Python: get datetime for '3 years ago today'

In Python, how do I get a datetime object for '3 years ago today'?

UPDATE: FWIW, I don't care hugely about accuracy... i.e. it's Feb 29th today, I don't care whether I'm given Feb 28th or March 1st in my answer. Concision is more important than configurability, in this case.

If you need to be exact use the dateutil module to calculate relative dates

from datetime import datetime
from dateutil.relativedelta import relativedelta

three_yrs_ago = datetime.now() - relativedelta(years=3)

import datetime
datetime.datetime.now() - datetime.timedelta(days=3*365)

Subtracting 365*3 days is wrong, of course--you're crossing a leap year more than half the time.

dt = datetime.now()
dt = dt.replace(year=dt.year-3)
# datetime.datetime(2008, 3, 1, 13, 2, 36, 274276)

ED: To get the leap-year issue right,

def subtract_years(dt, years):
    try:
        dt = dt.replace(year=dt.year-years)
    except ValueError:
        dt = dt.replace(year=dt.year-years, day=dt.day-1)
    return dt

def add_years(dt, years):
	try:
		result = datetime.datetime(dt.year + years, dt.month, dt.day, dt.hour, dt.minute, dt.second, dt.microsecond, dt.tzinfo)
	except ValueError:
		result = datetime.datetime(dt.year + years, dt.month, dt.day - 1, dt.hour, dt.minute, dt.second, dt.microsecond, dt.tzinfo)
	return result

>>> add_years(datetime.datetime.now(), -3)
datetime.datetime(2008, 3, 1, 12, 2, 35, 22000)
>>> add_years(datetime.datetime(2008, 2, 29), -3)
datetime.datetime(2005, 2, 28, 0, 0)

This works to cater for leap year corner cases and non-leap years too. Because, if day = 29 and month = 2 (Feb), a non-leap year would throw a value error because there is no 29th Feb and the last day of Feb would be 28th, thus doing a -1 on the date works in a try-except block.

from datetime import datetime

last_year = datetime.today().year - 1
month = datetime.today().month
day = datetime.today().day

try:
    # try returning same date last year
    last_year_date = datetime.strptime(f"{last_year}-{month}-{day}",'%Y-%m-%d').date()
except ValueError: 
    # incase of error due to leap year, return date - 1 in last year
    last_year_date = datetime.strptime(f"{last_year}-{month}-{day-1}",'%Y-%m-%d').date()

print(last_year_date)

In [3]: import datetime as dt

In [4]: today=dt.date.today()

In [5]: three_years_ago=today-dt.timedelta(days=3*365)

In [6]: three_years_ago
Out[6]: datetime.date(2008, 3, 1)

Why not simply do a check for leap year before replacing the year. This does not need any extra package or try:Except.

def years_ago(dt, years):
    if dt.month == 2 and dt.day == 29:
        dt = dt.replace(day=28)

    return dt.replace(year=dt.year - years)