blocks|key|1733206|text|您应该使用dateutil。有了这个，你可以使用relativedelta，它是时间增量的改进版本。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1733207|>>>+import+datetime+
>>>+import+dateutil.relativedelta
>>>+now+=+datetime.datetime.now()
>>>+print+now
2012-03-15+12:33:04.281248
>>>+print+now+%2B+dateutil.relativedelta.relativedelta(months=-1)
2012-02-15+12:33:04.281248|code-block|syntax|javascript|1733208|entityMap|0|LINK|mutability|MUTABLE|url|http://labix.org/python-dateutil^0|5|8|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]]]]

You should use <a href="http://labix.org/python-dateutil" rel="noreferrer">dateutil</a>.
With that, you can use relativedelta, it's an improved version of timedelta.

<pre><code>&gt;&gt;&gt; import datetime 
&gt;&gt;&gt; import dateutil.relativedelta
&gt;&gt;&gt; now = datetime.datetime.now()
&gt;&gt;&gt; print now
2012-03-15 12:33:04.281248
&gt;&gt;&gt; print now + dateutil.relativedelta.relativedelta(months=-1)
2012-02-15 12:33:04.281248
</code></pre>

blocks|key|1733228|text|在bgporter's+answer上构建。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1733229|def+prev_month_range(when+=+None):+
++++"""Return+(previous+month's+start+date,+previous+month's+end+date)."""
++++if+not+when:
++++++++#+Default+to+today.
++++++++when+=+datetime.datetime.today()
++++#+Find+previous+month:+https://stackoverflow.com/a/9725093/564514
++++#+Find+today.
++++first+=+datetime.date(day=1,+month=when.month,+year=when.year)
++++#+Use+that+to+find+the+first+day+of+this+month.
++++prev_month_end+=+first+-+datetime.timedelta(days=1)
++++prev_month_start+=+datetime.date(day=1,+month=+prev_month_end.month,+year=+prev_month_end.year)
++++#+Return+previous+month's+start+and+end+dates+in+YY-MM-DD+format.
++++return+(prev_month_start.strftime('%25Y-%25m-%25d'),+prev_month_end.strftime('%25Y-%25m-%25d'))|code-block|syntax|javascript|1733230|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/9725093/564514^0|1|H|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]]]]

Building on <a href="https://stackoverflow.com/a/9725093/564514">bgporter's answer</a>.

<pre><code>def prev_month_range(when = None): 
 """Return (previous month's start date, previous month's end date)."""
 if not when:
 # Default to today.
 when = datetime.datetime.today()
 # Find previous month: https://stackoverflow.com/a/9725093/564514
 # Find today.
 first = datetime.date(day=1, month=when.month, year=when.year)
 # Use that to find the first day of this month.
 prev_month_end = first - datetime.timedelta(days=1)
 prev_month_start = datetime.date(day=1, month= prev_month_end.month, year= prev_month_end.year)
 # Return previous month's start and end dates in YY-MM-DD format.
 return (prev_month_start.strftime('%Y-%m-%d'), prev_month_end.strftime('%Y-%m-%d'))
</code></pre>

blocks|key|1733239|text|from+datetime+import+date,+timedelta

first_day_of_current_month+=+date.today().replace(day=1)
last_day_of_previous_month+=+first_day_of_current_month+-+timedelta(days=1)

print+"Previous+month:",+last_day_of_previous_month.month|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1733240|或者：|unstyled|1733241|from+datetime+import+date,+timedelta

prev+=+date.today().replace(day=1)+-+timedelta(days=1)
print+prev.month|1733242|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

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

first_day_of_current_month = date.today().replace(day=1)
last_day_of_previous_month = first_day_of_current_month - timedelta(days=1)

print "Previous month:", last_day_of_previous_month.month
</code></pre>

Or:

