<p>这里有一个解决方案,支持查找差异,即使结束时间小于开始时间(超过午夜间隔),例如<code>23:55:00-00:25:00</code>(半小时持续时间):</p>
<pre><code>#!/usr/bin/env python
from datetime import datetime, time as datetime_time, timedelta
def time_diff(start, end):
if isinstance(start, datetime_time): # convert to datetime
assert isinstance(end, datetime_time)
start, end = [datetime.combine(datetime.min, t) for t in [start, end]]
if start <= end: # e.g., 10:33:26-11:15:49
return end - start
else: # end < start e.g., 23:55:00-00:25:00
end += timedelta(1) # +day
assert end > start
return end - start
for time_range in ['10:33:26-11:15:49', '23:55:00-00:25:00']:
s, e = [datetime.strptime(t, '%H:%M:%S') for t in time_range.split('-')]
print(time_diff(s, e))
assert time_diff(s, e) == time_diff(s.time(), e.time())
</code></pre>
<h3>输出</h3>
<pre><code>0:42:23
0:30:00
</code></pre>
<p><code>time_diff()</code>返回一个timedelta对象,您可以将它(作为序列的一部分)直接传递给一个<code>mean()</code>函数,例如:</p>
<pre><code>#!/usr/bin/env python
from datetime import timedelta
def mean(data, start=timedelta(0)):
"""Find arithmetic average."""
return sum(data, start) / len(data)
data = [timedelta(minutes=42, seconds=23), # 0:42:23
timedelta(minutes=30)] # 0:30:00
print(repr(mean(data)))
# -> datetime.timedelta(0, 2171, 500000) # days, seconds, microseconds
</code></pre>
<p><code>mean()</code>结果也是<code>timedelta()</code>对象,您可以将其转换为秒(<code>td.total_seconds()</code>方法(自Python 2.7以来))、小时(<code>td / timedelta(hours=1)</code>(Python 3))等</p>