I'm using the datetime
module, i.e.:
>>> import datetime
>>> today = datetime.datetime.now()
>>> print(today)
2009-03-06 13:24:58.857946
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:
Create a number_of_days_in_month = [31, 28, ...]
array, decide if it's a leap year and manually sum up the days.
Use datetime.timedelta
to make a guess & then binary search for the correct day of the year:
>>> import datetime
>>> YEAR = 2009
>>> DAY_OF_YEAR = 62
>>> d = datetime.date(YEAR, 1, 1) + datetime.timedelta(DAY_OF_YEAR - 1)
These both feel pretty clunky & I have a gut feeling that there's a more "Pythonic" way of calculating the day of the year. Any ideas/suggestions?
DZinX's answer is a great answer for the question. I found this question and used DZinX's answer while looking for the inverse function: convert dates with the julian day-of-year into the datetimes.
I found this to work:
import datetime
datetime.datetime.strptime('1936-077T13:14:15','%Y-%jT%H:%M:%S')
>>>> datetime.datetime(1936, 3, 17, 13, 14, 15)
datetime.datetime.strptime('1936-077T13:14:15','%Y-%jT%H:%M:%S').timetuple().tm_yday
>>>> 77
Or numerically:
import datetime
year,julian = [1936,77]
datetime.datetime(year, 1, 1)+datetime.timedelta(days=julian -1)
>>>> datetime.datetime(1936, 3, 17, 0, 0)
Or with fractional 1-based jdates popular in some domains:
jdate_frac = (datetime.datetime(1936, 3, 17, 13, 14, 15)-datetime.datetime(1936, 1, 1)).total_seconds()/86400+1
display(jdate_frac)
>>>> 77.5515625
year,julian = [1936,jdate_frac]
display(datetime.datetime(year, 1, 1)+datetime.timedelta(days=julian -1))
>>>> datetime.datetime(1936, 3, 17, 13, 14, 15)
I'm not sure of etiquette around here, but I thought a pointer to the inverse functionality might be useful for others like me.
Just subtract january 1 from the date:
import datetime
today = datetime.datetime.now()
day_of_year = (today - datetime.datetime(today.year, 1, 1)).days + 1
I want to present performance of different approaches, on Python 3.4, Linux x64. Excerpt from line profiler:
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
(...)
So most efficient is
yday = (period_end - date(period_end.year, 1, 1)).days
You could use strftime
with a %j
format string:
>>> import datetime
>>> today = datetime.datetime.now()
>>> today.strftime('%j')
'065'
but if you wish to do comparisons or calculations with this number, you would have to convert it to int()
because strftime()
returns a string. If that is the case, you are better off using DzinX's answer.
If you have reason to avoid the use of the datetime
module, then these functions will work.
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 < 32:
M = 1
D = N - int((275 * M) / 9.0) + K * int((M + 9) / 12.0) + 30
return Y, M, D
Source: Stackoverflow.com