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

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.

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