blocks|key|734852|text|我想这就是你要找的..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|734853|>>>+import+datetime
>>>+dt+=+datetime.datetime.now()
>>>+dt+=+dt.replace(hour=0,+minute=0,+second=0,+microsecond=0)+#+Returns+a+copy
>>>+dt
datetime.datetime(2011,+3,+29,+0,+0)|code-block|syntax|javascript|734854|但是如果你真的不关心时间方面的事情，那么你真的应该只传递date对象……|offset|length|style|CODE|734855|>>>+d_truncated+=+datetime.date(dt.year,+dt.month,+dt.day)
>>>+d_truncated
datetime.date(2011,+3,+29)|734856|entityMap^0|0|0|S|4|0|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|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I think this is what you're looking for...

<pre><code>&gt;&gt;&gt; import datetime
&gt;&gt;&gt; dt = datetime.datetime.now()
&gt;&gt;&gt; dt = dt.replace(hour=0, minute=0, second=0, microsecond=0) # Returns a copy
&gt;&gt;&gt; dt
datetime.datetime(2011, 3, 29, 0, 0)
</code></pre>

But if you really don't care about the time aspect of things, then you should really only be passing around <code>date</code> objects...

<pre><code>&gt;&gt;&gt; d_truncated = datetime.date(dt.year, dt.month, dt.day)
&gt;&gt;&gt; d_truncated
datetime.date(2011, 3, 29)
</code></pre>

blocks|key|734716|text|如果你不关心时间，那就使用date，而不是datetime。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|734717|>>>+now+=+datetime.now()
>>>+now.date()
datetime.date(2011,+3,+29)|code-block|syntax|javascript|734718|您可以像这样更新日期时间：|734719|>>>+now.replace(minute=0,+hour=0,+second=0,+microsecond=0)
datetime.datetime(2011,+3,+29,+0,+0)|734720|entityMap^0|D|4|L|8|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

Use a <code>date</code> not a <code>datetime</code> if you dont care about the time.

<pre><code>&gt;&gt;&gt; now = datetime.now()
&gt;&gt;&gt; now.date()
datetime.date(2011, 3, 29)
</code></pre>

You can update a datetime like this:

<pre><code>&gt;&gt;&gt; now.replace(minute=0, hour=0, second=0, microsecond=0)
datetime.datetime(2011, 3, 29, 0, 0)
</code></pre>

blocks|key|951001|text|不能截断datetime对象，因为it+is+immutable。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|951002|但是，这里有一种方法可以构造一个新的日期时间，其中包含0小时、分钟、秒和微秒字段，而不会丢弃原始的日期或tzinfo：|951003|newdatetime+=+now.replace(hour=0,+minute=0,+second=0,+microsecond=0)|code-block|syntax|javascript|951004|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/datetime.html#available-types^0|H|F|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]]]]

You cannot truncate a datetime object because <a href="http://docs.python.org/library/datetime.html#available-types" rel="noreferrer">it is immutable</a>.

However, here is one way to construct a new datetime with 0 hour, minute, second, and microsecond fields, without throwing away the original date or tzinfo:

<pre><code>newdatetime = now.replace(hour=0, minute=0, second=0, microsecond=0)
</code></pre>

blocks|key|735057|text|要获得与给定datetime对象对应的午夜，可以使用datetime.combine()+method|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|735058|>>>+from+datetime+import+datetime,+time
>>>+dt+=+datetime.utcnow()
>>>+dt.date()
datetime.date(2015,+2,+3)
>>>+datetime.combine(dt,+time.min)
datetime.datetime(2015,+2,+3,+0,+0)|code-block|syntax|javascript|735059|与the+.replace()+method相比，datetime.combine()-based解决方案的优势在于即使datetime+module+introduces+the+nanoseconds+support也能继续工作。|735060|如果需要，可以保留tzinfo，但是在午夜，例如由于DST转换，utc偏移可能不同，并且因此朴素的解决方案(设置tzinfo时间属性)可能失败。请参阅How+do+I+get+the+UTC+time+of+“midnight”+for+a+given+timezone?|735061|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/datetime.html#datetime.datetime.combine|1|https://stackoverflow.com/a/5476114/4279|2|http://bugs.python.org/issue15443|3|https://stackoverflow.com/a/11236372/4279^0|Q|I|Q|P|0|0|0|5|A|P|I|1O|8|1|L|1|1O|1E|2|0|9|6|1K|6|23|1P|3|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|F|3|G|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@$9|1G|A|1H|1|1I]|$9|1J|A|1K|1|1L]]|E|$]]|$1|M|3|N|5|6|7|1M|8|@$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]]|D|@$9|1R|A|1S|1|1T]]|E|$]]|$1|O|3|-4|5|6|7|1U|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]|Y|$5|R|S|T|E|$U|Z]]|10|$5|R|S|T|E|$U|11]]]]

