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

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:

1. Create a number_of_days_in_month = [31, 28, ...] array, decide if it’s a leap year and manually sum up the days.
2. 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?

Answers:

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Method 1

Use datetime.timetuple() to convert your datetime object to a time.struct_time object then get its tm_yday property:

from datetime import datetime
day_of_year = datetime.now().timetuple().tm_yday  # returns 1 for January 1st

Method 2

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.

Method 3

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.

Method 4

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

Method 5

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 + 1

Method 6

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

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0