blocks|key|3338077|text|您可以将strftime与%25j+format+string一起使用|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3338078|>>>+import+datetime
>>>+today+=+datetime.datetime.now()
>>>+today.strftime('%25j')
'065'|code-block|syntax|javascript|3338079|但是，如果您希望对此数字进行比较或计算，则必须将其转换为int()，因为strftime()返回一个字符串。如果是这样的话，您最好使用DzinX's+answer。|3338080|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/datetime.html#datetime.datetime.strftime|1|https://docs.python.org/3/library/datetime.html#strftime-and-strptime-format-codes|2|https://stackoverflow.com/questions/620305/convert-year-month-day-to-day-of-year-in-python/623312#623312^0|4|8|D|2|4|8|0|D|G|1|0|0|S|5|10|A|1V|7|2|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]|$9|16|A|17|1|18]]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|M|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]|W|$5|P|Q|R|E|$S|X]]]]

You could use <a href="https://docs.python.org/3/library/datetime.html#datetime.datetime.strftime" rel="noreferrer"><code>strftime</code></a> with a <a href="https://docs.python.org/3/library/datetime.html#strftime-and-strptime-format-codes" rel="noreferrer"><code>%j</code> format string</a>:
<pre><code>&gt;&gt;&gt; import datetime
&gt;&gt;&gt; today = datetime.datetime.now()
&gt;&gt;&gt; today.strftime('%j')
'065'
</code></pre>
but if you wish to do comparisons or calculations with this number, you would have to convert it to <code>int()</code> because <code>strftime()</code> returns a string. If that is the case, you are better off using <a href="https://stackoverflow.com/questions/620305/convert-year-month-day-to-day-of-year-in-python/623312#623312">DzinX's</a> answer.

blocks|key|347671|text|只需从日期中减去1月1日：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|347672|import+datetime
today+=+datetime.datetime.now()
day_of_year+=+(today+-+datetime.datetime(today.year,+1,+1)).days+%2B+1|code-block|syntax|javascript|347673|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 subtract january 1 from the date:

<pre><code>import datetime
today = datetime.datetime.now()
day_of_year = (today - datetime.datetime(today.year, 1, 1)).days + 1
</code></pre>

blocks|key|348496|text|使用datetime.timetuple()将datetime对象转换为time.struct_time对象，然后获取其tm_yday属性：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|348497|from+datetime+import+datetime
day_of_year+=+datetime.now().timetuple().tm_yday++#+returns+1+for+January+1st|code-block|syntax|javascript|348498|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/datetime.html#datetime.datetime.timetuple|1|https://docs.python.org/3/library/time.html#time.struct_time|2^0|2|K|N|8|10|G|1O|7|2|K|0|10|G|1|1O|7|2|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@$9|14|A|15|1|16]|$9|17|A|18|1|19]|$9|1A|A|1B|1|1C]]|E|$]]|$1|F|3|G|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]|S|$5|N|O|P|E|$Q|T]]|U|$5|N|O|P|E|$Q|T]]]]

Use <a href="https://docs.python.org/3/library/datetime.html#datetime.datetime.timetuple" rel="noreferrer"><code>datetime.timetuple()</code></a> to convert your <code>datetime</code> object to a <a href="https://docs.python.org/3/library/time.html#time.struct_time" rel="noreferrer"><code>time.struct_time</code></a> object then get its <a href="https://docs.python.org/3/library/time.html#time.struct_time" rel="noreferrer"><code>tm_yday</code></a> property:
<pre><code>from datetime import datetime
day_of_year = datetime.now().timetuple().tm_yday # returns 1 for January 1st
</code></pre>

blocks|key|3125223|text|对于这个问题，DZinX's+answer是一个很好的答案。我发现了这个问题，并在寻找反函数时使用了DZinX's+answer：将带有儒略日期的日期转换为日期时间。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3125224|我发现这是可行的：|3125225|import+datetime
datetime.datetime.strptime('1936-077T13:14:15','%25Y-%25jT%25H:%25M:%25S')

>>>>+datetime.datetime(1936,+3,+17,+13,+14,+15)

datetime.datetime.strptime('1936-077T13:14:15','%25Y-%25jT%25H:%25M:%25S').timetuple().tm_yday

>>>>+77|code-block|syntax|javascript|3125226|或者在数字上：|3125227|import+datetime
year,julian+=+[1936,77]
datetime.datetime(year,+1,+1)%2Bdatetime.timedelta(days=julian+-1)

>>>>+datetime.datetime(1936,+3,+17,+0,+0)|3125228|或者使用在某些域中流行的基于小数1的jdate：|3125229|jdate_frac+=+(datetime.datetime(1936,+3,+17,+13,+14,+15)-datetime.datetime(1936,+1,+1)).total_seconds()/86400%2B1
display(jdate_frac)