To get a midnight corresponding to a given datetime object, you could use <a href="https://docs.python.org/3/library/datetime.html#datetime.datetime.combine" rel="noreferrer"><code>datetime.combine()</code> method</a>:

<pre><code>&gt;&gt;&gt; from datetime import datetime, time
&gt;&gt;&gt; dt = datetime.utcnow()
&gt;&gt;&gt; dt.date()
datetime.date(2015, 2, 3)
&gt;&gt;&gt; datetime.combine(dt, time.min)
datetime.datetime(2015, 2, 3, 0, 0)
</code></pre>

The advantage compared to <a href="https://stackoverflow.com/a/5476114/4279">the <code>.replace()</code> method</a> is that <code>datetime.combine()</code>-based solution will continue to work even if <a href="http://bugs.python.org/issue15443" rel="noreferrer"><code>datetime</code> module introduces the nanoseconds support</a>.

<code>tzinfo</code> can be preserved if necessary but the utc offset may be different at midnight e.g., due to a DST transition and therefore a naive solution (setting <code>tzinfo</code> time attribute) may fail. See <a href="https://stackoverflow.com/a/11236372/4279">How do I get the UTC time of “midnight” for a given timezone?</a>

blocks|key|951374|text|为此，您可以使用pandas+(尽管这可能是该任务的开销)。你可以使用round，floor和ceil，就像来自offset-aliases的普通数字和任何熊猫频率一样|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|951375|import+pandas+as+pd
import+datetime+as+dt

now+=+dt.datetime.now()
pd_now+=+pd.Timestamp(now)

freq+=+'1d'
pd_round+=+pd_now.round(freq)
dt_round+=+pd_round.to_pydatetime()

print(now)
print(dt_round)

"""
2018-06-15+09:33:44.102292
2018-06-15+00:00:00
"""|code-block|syntax|javascript|951376|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/|1|https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Timestamp.round.html|2|https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Timestamp.floor.html|3|https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Timestamp.ceil.html|4|http://pandas.pydata.org/pandas-docs/stable/timeseries.html#offset-aliases^0|8|6|0|Z|5|1|15|5|2|1B|4|3|1K|E|4|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]|$A|12|B|13|1|14]|$A|15|B|16|1|17]|$A|18|B|19|1|1A]|$A|1B|B|1C|1|1D]]|C|$]]|$1|D|3|E|5|F|7|1E|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|1F|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]|Q|$5|L|M|N|C|$O|R]]|S|$5|L|M|N|C|$O|T]]|U|$5|L|M|N|C|$O|V]]|W|$5|L|M|N|C|$O|X]]]]

You could use <a href="https://pandas.pydata.org/" rel="noreferrer">pandas</a> for that (although it could be overhead for that task). You could use <a href="https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Timestamp.round.html" rel="noreferrer">round</a>, <a href="https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Timestamp.floor.html" rel="noreferrer">floor</a> and <a href="https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Timestamp.ceil.html" rel="noreferrer">ceil</a> like for usual numbers and any pandas frequency from <a href="http://pandas.pydata.org/pandas-docs/stable/timeseries.html#offset-aliases" rel="noreferrer">offset-aliases</a>:

<pre><code>import pandas as pd
import datetime as dt

now = dt.datetime.now()
pd_now = pd.Timestamp(now)

freq = '1d'
pd_round = pd_now.round(freq)
dt_round = pd_round.to_pydatetime()

print(now)
print(dt_round)

"""
2018-06-15 09:33:44.102292
2018-06-15 00:00:00
"""
</code></pre>

blocks|key|735470|text|欲了解更多信息，请访问https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.dt.floor.html|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|735471|现在是2019年，我认为最有效的方法是：|735472|df['truncate_date']+=+df['timestamp'].dt.floor('d')|code-block|syntax|javascript|735473|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.dt.floor.html^0|B|2E|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]]]]

See more at <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.dt.floor.html" rel="noreferrer">https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.dt.floor.html</a>

It's now 2019, I think the most efficient way to do it is:

<pre><code>df['truncate_date'] = df['timestamp'].dt.floor('d')
</code></pre>

blocks|key|734777|text|您可以使用datetime.strftime提取日、月、年……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|734778|示例：|734779|from+datetime+import+datetime
d+=+datetime.today()

#+Retrieves+the+day+and+the+year
print+d.strftime("%25d-%25Y")|code-block|syntax|javascript|734780|输出(今天)：|734781|29-2011|734782|如果您只想检索日期，则可以使用如下所示的day属性：|734783|from+datetime+import+datetime
d+=+datetime.today()

#+Retrieves+the+day
print+d.day|734784|734785|29|734786|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|X|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Y|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|Z|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|10|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|11|8|@]|9|@]|A|$G|H]]|$1|Q|3|J|5|6|7|12|8|@]|9|@]|A|$]]|$1|R|3|S|5|F|7|13|8|@]|9|@]|A|$G|H]]|$1|T|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|U|$]]

You can use datetime.strftime to extract the day, the month, the year...

