Add/deduct month taking into account unequal days in calendar month

Question

I am trying to come up with a way to create a list of dates n months back from given date dt. However, it seems to tricky based on what dt is. Below I am illustrating the dilemma through a few examples (esp. look at tricky case-3 below):

from datetime import datetime
from dateutil.relativedelta import relativedelta
# Simple case.
dt = datetime(2021, 2, 15)
dt - relativedelta(months=1)  # n=1 gives datetime.datetime(2021, 1, 15, 0, 0)
dt - relativedelta(months=2)  # n=2 gives datetime.datetime(2020, 12, 15, 0, 0)
# Simple case-2
dt = datetime(2021, 3, 31)
dt - relativedelta(months=1)  # n=1 gives datetime.datetime(2021, 2, 28, 0, 0)
dt - relativedelta(months=2)  # n=2 gives datetime.datetime(2021, 1, 31, 0, 0)
dt - relativedelta(months=3)  # n=3 gives datetime.datetime(2020, 12, 31, 0, 0)
dt - relativedelta(months=4)  # n=4 gives datetime.datetime(2020, 11, 30, 0, 0)
# Tricky case-3
dt = datetime(2021, 2, 28)
dt - relativedelta(months=1)  # n=1 gives datetime.datetime(2021, 1, 28, 0, 0) and not datetime.datetime(2021, 1, 31, 0, 0)
dt - relativedelta(months=2)  # n=2 gives datetime.datetime(2020, 12, 28, 0, 0) and not datetime.datetime(2020, 12, 31, 0, 0)
dt - relativedelta(months=3)  # n=3 gives datetime.datetime(2020, 11, 28, 0, 0) and not datetime.datetime(2020, 11, 30, 0, 0)
dt - relativedelta(months=4)  # n=4 gives datetime.datetime(2020, 10, 28, 0, 0) and not datetime.datetime(2020, 10, 31, 0, 0)

relativedelta in this application not only fails for Feb 28th, but anytime a month has less than 31 days and `dt` represents the end of the month. — FObersteiner, Aug 22 '21 at 14:15
Thanks @MrFuppes that is exactly the issue. I was thinking if there is a more natural way. — Gerry, Aug 22 '21 at 18:24

FObersteiner · Accepted Answer · 2023-04-13T08:13:07.657

relativedelta seems to give unexpected results on the corner case of date is end of month while month has less than 31 days. Here's a work-around:

check if date is end of month
if not, simply use relativedelta
if so, use relativedelta but make sure the day is the last of the month by setting the day attribute explicitly

from datetime import datetime, timedelta
from dateutil.relativedelta import relativedelta

# add_month adds n months to datetime object dt
def add_month(dt, n):
    # we can add a day without month changing - not end of month:
    if (dt + timedelta(1)).month == dt.month:
        return dt + relativedelta(months=n)
    # implicit else: end of month
    return (dt + relativedelta(months=n+1)).replace(day=1) - timedelta(1)

Examples:

d = datetime(2021, 3, 15)
print(add_month(d, -1).date(), d.date(), add_month(d, 1).date())
# 2021-02-15 2021-03-15 2021-04-15

d = datetime(2021, 3, 31)
print(add_month(d, -1).date(), d.date(), add_month(d, 1).date())
# 2021-02-28 2021-03-31 2021-04-30

d = datetime(2021,2,28)
print(add_month(d, -1).date(), d.date(), add_month(d, 1).date())
# 2021-01-31 2021-02-28 2021-03-31

d = datetime(2021,11,30)
print(add_month(d, -1).date(), d.date(), add_month(d, 1).date())
# 2021-10-31 2021-11-30 2021-12-31

@dataviews what do you mean by "it misses"? As I've commented on your (deleted) question, I think this approach *will* create duplicate dates and miss dates. However, here, it was not the question to avoid that. — FObersteiner, Apr 13 '23 at 08:21

Add/deduct month taking into account unequal days in calendar month

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