Add one month to a given date (rounded day after) with Python

Question

I'd like to add one month to a given date

import datetime
dt = datetime.datetime(year=2014, month=5, day=2)

so I should get

datetime.datetime(year=2014, month=6, day=2)

but with

dt = datetime.datetime(year=2015, month=1, day=31)

I should get

datetime.datetime(year=2015, month=3, day=1)

because there is no 2015-02-31 (and I want my result being round one day after)

Some months have 31 days, some other 30, some 29, some 28 !

so adding a datetime.timedelta is probably not a good manner of doing (because we don't know the number of days to add)

I noticed that Pandas have an interesting concept of DateOffset

http://pandas.pydata.org/pandas-docs/stable/timeseries.html#dateoffset-objects

but I didn't find a Month offset, just MonthBegin or MonthEnd

I see also this post How do I calculate the date six months from the current date using the datetime Python module?

so I tried dateutil.relativedelta but

from dateutil.relativedelta import relativedelta
datetime.datetime(year=2015, month=1, day=31)+relativedelta(months=1)

returns

datetime.datetime(2015, 2, 28, 0, 0)

so result was rounded one day before.

Is there a (clean) way to round day after ?

edit: I gave an example with one month to add but I also want to be able to add for example : 2 years and 6 months (using a relativedelta(years=2, months=6))

Answer 1

You can use dateutil.relativedelta.relativedelta and manually check the datetime.day attribute, if the original day is greater than the new day, then add a day.

The function below accepts a datetime object and relativedelta object. Note that the code below only works for years and months, I don't believe it'll work if you use anything below that (days, hours, etc). You could easily modify this function to take years and months as arguments and then construct the relativedelta inside the function itself.

from datetime import datetime
from dateutil.relativedelta import relativedelta

def add_time(d, rd):
    day = relativedelta(days=+1)

    out = d + rd
    if d.day > out.day:
        out = out + day

    return out    

# Check that it "rolls over"
print(add_time(datetime(year=2015, month=1, day=29), relativedelta(years=+4, months=+1))) # 2019-03-01 00:00:00
print(add_time(datetime(year=2015, month=3, day=31), relativedelta(years=+0, months=+2))) # 2015-05-01 00:00:00

# Check that it handles "normal" scenarios
print(add_time(datetime(year=2015, month=6, day=19), relativedelta(months=+1))) # 2015-07-19 00:00:00
print(add_time(datetime(year=2015, month=6, day=30), relativedelta(years=+2, months=+1))) # 2017-07-30 00:00:00

# Check across years
print(add_time(datetime(year=2015, month=12, day=25), relativedelta(months=+1))) # 2016-01-25 00:00:00

# Check leap years
print(add_time(datetime(year=2016, month=1, day=29), relativedelta(years=+4, months=+1))) # 2020-02-29 00:00:00

Answer 2

This seems to work. It is quite clean, but not beautiful:

def add_month(now):
    try:
        then = (now + relativedelta(months=1)).replace(day=now.day)
    except ValueError:
        then = (now + relativedelta(months=2)).replace(day=1)
    return then

for now in [datetime(2015, 1, 20), datetime(2015, 1, 31), datetime(2015, 2, 28)]:
    print now, add_month(now)

prints:

2015-01-20 00:00:00 2015-02-20 00:00:00
2015-01-31 00:00:00 2015-03-01 00:00:00
2015-02-28 00:00:00 2015-03-28 00:00:00

It adds one month and tries to replace the day with the original day. If it succeeds, it is no special case. If it fails (ValueError), we have to add another month and go to its first day.