[Solved] How to increment datetime by custom months in python without using library [duplicate]

I need to increment the month of a datetime value

next_month = datetime.datetime(mydate.year, mydate.month+1, 1)

when the month is 12, it becomes 13 and raises error “month must be in 1..12”. (I expected the year would increment)

I wanted to use timedelta, but it doesn’t take month argument.
There is relativedelta python package, but i don’t want to install it just only for this.
Also there is a solution using strtotime.

time = strtotime(str(mydate));
next_month = date("Y-m-d", strtotime("+1 month", time));

I don’t want to convert from datetime to str then to time, and then to datetime; therefore, it’s still a library too

Does anyone have any good and simple solution just like using timedelta?

Enquirer: jargalan

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Solution #1:

Edit – based on your comment of dates being needed to be rounded down if there are fewer days in the next month, here is a solution:

import datetime
import calendar

def add_months(sourcedate, months):
    month = sourcedate.month - 1 + months
    year = sourcedate.year + month // 12
    month = month % 12 + 1
    day = min(sourcedate.day, calendar.monthrange(year,month)[1])
    return datetime.date(year, month, day)

In use:

>>> somedate = datetime.date.today()
>>> somedate
datetime.date(2010, 11, 9)
>>> add_months(somedate,1)
datetime.date(2010, 12, 9)
>>> add_months(somedate,23)
datetime.date(2012, 10, 9)
>>> otherdate = datetime.date(2010,10,31)
>>> add_months(otherdate,1)
datetime.date(2010, 11, 30)

Also, if you’re not worried about hours, minutes and seconds you could use date rather than datetime. If you are worried about hours, minutes and seconds you need to modify my code to use datetime and copy hours, minutes and seconds from the source to the result.

Respondent: jargalan

Solution #2:

This is short and sweet method to add a month to a date using dateutil’s relativedelta.

from datetime import datetime
from dateutil.relativedelta import relativedelta

date_after_month = datetime.today()+ relativedelta(months=1)
print 'Today: ',datetime.today().strftime('%d/%m/%Y')
print 'After Month:', date_after_month.strftime('%d/%m/%Y')

Output:

Today: 01/03/2013

After Month: 01/04/2013

A word of warning: relativedelta(months=1) and relativedelta(month=1) have different meanings. Passing month=1 will replace the month in original date to January whereas passing months=1 will add one month to original date.

Note: this will requires python-dateutil. To install it you need to run in Linux terminal.

sudo apt-get update && sudo apt-get install python-dateutil

Explanation : Add month value in python

Respondent: Dave Webb

Solution #3:

Here’s my salt :

current = datetime.datetime(mydate.year, mydate.month, 1)
next_month = datetime.datetime(mydate.year + int(mydate.month / 12), ((mydate.month % 12) + 1), 1)

Quick and easy 🙂

Respondent: Atul Arvind

Solution #4:

since no one suggested any solution, here is how i solved so far

year, month= divmod(mydate.month+1, 12)
if month == 0: 
      month = 12
      year = year -1
next_month = datetime.datetime(mydate.year + year, month, 1)
Respondent: Cyril N.

Solution #5:

Use the monthdelta package, it works just like timedelta but for calendar months rather than days/hours/etc.

Here’s an example:

from monthdelta import MonthDelta

def prev_month(date):
    """Back one month and preserve day if possible"""
    return date + MonthDelta(-1)

Compare that to the DIY approach:

def prev_month(date):
    """Back one month and preserve day if possible"""
   day_of_month = date.day
   if day_of_month != 1:
           date = date.replace(day=1)
   date -= datetime.timedelta(days=1)
   while True:
           try:
                   date = date.replace(day=day_of_month)
                   return date
           except ValueError:
                   day_of_month -= 1               
Respondent: jargalan

Solution #6:

from datetime import timedelta
try:
    next = (x.replace(day=1) + timedelta(days=31)).replace(day=x.day)
except ValueError:  # January 31 will return last day of February.
    next = (x + timedelta(days=31)).replace(day=1) - timedelta(days=1)

