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()
>>> o         


        

Answer 1

After implementing the accepted solution, I had the revelation that this modern pandas version is identical, and much simpler:

dat['decimal_date']=dat.index.year+ (dat.index.dayofyear -1)/365

Must be used on a date-time index Pandas dataframe. Adding as this solution post comes up in the top of my google search for this issue.

Answer 2

If you want to include the minutes and seconds use this:

YearF=[(x.timetuple().tm_yday-1+x.timetuple().tm_hour/24+x.timetuple().tm_min/(60*24)+x.timetuple().tm_sec/(24*3600))/(365+((x.timetuple().tm_year//4)==(x.timetuple().tm_year/4)))+x.timetuple().tm_year for x in DateArray]

Answer 3

I'm assuming that you are using this to compare datetime values. To do that, please use the the timedelta objects instead of reiniventing the wheel.

Example:

>>> from datetime import timedelta
>>> from datetime import datetime as dt
>>> d = dt.now()
>>> year = timedelta(days=365)
>>> tomorrow = d + timedelta(days=1)
>>> tomorrow + year > d + year
True

If for some reason you truly need decimal years, datetime objects method strftime() can give you an integer representation of day of the year if asked for %j - if this is what you are looking for, see below for a simple sample (only on 1 day resolution):

>>> from datetime import datetime
>>> d = datetime(2007, 4, 14, 11, 42, 50)
>>> (float(d.strftime("%j"))-1) / 366 + float(d.strftime("%Y"))
2007.2814207650274

Answer 4

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.

Answer 5

Surprised no one has mentioned this... but the datetime.timedelta objects that result from subtracting datetime.datetime objects have a division method. So, you could use the simple function

from datetime import datetime
def datetime2year(dt): 
    year_part = dt - datetime(year=dt.year, month=1, day=1)
    year_length = (
        datetime(year=dt.year + 1, month=1, day=1)
        - datetime(year=dt.year, month=1, day=1)
    )
    return dt.year + year_part / year_length

where the division is between datetime.timedelta objects.

Answer 6

It's possible to calculate decimal date by using Pandas's julian date and the following formulas.

In the case where your pandas dataframe has an index that is date-time:

JD=dat.index.to_julian_date() #create julian date
L= JD+68569
N= 4*L/146097
L= L-(146097*N+3)/4
I= 4000*(L+1)/1461001
L= L-1461*I/4+31
J= 80*L/2447
K= L-2447*J/80
L= J/11
J= J+2-12*L
decimal_date= 100*(N-49)+I+L

decimal_date is a series of your date (in the same TZ as the dataframe index) in form of something like 2007.123452.

Adapted from this post.