>>>>+77.5515625

year,julian+=+[1936,jdate_frac]
display(datetime.datetime(year,+1,+1)%2Bdatetime.timedelta(days=julian+-1))

>>>>+datetime.datetime(1936,+3,+17,+13,+14,+15)|3125230|我不确定这里的礼仪，但我认为指向反向功能的指针可能对像我这样的人有用。|3125231|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/623312/40785|1^0|7|E|0|1E|E|1|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@$A|14|B|15|1|16]|$A|17|B|18|1|19]]|C|$]]|$1|D|3|E|5|6|7|1A|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|1B|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|1C|8|@]|9|@]|C|$]]|$1|M|3|N|5|H|7|1D|8|@]|9|@]|C|$I|J]]|$1|O|3|P|5|6|7|1E|8|@]|9|@]|C|$]]|$1|Q|3|R|5|H|7|1F|8|@]|9|@]|C|$I|J]]|$1|S|3|T|5|6|7|1G|8|@]|9|@]|C|$]]|$1|U|3|-4|5|6|7|1H|8|@]|9|@]|C|$]]]|V|$W|$5|X|Y|Z|C|$10|11]]|12|$5|X|Y|Z|C|$10|11]]]]

<a href="https://stackoverflow.com/a/623312/40785">DZinX's answer</a> is a great answer for the question. I found this question and used <a href="https://stackoverflow.com/a/623312/40785">DZinX's answer</a> while looking for the inverse function: convert dates with the julian day-of-year into the datetimes.
I found this to work:
<pre><code>import datetime
datetime.datetime.strptime('1936-077T13:14:15','%Y-%jT%H:%M:%S')

&gt;&gt;&gt;&gt; datetime.datetime(1936, 3, 17, 13, 14, 15)

datetime.datetime.strptime('1936-077T13:14:15','%Y-%jT%H:%M:%S').timetuple().tm_yday

&gt;&gt;&gt;&gt; 77
</code></pre>
Or numerically:
<pre><code>import datetime
year,julian = [1936,77]
datetime.datetime(year, 1, 1)+datetime.timedelta(days=julian -1)

&gt;&gt;&gt;&gt; datetime.datetime(1936, 3, 17, 0, 0)
</code></pre>
Or with fractional 1-based jdates popular in some domains:
<pre><code>jdate_frac = (datetime.datetime(1936, 3, 17, 13, 14, 15)-datetime.datetime(1936, 1, 1)).total_seconds()/86400+1
display(jdate_frac)

&gt;&gt;&gt;&gt; 77.5515625

year,julian = [1936,jdate_frac]
display(datetime.datetime(year, 1, 1)+datetime.timedelta(days=julian -1))

&gt;&gt;&gt;&gt; datetime.datetime(1936, 3, 17, 13, 14, 15)
</code></pre>
I'm not sure of etiquette around here, but I thought a pointer to the inverse functionality might be useful for others like me.

blocks|key|3338169|text|我想在Python3.4，Linux+x64上展示不同方法的性能。摘自line+profiler：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3338170|++++++Line+#++++++Hits+++++++++Time++Per+Hit+++%25+Time++Line+Contents
++++++==============================================================
+++++++++(...)
+++++++++823++++++1508++++++++11334++++++7.5+++++41.6++++++++++yday+=+int(period_end.strftime('%25j'))
+++++++++824++++++1508+++++++++2492++++++1.7++++++9.1++++++++++yday+=+period_end.toordinal()+-+date(period_end.year,+1,+1).toordinal()+%2B+1
+++++++++825++++++1508+++++++++1852++++++1.2++++++6.8++++++++++yday+=+(period_end+-+date(period_end.year,+1,+1)).days+%2B+1
+++++++++826++++++1508+++++++++5078++++++3.4+++++18.6++++++++++yday+=+period_end.timetuple().tm_yday
+++++++++(...)|code-block|syntax|javascript|3338171|所以最有效的是|3338172|yday+=+(period_end+-+date(period_end.year,+1,+1)).days|3338173|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|$]]

I want to present performance of different approaches, on Python 3.4, Linux x64. Excerpt from line profiler:

<pre><code> Line # Hits Time Per Hit % Time Line Contents
 ==============================================================
 (...)
 823 1508 11334 7.5 41.6 yday = int(period_end.strftime('%j'))
 824 1508 2492 1.7 9.1 yday = period_end.toordinal() - date(period_end.year, 1, 1).toordinal() + 1
 825 1508 1852 1.2 6.8 yday = (period_end - date(period_end.year, 1, 1)).days + 1
 826 1508 5078 3.4 18.6 yday = period_end.timetuple().tm_yday
 (...)
</code></pre>

