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

Question

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

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.

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.

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