Answer 3

例如，可以操纵timedelta的输入，而不是操纵timedelta

import datetime

percent = 0.06    
sunrise = datetime.datetime.fromtimestamp(/*a timestamp*/)
sunset = datetime.datetime.fromtimestamp(/*a timestamp*/)

sunTime = datetime.timedelta(hours=(sunset.hour-sunrise.hour)*percent,
                             minutes=(sunset.minute-sunrise.minute)*percent,
                             seconds=(sunset.second-sunrise.second)*percent)
print sunTime

达到相同的效果，您只创建一个timedelta对象。