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

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.

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

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

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.