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

prev = date.today().replace(day=1) - timedelta(days=1)
print prev.month
</code></pre>

blocks|key|1946738|text|只是为了好玩，一个使用divmod的纯数学答案。由于乘法的不足，可以做一个简单的月数检查(如果等于12，则增加年份等)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1946739|year+=+today.year
month+=+today.month

nm+=+list(divmod(year+*+12+%2B+month+%2B+1,+12))
if+nm[1]+==+0:
++++nm[1]+=+12
++++nm[0]+-=+1
pm+=+list(divmod(year+*+12+%2B+month+-+1,+12))
if+pm[1]+==+0:
++++pm[1]+=+12
++++pm[0]+-=+1

next_month+=+nm
previous_month+=+pm|code-block|syntax|javascript|1946740|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|$]]

Just for fun, a pure math answer using divmod. Pretty inneficient because of the multiplication, could do just as well a simple check on the number of month (if equal to 12, increase year, etc)

<pre><code>year = today.year
month = today.month

nm = list(divmod(year * 12 + month + 1, 12))
if nm[1] == 0:
 nm[1] = 12
 nm[0] -= 1
pm = list(divmod(year * 12 + month - 1, 12))
if pm[1] == 0:
 pm[1] = 12
 pm[0] -= 1

next_month = nm
previous_month = pm
</code></pre>

blocks|key|1733281|text|基于@J.F.Sebastian的注释，您可以将replace()函数链接到前一个“月”。由于一个月不是一个恒定的时间段，此解决方案尝试返回到上个月的相同日期，这当然不适用于所有月份。在这种情况下，此算法默认为上个月的最后一天。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1733282|from+datetime+import+datetime,+timedelta

d+=+datetime(2012,+3,+31)+#+A+problem+date+as+an+example

#+last+day+of+last+month
one_month_ago+=+(d.replace(day=1)+-+timedelta(days=1))
try:
++++#+try+to+go+back+to+same+day+last+month
++++one_month_ago+=+one_month_ago.replace(day=d.day)
except+ValueError:
++++pass
print("one_month_ago:+{0}".format(one_month_ago))|code-block|syntax|javascript|1733283|输出：|1733284|one_month_ago:+2012-02-29+00:00:00|1733285|entityMap^0|O|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

Building off the comment of @J.F. Sebastian, you can chain the <code>replace()</code> function to go back one "month". Since a month is not a constant time period, this solution tries to go back to the same date the previous month, which of course does not work for all months. In such a case, this algorithm defaults to the last day of the prior month.

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

d = datetime(2012, 3, 31) # A problem date as an example

# last day of last month
one_month_ago = (d.replace(day=1) - timedelta(days=1))
try:
 # try to go back to same day last month
 one_month_ago = one_month_ago.replace(day=d.day)
except ValueError:
 pass
print("one_month_ago: {0}".format(one_month_ago))
</code></pre>

Output:

<pre><code>one_month_ago: 2012-02-29 00:00:00
</code></pre>

blocks|key|1946759|text|def+prev_month(date=datetime.datetime.today()):
++++if+date.month+==+1:
++++++++return+date.replace(month=12,year=date.year-1)
++++else:
++++++++try:
++++++++++++return+date.replace(month=date.month-1)
++++++++except+ValueError:
++++++++++++return+prev_month(date=date.replace(day=date.day-1))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1946760|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>def prev_month(date=datetime.datetime.today()):
 if date.month == 1:
 return date.replace(month=12,year=date.year-1)
 else:
 try:
 return date.replace(month=date.month-1)
 except ValueError:
 return prev_month(date=date.replace(day=date.day-1))
</code></pre>

blocks|key|1733308|text|这非常容易和简单。执行此操作|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1733309|from+dateutil.relativedelta+import+relativedelta
from+datetime+import+datetime

today_date+=+datetime.today()
print+"todays+date+time:+%25s"+%25today_date

one_month_ago+=+today_date+-+relativedelta(months=1)
print+"one+month+ago+date+time:+%25s"+%25+one_month_ago
print+"one+month+ago+date:+%25s"+%25+one_month_ago.date()|code-block|syntax|javascript|1733310|下面是输出：$python2.7+main.py|1733311|todays+date+time:+2016-09-06+02:13:01.937121
one+month+ago+date+time:+2016-08-06+02:13:01.937121
one+month+ago+date:+2016-08-06|1733312|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Its very easy and simple. Do this