Example :

<pre><code>from datetime import datetime
d = datetime.today()

# Retrieves the day and the year
print d.strftime("%d-%Y")
</code></pre>

Output (for today):

<pre><code>29-2011
</code></pre>

If you just want to retrieve the day, you can use day attribute like :

<pre><code>from datetime import datetime
d = datetime.today()

# Retrieves the day
print d.day
</code></pre>

Ouput (for today):

<pre><code>29
</code></pre>

blocks|key|735137|text|有一个很棒的用于操作日期的库：Delorean|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|735138|import+datetime
from+delorean+import+Delorean
now+=+datetime.datetime.now()
d+=+Delorean(now,+timezone='US/Pacific')

>>>+now++++
datetime.datetime(2015,+3,+26,+19,+46,+40,+525703)

>>>+d.truncate('second')
Delorean(datetime=2015-03-26+19:46:40-07:00,+timezone='US/Pacific')

>>>+d.truncate('minute')
Delorean(datetime=2015-03-26+19:46:00-07:00,+timezone='US/Pacific')

>>>+d.truncate('hour')
Delorean(datetime=2015-03-26+19:00:00-07:00,+timezone='US/Pacific')

>>>+d.truncate('day')
Delorean(datetime=2015-03-26+00:00:00-07:00,+timezone='US/Pacific')

>>>+d.truncate('month')
Delorean(datetime=2015-03-01+00:00:00-07:00,+timezone='US/Pacific')

>>>+d.truncate('year')
Delorean(datetime=2015-01-01+00:00:00-07:00,+timezone='US/Pacific')|code-block|syntax|javascript|735139|如果你想取回日期时间值：|735140|>>>+d.truncate('year').datetime
datetime.datetime(2015,+1,+1,+0,+0,+tzinfo=<DstTzInfo+'US/Pacific'+PDT-1+day,+17:00:00+DST>)|735141|entityMap|0|LINK|mutability|MUTABLE|url|http://delorean.readthedocs.org/en/latest/quickstart.html^0|F|8|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

There is a great library used to manipulate dates: <a href="http://delorean.readthedocs.org/en/latest/quickstart.html" rel="nofollow noreferrer">Delorean</a>

<pre><code>import datetime
from delorean import Delorean
now = datetime.datetime.now()
d = Delorean(now, timezone='US/Pacific')

&gt;&gt;&gt; now 
datetime.datetime(2015, 3, 26, 19, 46, 40, 525703)

&gt;&gt;&gt; d.truncate('second')
Delorean(datetime=2015-03-26 19:46:40-07:00, timezone='US/Pacific')

&gt;&gt;&gt; d.truncate('minute')
Delorean(datetime=2015-03-26 19:46:00-07:00, timezone='US/Pacific')

&gt;&gt;&gt; d.truncate('hour')
Delorean(datetime=2015-03-26 19:00:00-07:00, timezone='US/Pacific')

&gt;&gt;&gt; d.truncate('day')
Delorean(datetime=2015-03-26 00:00:00-07:00, timezone='US/Pacific')

&gt;&gt;&gt; d.truncate('month')
Delorean(datetime=2015-03-01 00:00:00-07:00, timezone='US/Pacific')

&gt;&gt;&gt; d.truncate('year')
Delorean(datetime=2015-01-01 00:00:00-07:00, timezone='US/Pacific')
</code></pre>

and if you want to get datetime value back:

<pre><code>&gt;&gt;&gt; d.truncate('year').datetime
datetime.datetime(2015, 1, 1, 0, 0, tzinfo=&lt;DstTzInfo 'US/Pacific' PDT-1 day, 17:00:00 DST&gt;)
</code></pre>

blocks|key|735332|text|如果您处理的是DateTime类型的Series，那么有一种更有效的方法来截断它们，特别是当Series对象有很多行的时候。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|735333|您可以使用floor函数|offset|length|735334|例如，如果要将其截断为小时数：|735335|生成一系列日期|735336|times+=+pd.Series(pd.date_range(start='1/1/2018+04:00:00',+end='1/1/2018+22:00:00',+freq='s'))|code-block|syntax|javascript|735337|我们可以通过比较replace函数和floor函数的运行时间来检查它。|735338|%25timeit+times.apply(lambda+x+:+x.replace(minute=0,+second=0,+microsecond=0))
>>>+341+ms+±+18.2+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+1+loop+each)

