blocks|key|3196536|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3196537|from+datetime+import+datetime,+date

datetime.combine(date.today(),+exit)+-+datetime.combine(date.today(),+enter)|code-block|syntax|javascript|3196538|combine构建了一个可以减去的datetime。|offset|length|style|CODE|3196539|entityMap^0|0|0|0|7|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]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Try this:

<pre><code>from datetime import datetime, date

datetime.combine(date.today(), exit) - datetime.combine(date.today(), enter)
</code></pre>

<code>combine</code> builds a datetime, that can be subtracted.

blocks|key|276831|text|使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|276832|from+datetime+import+datetime,+date

duration+=+datetime.combine(date.min,+end)+-+datetime.combine(date.min,+beginning)|code-block|syntax|javascript|276833|使用date.min更加简洁，即使在午夜也能正常工作。|offset|length|style|CODE|276834|如果第一次调用发生在23:59:59，下一次调用发生在00:00:00，则date.today()可能不会返回意外结果。|276835|entityMap^0|0|0|2|8|0|11|C|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@$I|W|J|X|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

Use:

<pre><code>from datetime import datetime, date

duration = datetime.combine(date.min, end) - datetime.combine(date.min, beginning)
</code></pre>

Using <code>date.min</code> is a bit more concise and works even at midnight.

This might not be the case with <code>date.today()</code> that might return unexpected results if the first call happens at 23:59:59 and the next one at 00:00:00.

blocks|key|2983237|text|不要使用时间，而是尝试使用timedelta：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2983238|from+datetime+import+timedelta

t1+=+timedelta(hours=7,+minutes=36)
t2+=+timedelta(hours=11,+minutes=32)
t3+=+timedelta(hours=13,+minutes=7)
t4+=+timedelta(hours=21,+minutes=0)

arrival+=+t2+-+t1
lunch+=+(t3+-+t2+-+timedelta(hours=1))
departure+=+t4+-+t3

print(arrival,+lunch,+departure)|code-block|syntax|javascript|2983239|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|$]]

instead of using time try timedelta:

<pre><code>from datetime import timedelta

t1 = timedelta(hours=7, minutes=36)
t2 = timedelta(hours=11, minutes=32)
t3 = timedelta(hours=13, minutes=7)
t4 = timedelta(hours=21, minutes=0)

arrival = t2 - t1
lunch = (t3 - t2 - timedelta(hours=1))
departure = t4 - t3

print(arrival, lunch, departure)
</code></pre>

blocks|key|3196725|text|python+timedelta库应该可以做您需要的事情。减去两个datetime实例会返回一个timedelta。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3196726|import+datetime
dt_started+=+datetime.datetime.utcnow()

#+do+some+stuff

dt_ended+=+datetime.datetime.utcnow()
print((dt_ended+-+dt_started).total_seconds())|code-block|syntax|javascript|3196727|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/datetime.html^0|X|8|1C|9|7|9|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

The python <a href="https://docs.python.org/2/library/datetime.html" rel="noreferrer">timedelta</a> library should do what you need. A <code>timedelta</code> is returned when you subtract two <code>datetime</code> instances.

<pre><code>import datetime
dt_started = datetime.datetime.utcnow()

# do some stuff

dt_ended = datetime.datetime.utcnow()
print((dt_ended - dt_started).total_seconds())
</code></pre>

blocks|key|274856|text|您有两个datetime.time对象，因此您只需使用datetime.timedetla创建两个时间增量，然后使用"-“操作数像现在这样减去。下面是不使用datetime进行两次减法运算的示例。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|274857|enter+=+datetime.time(hour=1)++#+Example+enter+time
exit+=+datetime.time(hour=2)++#+Example+start+time
enter_delta+=+datetime.timedelta(hours=enter.hour,+minutes=enter.minute,+seconds=enter.second)
exit_delta+=+datetime.timedelta(hours=exit.hour,+minutes=exit.minute,+seconds=exit.second)
difference_delta+=+exit_delta+-+enter_delta|code-block|syntax|javascript|274858|difference_delta是你的不同之处，你可以出于你的原因使用它。|offset|length|style|BOLD|274859|entityMap^0|0|0|0|G|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]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You have two datetime.time objects so for that you just create two timedelta using datetime.timedetla and then substract as you do right now using "-" operand. Following is the example way to substract two times without using datetime.

<pre><code>enter = datetime.time(hour=1) # Example enter time
exit = datetime.time(hour=2) # Example start time
enter_delta = datetime.timedelta(hours=enter.hour, minutes=enter.minute, seconds=enter.second)
exit_delta = datetime.timedelta(hours=exit.hour, minutes=exit.minute, seconds=exit.second)
difference_delta = exit_delta - enter_delta
</code></pre>

difference_delta is your difference which you can use for your reasons.

blocks|key|2983135|text|datetime.time不支持这一点，因为以这种方式减去时间几乎没有意义。如果要执行此操作，请使用完整的datetime.datetime。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2983136|entityMap^0|0|D|1H|H|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

<code>datetime.time</code> does not support this, because it's nigh meaningless to subtract times in this manner. Use a full <code>datetime.datetime</code> if you want to do this.

