blocks|key|34316|text|获取timedelta并删除它自己的微秒，作为微秒和只读属性：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|34317|avg+=+sum(datetimes,+datetime.timedelta(0))+/+len(datetimes)
avg+=+avg+-+datetime.timedelta(microseconds=avg.microseconds)|code-block|syntax|javascript|34318|如果经常需要，您可以创建自己的小函数：|34319|import+datetime

def+chop_microseconds(delta):
++++return+delta+-+datetime.timedelta(microseconds=delta.microseconds)|34320|我还没有找到更好的解决方案。|34321|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Take the timedelta and remove its own microseconds, as microseconds and read-only attribute:

<pre><code>avg = sum(datetimes, datetime.timedelta(0)) / len(datetimes)
avg = avg - datetime.timedelta(microseconds=avg.microseconds)
</code></pre>

You can make your own little function if it is a recurring need:

<pre><code>import datetime

def chop_microseconds(delta):
 return delta - datetime.timedelta(microseconds=delta.microseconds)
</code></pre>

I have not found a better solution.

blocks|key|34348|text|另一种选择是，给定timedelta，您可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|34349|avg+=+datetime.timedelta(seconds=math.ceil(avg.total_seconds()))|code-block|syntax|javascript|34350|您可以根据具体情况，将math.ceil()替换为math.round()或math.floor()。|offset|length|style|CODE|34351|entityMap^0|0|0|B|B|P|C|12|C|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|N|$]]

another option, given timedelta you can do:

<pre><code>avg = datetime.timedelta(seconds=math.ceil(avg.total_seconds()))
</code></pre>

You can replace the <code>math.ceil()</code>, with <code>math.round()</code> or <code>math.floor()</code>, depending on the situation.

blocks|key|250783|text|我发现这对我很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|250784|start_time+=+datetime.now()
some_work()
end+time+=+datetime.now()
print+str(end_time+-+start_time)[:-3]|code-block|syntax|javascript|250785|输出：|250786|0:00:01.955|250787|我在https://docs.python.org/3.2/library/datetime.html#timedelta-objects上搜索后想到了这一点|offset|length|250788|希望这能有所帮助！|250789|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.2/library/datetime.html#timedelta-objects^0|0|0|0|0|2|1V|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@$M|13|N|14|1|15]]|A|$]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

I found this to work for me:

<pre><code>start_time = datetime.now()
some_work()
end time = datetime.now()
print str(end_time - start_time)[:-3]
</code></pre>

Output:

<pre><code>0:00:01.955
</code></pre>

I thought about this after searching in <a href="https://docs.python.org/3.2/library/datetime.html#timedelta-objects" rel="nofollow noreferrer">https://docs.python.org/3.2/library/datetime.html#timedelta-objects</a>

Hope this helps!

blocks|key|250819|text|考虑到timedelta内部只存储天、秒和微秒，您可以构造一个不带微秒的新timedelta：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|250820|from+datetime+import+timedelta
datetimes+=+(timedelta(hours=1,+minutes=5,+seconds=30,+microseconds=9999),+
+++++++++++++timedelta(minutes=35,+seconds=17,+microseconds=55),+
+++++++++++++timedelta(hours=2,+minutes=17,+seconds=3,+microseconds=1234))
avg+=+sum(datetimes,+timedelta(0))+/+len(datetimes)+#+timedelta(0,+4756,+670429)
avg+=+timedelta(days=avg.days,+seconds=avg.seconds)+#+timedelta(0,+4756)|code-block|syntax|javascript|250821|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/datetime.html#datetime.timedelta^0|3|9|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Given that <a href="https://docs.python.org/3/library/datetime.html#datetime.timedelta" rel="nofollow noreferrer">timedelta</a> only stores days, seconds and microseconds internally, you can construct a new timedelta without microseconds:

<pre><code>from datetime import timedelta
datetimes = (timedelta(hours=1, minutes=5, seconds=30, microseconds=9999), 
 timedelta(minutes=35, seconds=17, microseconds=55), 
 timedelta(hours=2, minutes=17, seconds=3, microseconds=1234))
avg = sum(datetimes, timedelta(0)) / len(datetimes) # timedelta(0, 4756, 670429)
avg = timedelta(days=avg.days, seconds=avg.seconds) # timedelta(0, 4756)
</code></pre>

blocks|key|34430|text|c+-=+timedelta(microseconds=c.microseconds)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|34431|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>c -= timedelta(microseconds=c.microseconds)
</code></pre>

blocks|key|250865|text|我认为这是获得没有微秒的最后几分钟的最好方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|250866|import+time
import+datetime
NOW+=+datetime.datetime.now()
LASTMINUTE+=++NOW+-+datetime.timedelta(minutes=1,+seconds+=+NOW.second,+microseconds+=+NOW.microsecond)|code-block|syntax|javascript|250867|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I think this is the best way to get the last minutes without microseconds:
<pre><code>import time
import datetime
NOW = datetime.datetime.now()
LASTMINUTE = NOW - datetime.timedelta(minutes=1, seconds = NOW.second, microseconds = NOW.microsecond)
</code></pre>

I do a calculation of average time, and I would like to display the resulted average without microseconds.

<pre><code>avg = sum(datetimes, datetime.timedelta(0)) / len(datetimes)
</code></pre>

How do I remove the microseconds from a timedelta object?

我计算平均时间，我想显示没有微秒的结果平均值。avg = sum(datetimes, datetime.timedelta(0)) / len(datetimes)

问如何从timedelta对象中删除微秒？
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