How to convert Python datetime dates to decimal/float years

Question

I am looking for a way to convert datetime objects to decimal(/float) year, including fractional part. Example:

>>> obj = SomeObjet()
>>> obj.DATE_OBS
datetime.datetime(2007, 4, 14, 11, 42, 50)

How do I convert datetime.datetime(2007, 4, 14, 11, 42, 50) to decimal years. By decimal format I mean the float value 2007.4523, where the fractional part is the number of seconds from the beginning of the year (2007-01-01 till 2007-04-14), divided by the total number of seconds in that year (2007-01-01 till 2008-01-01).

(NOTE: in statistical modeling (e.g. for linear regression), this is called "time index")

Answer 1

from datetime import datetime as dt
import time

def toYearFraction(date):
    def sinceEpoch(date): # returns seconds since epoch
        return time.mktime(date.timetuple())
    s = sinceEpoch

    year = date.year
    startOfThisYear = dt(year=year, month=1, day=1)
    startOfNextYear = dt(year=year+1, month=1, day=1)

    yearElapsed = s(date) - s(startOfThisYear)
    yearDuration = s(startOfNextYear) - s(startOfThisYear)
    fraction = yearElapsed/yearDuration

    return date.year + fraction

Demo:

>>> toYearFraction(dt.today())
2011.47447514

This method is probably accurate to within the second (or the hour if daylight savings or other strange regional things are in effect). It also works correctly during leapyears. If you need drastic resolution (such as due to changes in the Earth's rotation) you are better off querying a net service.

Answer 2

This is a little simpler way than the other solutions:

import datetime
def year_fraction(date):
    start = datetime.date(date.year, 1, 1).toordinal()
    year_length = datetime.date(date.year+1, 1, 1).toordinal() - start
    return date.year + float(date.toordinal() - start) / year_length

>>> print year_fraction(datetime.datetime.today())
2016.32513661

Note that this calculates the fraction based on the start of the day, so December 31 will be 0.997, not 1.0.