Pythonic difference between two dates in years?

Question

Is there a more efficient way of doing this below? I want to have the difference in years between two dates as a single scalar. Any suggestions are welcome.

Answer 1

I think what you're looking for is:

difference_in_years = difference.dt.days / 365.25

Answer 2

Before install library :

choco upgrade Python -y
python pip install python-dateutil

In One line in Python in Cmder (windows) :

python -c "import datetime; from dateutil.relativedelta import relativedelta; myBirthday = datetime.datetime(2019,2,6,11,0,0,0); now = datetime.datetime.utcnow(); diff = relativedelta(now, myBirthday); print ("'My'+'" "'+'year'+'" "'+':'+'" "'+'%d''" "''and''" "''%d''" "''microseconds'" % (diff.years, diff.microseconds))"

In Batch escape percent with %% :

python -c "import datetime; from dateutil.relativedelta import relativedelta; myBirthday = datetime.datetime(2019,2,6,11,0,0,0); now = datetime.datetime.utcnow(); diff = relativedelta(now, myBirthday); print ("'My'+'" "'+'year'+'" "'+':'+'" "'+'%%d''" "''and''" "''%%d''" "''microseconds'" %% (diff.years, diff.microseconds))"

Answer 3

More robust function - calculates difference in years (age) and days:

def get_diff_in_years_and_days(from_date, to_date):
    try:
        from_in_this_year = date(to_date.year, from_date.month, from_date.day)
    except:
        from_in_this_year = date(to_date.year, from_date.month, from_date.day-1) # today is feb in leap year

    if from_in_this_year <= to_date:
        years = to_date.year - from_date.year
        days = (to_date - from_in_this_year).days
    else:
        years = to_date.year - from_date.year - 1
        try:
            from_in_prev_year = date(to_date.year-1, from_date.month, from_date.day)
        except:
            from_in_prev_year = date(to_date.year-1, from_date.month, from_date.day-1) # today is feb in leap year
        days = (to_date - from_in_prev_year).days

    assert days>=0 and days<=365, days
    assert years>=0, years

    return years, days

some unit-tests:

self.assertEqual((0,  0), get_diff_in_years_and_days(date(2018,1, 1), date(2018,1, 1)))
self.assertEqual((1,  0), get_diff_in_years_and_days(date(2017,1, 1), date(2018,1, 1)))
self.assertEqual((1,  1), get_diff_in_years_and_days(date(2017,1, 1), date(2018,1, 2)))
self.assertEqual((2,  0), get_diff_in_years_and_days(date(2016,2,29), date(2018,2,28)))
self.assertEqual((2,  1), get_diff_in_years_and_days(date(2014,2,28), date(2016,2,29)))
self.assertEqual((1,364), get_diff_in_years_and_days(date(2014,2,28), date(2016, 2,27)))
self.assertEqual((3,30) , get_diff_in_years_and_days(date(2015,10,1), date(2018,10,31)))
self.assertEqual((10,30), get_diff_in_years_and_days(date(2010,10,1), date(2020,10,31)))
self.assertEqual((3,31) , get_diff_in_years_and_days(date(2015,10,1), date(2018,11, 1)))
self.assertEqual((2,364), get_diff_in_years_and_days(date(2015,10,1), date(2018, 9,30)))

Answer 4

Here's what I came up with, without using an external dependency:

def year_diff(d1, d2):
    """Returns the number of years between the dates as a positive integer."""
    later = max(d1, d2)
    earlier = min(d1, d2)

    result = later.year - earlier.year
    if later.month < earlier.month or (later.month == earlier.month and later.day < earlier.day):
        result -= 1

    return result

Answer 5

If you want precise results, I recommend using the dateutil library.

from dateutil.relativedelta import relativedelta
difference_in_years = relativedelta(end_date, start_date).years

This is for complete years (e.g. a person's age). If you want fractional years, then add months, days, hours, ... up to the desired precision.

Answer 6

Just do this:

from dateutil.relativedelta import relativedelta

myBirthday = datetime.datetime(1983,5,20,0,0,0,0)
now = datetime.datetime.now()



difference = relativedelta(now, myBirthday)
print("My years: "+str(difference.years))