%25timeit+times.dt.floor('h')
>>>>2.26+ms+±+451+µs+per+loop+(mean+±+std.+dev.+of+7+runs,+100+loops+each)|735339|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.dt.floor.html^0|0|5|5|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@$D|12|E|13|1|14]]|A|$]]|$1|F|3|G|5|6|7|15|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|16|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|17|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|18|8|@]|9|@]|A|$]]|$1|Q|3|R|5|L|7|19|8|@]|9|@]|A|$M|N]]|$1|S|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

If you are dealing with a Series of type DateTime there is a more efficient way to truncate them, specially when the Series object has a lot of rows. 

You can use the <a href="https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.dt.floor.html" rel="nofollow noreferrer">floor</a> function 

For example, if you want to truncate it to hours: 

Generate a range of dates

<pre><code>times = pd.Series(pd.date_range(start='1/1/2018 04:00:00', end='1/1/2018 22:00:00', freq='s'))
</code></pre>

We can check it comparing the running time between the replace and the floor functions.

<pre><code>%timeit times.apply(lambda x : x.replace(minute=0, second=0, microsecond=0))
&gt;&gt;&gt; 341 ms ± 18.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit times.dt.floor('h')
&gt;&gt;&gt;&gt;2.26 ms ± 451 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
</code></pre>

blocks|key|951043|text|有一个模块datetime_truncate可以帮你处理这个问题。它只是调用datetime.replace。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|951044|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

There is a module datetime_truncate which handlers this for you. It just calls datetime.replace.

blocks|key|951274|text|六年后...我找到了这篇文章，我更喜欢这种麻木的方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|951275|import+numpy+as+np
dates_array+=+np.array(['2013-01-01',+'2013-01-15',+'2013-01-30']).astype('datetime64[ns]')
truncated_dates+=+dates_array.astype('datetime64[D]')|code-block|syntax|javascript|951276|干杯|951277|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

6 years later... I found this post and I liked more the numpy aproach:

<pre><code>import numpy as np
dates_array = np.array(['2013-01-01', '2013-01-15', '2013-01-30']).astype('datetime64[ns]')
truncated_dates = dates_array.astype('datetime64[D]')
</code></pre>

cheers

blocks|key|735426|text|>>>+import+datetime
>>>+dt+=+datetime.datetime.now()
>>>+datetime.datetime.date(dt)
datetime.date(2019,+4,+2)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|735427|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>&gt;&gt;&gt; import datetime
&gt;&gt;&gt; dt = datetime.datetime.now()
&gt;&gt;&gt; datetime.datetime.date(dt)
datetime.date(2019, 4, 2)
</code></pre>

blocks|key|735496|text|下面是另一种方法，它适合于一行，但不是特别优雅：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|735497|dt+=+datetime.datetime.fromordinal(datetime.date.today().toordinal())|code-block|syntax|javascript|735498|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|$]]

Here is yet another way which fits in one line but is not particularly elegant:

<pre><code>dt = datetime.datetime.fromordinal(datetime.date.today().toordinal())
</code></pre>

blocks|key|951493|text|你可以直接使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|951494|datetime.date.today()|code-block|syntax|javascript|951495|它很轻，返回的正是你想要的。|951496|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can just use
<pre><code>datetime.date.today()
</code></pre>
It's light and returns exactly what you want.

blocks|key|734677|text|截断是什么意思？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|734678|通过使用strftime()方法和适当的格式字符串，您可以完全控制格式设置。|734679|http://docs.python.org/library/datetime.html#strftime-strptime-behavior|offset|length|734680|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|1Z|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@$F|R|G|S|1|T]]|A|$]]|$1|H|3|-4|5|6|7|U|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|E]]]]

What does truncate mean?

You have full control over the formatting by using the strftime() method and using an appropriate format string.

<a href="http://docs.python.org/library/datetime.html#strftime-strptime-behavior" rel="nofollow">http://docs.python.org/library/datetime.html#strftime-strptime-behavior</a>

What is a classy way to way truncate a python <code>datetime</code> object?
In this particular case, to the day. So basically setting hour, minute, seconds, and microseconds to 0.
I would like the output to also be a <code>datetime</code> object, not a string.

How to truncate the time on a datetime object?

截断python datetime对象的最好方法是什么？在这种特殊情况下，到今天为止。因此，基本上将小时、分钟、秒和微秒设置为0。我希望输出也是一个datetime对象，而不是字符串。

问如何截断datetime对象上的时间？
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