blocks|key|280382|text|import+datetime

def+diff_times_in_seconds(t1,+t2):
++++#+caveat+emptor+-+assumes+t1+&+t2+are+python+times,+on+the+same+day+and+t2+is+after+t1
++++h1,+m1,+s1+=+t1.hour,+t1.minute,+t1.second
++++h2,+m2,+s2+=+t2.hour,+t2.minute,+t2.second
++++t1_secs+=+s1+%2B+60+*+(m1+%2B+60*h1)
++++t2_secs+=+s2+%2B+60+*+(m2+%2B+60*h2)
++++return(+t2_secs+-+t1_secs)

#+using+it
diff_times_in_seconds(+datetime.datetime.strptime(+"13:23:34",+'%25H:%25M:%25S').time(),datetime.datetime.strptime(+"14:02:39",+'%25H:%25M:%25S').time())|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|280383|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>import datetime

def diff_times_in_seconds(t1, t2):
 # caveat emptor - assumes t1 &amp; t2 are python times, on the same day and t2 is after t1
 h1, m1, s1 = t1.hour, t1.minute, t1.second
 h2, m2, s2 = t2.hour, t2.minute, t2.second
 t1_secs = s1 + 60 * (m1 + 60*h1)
 t2_secs = s2 + 60 * (m2 + 60*h2)
 return( t2_secs - t1_secs)

# using it
diff_times_in_seconds( datetime.datetime.strptime( "13:23:34", '%H:%M:%S').time(),datetime.datetime.strptime( "14:02:39", '%H:%M:%S').time())
</code></pre>

blocks|key|3196747|text|datetime.time不能做到这一点--但是你可以使用datetime.datetime.now()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3196748|start+=+datetime.datetime.now()
sleep(10)
end+=+datetime.datetime.now()
duration+=+end+-+start|code-block|syntax|javascript|3196749|entityMap^0|0|D|T|N|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

<code>datetime.time</code> can not do it - But you could use <code>datetime.datetime.now()</code>

<pre><code>start = datetime.datetime.now()
sleep(10)
end = datetime.datetime.now()
duration = end - start
</code></pre>

blocks|key|2983251|text|我遇到了类似的情况，因为你和我最终都使用了名为arrow的外部库。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2983252|下面是它看起来的样子：|2983253|>>>+import+arrow
>>>+enter+=+arrow.get('12:30:45',+'HH:mm:ss')
>>>+exit+=+arrow.now()
>>>+duration+=+exit+-+enter
>>>+duration
datetime.timedelta(736225,+14377,+757451)|code-block|syntax|javascript|2983254|entityMap|0|LINK|mutability|MUTABLE|url|https://arrow.readthedocs.io/en/stable/^0|N|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

I had similar situation as you and I ended up with using external library called <a href="https://arrow.readthedocs.io/en/stable/" rel="nofollow noreferrer">arrow</a>.
Here is what it looks like:
<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; import arrow
&gt;&gt;&gt; enter = arrow.get('12:30:45', 'HH:mm:ss')
&gt;&gt;&gt; exit = arrow.now()
&gt;&gt;&gt; duration = exit - enter
&gt;&gt;&gt; duration
datetime.timedelta(736225, 14377, 757451)
</code></pre>

blocks|key|2983350|text|timedelta接受负(-)时间值。所以它可以像下面这样简单(单行)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2983351|duration+=+datetime.timedelta(hours=exit.hour-enter.hour,+minutes=exit.minute-enter.minute)|code-block|syntax|javascript|2983352|2983353|运行测试|BOLD|2983354|import+datetime

enter+=+datetime.time(hour=1,+minute=30)
exit+=+datetime.time(hour=2,+minute=0)
duration+=+datetime.timedelta(hours=exit.hour-enter.hour,+minutes=exit.minute-enter.minute)

>>>+duration
datetime.timedelta(seconds=1800)|2983355|entityMap^0|0|9|0|0|0|0|4|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|X|8|@$9|Y|A|Z|B|N]]|D|@]|E|$]]|$1|O|3|P|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

<code>timedelta</code> accepts negative(-) time values. so it could be simple as below (single line)
<pre><code>duration = datetime.timedelta(hours=exit.hour-enter.hour, minutes=exit.minute-enter.minute)
</code></pre>
<hr />
run test
<pre><code>import datetime

enter = datetime.time(hour=1, minute=30)
exit = datetime.time(hour=2, minute=0)
duration = datetime.timedelta(hours=exit.hour-enter.hour, minutes=exit.minute-enter.minute)

&gt;&gt;&gt; duration
datetime.timedelta(seconds=1800)
</code></pre>

I have two <code>datetime.time</code> values, <code>exit</code> and <code>enter</code> and I want to do something like:

<pre><code>duration = exit - enter
</code></pre>

However, I get this error:

<blockquote>
 TypeError: unsupported operand type(s) for -: 'datetime.time' and
 'datetime.time
</blockquote>

How do I do this correctly? One possible solution is converting the <code>time</code> variables to <code>datetime</code> variables and then subtruct, but I'm sure you guys must have a better and cleaner way.

subtract two times in python

我有两个datetime.time值，exit和enter，我想这样做：duration = exit - enter然而，我得到了这个错误： TypeError：-不支持的操作数类型：'datetime.time‘和'datetime.time我怎样才能正确地做到这一点？一种可能的解决方案是将time变量转换为datetime变量，然后再转包，但我相信你们肯定有更好、更干净的方法。

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