<pre><code>from dateutil.relativedelta import relativedelta
from datetime import datetime

today_date = datetime.today()
print "todays date time: %s" %today_date

one_month_ago = today_date - relativedelta(months=1)
print "one month ago date time: %s" % one_month_ago
print "one month ago date: %s" % one_month_ago.date()
</code></pre>

Here is the output:
$python2.7 main.py

<pre><code>todays date time: 2016-09-06 02:13:01.937121
one month ago date time: 2016-08-06 02:13:01.937121
one month ago date: 2016-08-06
</code></pre>

blocks|key|1946784|text|对于来到这里并希望同时获得上个月的第一天和最后一天的人：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1946785|from+datetime+import+date,+timedelta

last_day_of_prev_month+=+date.today().replace(day=1)+-+timedelta(days=1)

start_day_of_prev_month+=+date.today().replace(day=1)+-+timedelta(days=last_day_of_prev_month.day)

#+For+printing+results
print("First+day+of+prev+month:",+start_day_of_prev_month)
print("Last+day+of+prev+month:",+last_day_of_prev_month)|code-block|syntax|javascript|1946786|输出：|1946787|First+day+of+prev+month:+2019-02-01
Last+day+of+prev+month:+2019-02-28|1946788|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

For someone who got here and looking to get both the first and last day of the previous month:

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

last_day_of_prev_month = date.today().replace(day=1) - timedelta(days=1)

start_day_of_prev_month = date.today().replace(day=1) - timedelta(days=last_day_of_prev_month.day)

# For printing results
print("First day of prev month:", start_day_of_prev_month)
print("Last day of prev month:", last_day_of_prev_month)
</code></pre>

Output:

<pre><code>First day of prev month: 2019-02-01
Last day of prev month: 2019-02-28
</code></pre>

blocks|key|1946800|text|有了Pendulum非常完整的库，我们就有了subtract方法(而不是"subStract")：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1946801|import+pendulum
today+=+pendulum.datetime.today()++#+2020,+january
lastmonth+=+today.subtract(months=1)
lastmonth.strftime('%25Y%25m')
#+'201912'|code-block|syntax|javascript|1946802|我们看到它可以处理跳跃的年份。|1946803|与之相反的等价物是add。|1946804|https://pendulum.eustace.io/docs/#addition-and-subtraction|1946805|entityMap|0|LINK|mutability|MUTABLE|url|https://pypi.org/project/pendulum/|1^0|M|8|2|8|0|0|0|0|9|3|0|0|1M|1|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|1A|8|@]|D|@$9|1B|A|1C|1|1D]]|E|$]]|$1|Q|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|T|U|V|E|$W|P]]]]

With the <a href="https://pypi.org/project/pendulum/" rel="noreferrer">Pendulum</a> very complete library, we have the <code>subtract</code> method (and not "subStract"):

<pre><code>import pendulum
today = pendulum.datetime.today() # 2020, january
lastmonth = today.subtract(months=1)
lastmonth.strftime('%Y%m')
# '201912'
</code></pre>

We see that it handles jumping years.

The reverse equivalent is <code>add</code>.