So most efficient is

<pre><code>yday = (period_end - date(period_end.year, 1, 1)).days
</code></pre>

blocks|key|3338204|text|如果您有理由避免使用datetime模块，那么这些函数将会起作用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3338205|def+is_leap_year(year):
++++"""+if+year+is+a+leap+year+return+True
++++++++else+return+False+"""
++++if+year+%25+100+==+0:
++++++++return+year+%25+400+==+0
++++return+year+%25+4+==+0

def+doy(Y,M,D):
++++"""+given+year,+month,+day+return+day+of+year
++++++++Astronomical+Algorithms,+Jean+Meeus,+2d+ed,+1998,+chap+7+"""
++++if+is_leap_year(Y):
++++++++K+=+1
++++else:
++++++++K+=+2
++++N+=+int((275+*+M)+/+9.0)+-+K+*+int((M+%2B+9)+/+12.0)+%2B+D+-+30
++++return+N

def+ymd(Y,N):
++++"""+given+year+=+Y+and+day+of+year+=+N,+return+year,+month,+day
++++++++Astronomical+Algorithms,+Jean+Meeus,+2d+ed,+1998,+chap+7+"""++++
++++if+is_leap_year(Y):
++++++++K+=+1
++++else:
++++++++K+=+2
++++M+=+int((9+*+(K+%2B+N))+/+275.0+%2B+0.98)
++++if+N+<+32:
++++++++M+=+1
++++D+=+N+-+int((275+*+M)+/+9.0)+%2B+K+*+int((M+%2B+9)+/+12.0)+%2B+30
++++return+Y,+M,+D|code-block|syntax|javascript|3338206|entityMap^0|A|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

If you have reason to avoid the use of the <code>datetime</code> module, then these functions will work.

<pre><code>def is_leap_year(year):
 """ if year is a leap year return True
 else return False """
 if year % 100 == 0:
 return year % 400 == 0
 return year % 4 == 0

def doy(Y,M,D):
 """ given year, month, day return day of year
 Astronomical Algorithms, Jean Meeus, 2d ed, 1998, chap 7 """
 if is_leap_year(Y):
 K = 1
 else:
 K = 2
 N = int((275 * M) / 9.0) - K * int((M + 9) / 12.0) + D - 30
 return N

def ymd(Y,N):
 """ given year = Y and day of year = N, return year, month, day
 Astronomical Algorithms, Jean Meeus, 2d ed, 1998, chap 7 """ 
 if is_leap_year(Y):
 K = 1
 else:
 K = 2
 M = int((9 * (K + N)) / 275.0 + 0.98)
 if N &lt; 32:
 M = 1
 D = N - int((275 * M) / 9.0) + K * int((M + 9) / 12.0) + 30
 return Y, M, D
</code></pre>

I'm using the <a href="https://docs.python.org/3/library/datetime.html" rel="noreferrer"><code>datetime</code></a> module, i.e.:
<pre><code>&gt;&gt;&gt; import datetime
&gt;&gt;&gt; today = datetime.datetime.now()
&gt;&gt;&gt; print(today)
2009-03-06 13:24:58.857946
</code></pre>
and I would like to compute the day of year that takes leap years into account. e.g. today (March 6, 2009) is the 65th day of 2009.
I see a two options:
<ol>
<li>Create a <code>number_of_days_in_month = [31, 28, ...]</code> array, decide if it's a leap year and manually sum up the days.
</li>
<li>Use <code>datetime.timedelta</code> to make a guess &amp; then binary search for the correct day of the year:
<pre><code>&gt;&gt;&gt; import datetime
&gt;&gt;&gt; YEAR = 2009
&gt;&gt;&gt; DAY_OF_YEAR = 62
&gt;&gt;&gt; d = datetime.date(YEAR, 1, 1) + datetime.timedelta(DAY_OF_YEAR - 1)
</code></pre>
</li>
</ol>
These both feel pretty clunky &amp; I have a gut feeling that there's a more &quot;Pythonic&quot; way of calculating the day of the year. Any ideas/suggestions?

Convert Year/Month/Day to Day of Year in Python

Linux

我使用的是模块，即：>>> import datetime>>> today = datetime.datetime.now()>>> print(today)2009-03-06 13:24:58.857946我想计算一下一年中的哪一天，考虑到了闰年。今天(2009年3月6日)是2009年的第六十五天。我看到了两个选...

问在Python中将年/月/日转换为年中的日
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问在Python中将年/月/日转换为年中的日EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问在Python中将年/月/日转换为年中的日
EN