Add and Subtract Dates Python

Question

I only have datetime, timedelta and date modules in my python. Unfortunately I’ll not be able to use relativedelta to simply add months and years easily.

I would need some advice in adding months and years to a date. Have been trying but can’t think of a better way to take into account for those have >30 days, Leap year.

30days = Date.today() + timedelta(days=30)
Dformat = 30days.strftime(“%Y-%m-%d”)

Anyone has a way to add different months and years in a better way given I only have these limited modules?

So are you trying to add 30 days to lets say, jan, to get output as feb ? — Shabari nath k
– Shabari nath k, Commented Mar 26, 2021 at 7:00

Serge Ballesta · Accepted Answer · 2021-03-26 16:46:32Z

1

I know no builtin way. But it can easily done by hand:

def add_year(dt, years):
    """
    Add years years to dt and return the new value.

    dt can be a date or datetime, years must be an integer value (may be negative)
    """
    try:
        return dt.replace(year=dt.year + years)
    except ValueError:
        # the day that does not exist in new month: return last day of month
        return dt.replace(year=dt.year + years, month=dt.month + 1, day=1
              ) - timedelta(days=1)

def add_month(dt, months):
    """
    Add months months to dt and return the new value.

    dt can be a date or datetime, months must be an integer value (may be negative)
    """
    y, m = divmod(months + dt.month, 12)
    try:
        return dt.replace(year=dt.year + y, month=m)
    except ValueError:
        # the day that does not exist in new month: return last day of month
        return dt.replace(year=dt.year + y, month=m + 1, day=1
                          ) - timedelta(days=1)

Demo:

>>> d = date(2020, 12, 10)
>>> add_month(d, 3)
datetime.date(2021, 3, 10)
>>> add_month(d, -13)
datetime.date(2019, 11, 10)

It even handles shorter months:

>>> d =date(2017,1,30)
>>> add_month(d, 1)
datetime.date(2017, 2, 28)

edited Mar 26, 2021 at 16:46

answered Mar 26, 2021 at 8:32

Serge Ballesta

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4 Comments

FObersteiner Over a year ago

note that this will fail if the initial date is Feb 29th in a leap year and you try to add a number of years that is not a multiple of 4 etc. - dateutil.relativedelta might offer a safer way to go.

Serge Ballesta Over a year ago

@MrFuppes: You are right, I was not cautious enough on my first code. Should be better now. BTW OP explicitely declared I’ll not be able to use relativedelta.

FObersteiner Over a year ago

Ok seems like I should be more cautious reading the question ^^ Actually reads like a homework assignment; "not able to use package x" sounds like "use the hard way and learn something"

MooMay Over a year ago

Managed to figure out with relativddelta installation. But the solution works too.

Gilles Criton · Accepted Answer · 2021-03-26 07:47:16Z

There are a lot of different possibilities. I give you an example which take a date and return the last day of quarter but you can, obviously, apply this methodology to any frequency, So:

If you have datetime, you can have a object dt defines as a datetime.date

You can build a function get_quarter_end which take as input a datetime.date object and return the last day of quarter.

def get_quarter_end(dt):
    # dt: datetime.date
    # firts we get the year of the next quarter
    nextQtYr = dt.year + (1 if dt.month > 9 else 0)
    # then the month
    nextQtFirstMo = (dt.month - 1) // 3 * 3 + 4
    nextQtFirstMo = 1 if nextQtFirstMo == 13 else nextQtFirstMo
    # finally the day (first day of the next quarter)
    nextQtFirstDy = date(nextQtYr, nextQtFirstMo, 1)
    # so we have all information about the next quarter, because we are  
    # interested by the date at the end of the quarter, we just need to 
    # substract 1 day: 
    return nextQtFirstDy - timedelta(days=1)

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