<a href="https://pendulum.eustace.io/docs/#addition-and-subtraction" rel="noreferrer">https://pendulum.eustace.io/docs/#addition-and-subtraction</a>

blocks|key|1946915|text|有一个高级库dateparser，它可以确定给定自然语言的过去日期，并返回相应的Python+datetime对象|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1946916|from+dateparser+import+parse
parse('4+months+ago')|code-block|syntax|javascript|1946917|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/scrapinghub/dateparser^0|6|A|1B|8|6|A|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]]]]

There is a high level library <a href="https://github.com/scrapinghub/dateparser" rel="nofollow noreferrer"><code>dateparser</code></a> that can determine the past date given natural language, and return the corresponding Python <code>datetime</code> object

<pre><code>from dateparser import parse
parse('4 months ago')
</code></pre>

blocks|key|1946928|text|您可以按如下方式操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1946929|from+datetime+import+datetime,+timedelta++++
last_month+=+(datetime.now()+-+timedelta(days=32)).strftime("%25Y%25m")|code-block|syntax|javascript|1946930|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|$]]

You can do it as below:
<pre><code>from datetime import datetime, timedelta 
last_month = (datetime.now() - timedelta(days=32)).strftime(&quot;%Y%m&quot;)
</code></pre>

blocks|key|1946991|text|显式方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1946992|import+datetime
result+=+(datetime.datetime.today().month+-+2)+%25+12+%2B+1|code-block|syntax|javascript|1946993|问题是如何将month+1，2，3，...，12转换为12，1，2，...，11。|offset|length|style|CODE|1946994|Step1：month+=+month+-+1+transfer+1，2，3，...，12+to+0，1，2，...，11。|1946995|Step2：month+=+(month+-+1)+%25+12+transfer+0，1，2，...，11+to+11，0，1，...，10。|1946996|Step3：month+=+month+%2B+1+transfer+11，0，1，...，10+to+12，1，2，...，11。|1946997|因此，结果是result+=+(month+-+2)+%25+12+%2B+1|1946998|entityMap^0|0|0|6|5|0|6|H|0|6|O|0|6|H|0|6|T|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@$I|12|J|13|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|14|8|@$I|15|J|16|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|17|8|@$I|18|J|19|K|L]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1A|8|@$I|1B|J|1C|K|L]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|V|$]]

Explicit way:
<pre><code>import datetime
result = (datetime.datetime.today().month - 2) % 12 + 1
</code></pre>
The problem is how to transfer <code>month</code> [1, 2, 3, ..., 12] to [12, 1, 2, ..., 11].
Step1: <code>month = month - 1</code> transfer [1, 2, 3, ..., 12] to [0, 1, 2, ..., 11].
Step2: <code>month = (month - 1) % 12</code> transfer [0, 1, 2, ..., 11] to [11, 0, 1, ..., 10].
Step3: <code>month = month + 1</code> transfer [11, 0, 1, ..., 10] to [12, 1, 2, ..., 11].
So, the result is <code>result = (month - 2) % 12 + 1</code>

I am trying to get the date of the previous month with python.
Here is what i've tried:
<pre><code>str( time.strftime('%Y') ) + str( int(time.strftime('%m'))-1 )
</code></pre>
However, this way is bad for 2 reasons: First it returns 20122 for the February of 2012 (instead of 201202) and secondly it will return 0 instead of 12 on January.
I have solved this trouble in bash with
<pre><code>echo $(date -d&quot;3 month ago&quot; &quot;+%G%m%d&quot;)
</code></pre>
I think that if bash has a built-in way for this purpose, then python, much more equipped, should provide something better than forcing writing one's own script to achieve this goal. Of course i could do something like:
<pre><code>if int(time.strftime('%m')) == 1:
 return '12'
else:
 if int(time.strftime('%m')) &lt; 10:
 return '0'+str(time.strftime('%m')-1)
 else:
 return str(time.strftime('%m') -1)
</code></pre>
I have not tested this code and i don't want to use it anyway (unless I can't find any other way:/)
Thanks for your help!

python date of the previous month

Python

我正在尝试用python获取上个月的日期。下面是我尝试过的：str( time.strftime('%Y') ) + str( int(time.strftime('%m'))-1 )然而，这种方式很糟糕，原因有两个:第一，它在2012年2月返回20122 (而不是201202)；第二，它在1月份返回0而不是12。我已...

问上个月的python日期
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