[python] Is there a simple way to increment a datetime object one month in Python?

So I am trying to find a way to increment a datetime object by one month. However, it seems this is not so simple, according to this question.

I was hoping for something like:

import datetime as dt

now = dt.datetime.now()
later = now + dt.timedelta(months=1)

But that doesn't work. I was also hoping to be able to go to the same day (or the closest alternative) in the next month if possible. For example, a datetime object set at January 1st would increment to Feb 1st whereas a datetime object set at February 28th would increment to March 31st as opposed to March 28th or something.

To be clear, February 28th would (typically) map to March 31st because it is the last day of the month, and thus it should go to the last day of the month for the next month. Otherwise it would be a direct link: the increment should go to the day in the next month with the same numbered day.

Is there a simple way to do this in the current release of Python?

Note: This answer shows how to achieve this using only the datetime and calendar standard library (stdlib) modules - which is what was explicitly asked for. The accepted answer shows how to better achieve this with one of the many dedicated non-stdlib libraries. If you can use non-stdlib libraries, by all means do so for these kinds of date/time manipulations!

How about this?

def add_one_month(orig_date):
    # advance year and month by one month
    new_year = orig_date.year
    new_month = orig_date.month + 1
    # note: in datetime.date, months go from 1 to 12
    if new_month > 12:
        new_year += 1
        new_month -= 12

    new_day = orig_date.day
    # while day is out of range for month, reduce by one
    while True:
        try:
            new_date = datetime.date(new_year, new_month, new_day)
        except ValueError as e:
            new_day -= 1
        else:
            break

    return new_date

EDIT:

Improved version which:

1. keeps the time information if given a datetime.datetime object
2. doesn't use try/catch, instead using calendar.monthrange from the calendar module in the stdlib:

import datetime
import calendar

def add_one_month(orig_date):
    # advance year and month by one month
    new_year = orig_date.year
    new_month = orig_date.month + 1
    # note: in datetime.date, months go from 1 to 12
    if new_month > 12:
        new_year += 1
        new_month -= 12

    last_day_of_month = calendar.monthrange(new_year, new_month)[1]
    new_day = min(orig_date.day, last_day_of_month)

    return orig_date.replace(year=new_year, month=new_month, day=new_day)

>>> now
datetime.datetime(2016, 1, 28, 18, 26, 12, 980861)
>>> later = now.replace(month=now.month+1)
>>> later
datetime.datetime(2016, 2, 28, 18, 26, 12, 980861)

EDIT: Fails on

y = datetime.date(2016, 1, 31); y.replace(month=2) results in ValueError: day is out of range for month

Ther is no simple way to do it, but you can use your own function like answered below.

Question: Is there a simple way to do this in the current release of Python?

Answer: There is no simple (direct) way to do this in the current release of Python.

Reference: Please refer to docs.python.org/2/library/datetime.html, section 8.1.2. timedelta Objects. As we may understand from that, we cannot increment month directly since it is not a uniform time unit.

Plus: If you want first day -> first day and last day -> last day mapping you should handle that separately for different months.