If you simply want the first day of the next month:

next = (x.replace(day=1) + timedelta(days=31)).replace(day=1)
Respondent: Eric Clack

Solution #7:

To calculate the current, previous and next month:

import datetime
this_month = datetime.date.today().month
last_month = datetime.date.today().month - 1 or 12
next_month = (datetime.date.today().month + 1) % 12 or 12
Respondent: Collin Anderson

Solution #8:

Perhaps add the number of days in the current month using calendar.monthrange()?

import calendar, datetime

def increment_month(when):
    days = calendar.monthrange(when.year, when.month)[1]
    return when + datetime.timedelta(days=days)

now = datetime.datetime.now()
print 'It is now %s' % now
print 'In a month, it will be %s' % increment_month(now)
Respondent: Darren Weber

Solution #9:

Similar in ideal to Dave Webb’s solution, but without all of that tricky modulo arithmetic:

import datetime, calendar

def increment_month(date):
    # Go to first of this month, and add 32 days to get to the next month
    next_month = date.replace(day=1) + datetime.timedelta(32)
    # Get the day of month that corresponds
    day = min(date.day, calendar.monthrange(next_month.year, next_month.month)[1])
    return next_month.replace(day=day)
Respondent: uzi

Solution #10:

What about this one? (doesn’t require any extra libraries)

from datetime import date, timedelta
from calendar import monthrange

today = date.today()
month_later = date(today.year, today.month, monthrange(today.year, today.month)[1]) + timedelta(1)
Respondent: Matthew Schinckel

Solution #11:

This implementation might have some value for someone who is working with billing.

If you are working with billing, you probably want to get “the same date next month (if possible)” as opposed to “add 1/12 of one year”.

What is so confusing about this is you actually need take into account two values if you are doing this continuously. Otherwise for any dates past the 27th, you’ll keep losing a few days until you end up at the 27th after leap year.

The values you need to account for:

  • The value you want to add a month to
  • The day you started with

This way if you get bumped from the 31st down to the 30th when you add one month, you’ll get bumped back up to the 31st for the next month that has that day.

This is how I did it:

def closest_date_next_month(year, month, day):
    month = month + 1
    if month == 13:
        month = 1
        year  = year + 1


    condition = True
    while condition:
        try:
            return datetime.datetime(year, month, day)
        except ValueError:
            day = day-1
        condition = day > 26

    raise Exception('Problem getting date next month')

paid_until = closest_date_next_month(
                 last_paid_until.year, 
                 last_paid_until.month, 
                 original_purchase_date.day)  # The trick is here, I'm using the original date, that I started adding from, not the last one
Respondent: Micha? Tabor

Solution #12:

Well with some tweaks and use of timedelta here we go:

from datetime import datetime, timedelta


def inc_date(origin_date):
    day = origin_date.day
    month = origin_date.month
    year = origin_date.year
    if origin_date.month == 12:
        delta = datetime(year + 1, 1, day) - origin_date
    else:
        delta = datetime(year, month + 1, day) - origin_date
    return origin_date + delta

final_date = inc_date(datetime.today())
print final_date.date()
Respondent: Chris Dutrow

Solution #13:

Simplest solution is to go at the end of the month (we always know that months have at least 28 days) and add enough days to move to the next moth:

>>> from datetime import datetime, timedelta
>>> today = datetime.today()
>>> today
datetime.datetime(2014, 4, 30, 11, 47, 27, 811253)
>>> (today.replace(day=28) + timedelta(days=10)).replace(day=today.day)
datetime.datetime(2014, 5, 30, 11, 47, 27, 811253)

Also works between years:

>>> dec31
datetime.datetime(2015, 12, 31, 11, 47, 27, 811253)
>>> today = dec31
>>> (today.replace(day=28) + timedelta(days=10)).replace(day=today.day)
datetime.datetime(2016, 1, 31, 11, 47, 27, 811253)

Just keep in mind that it is not guaranteed that the next month will have the same day, for example when moving from 31 Jan to 31 Feb it will fail:

>>> today
datetime.datetime(2016, 1, 31, 11, 47, 27, 811253)
>>> (today.replace(day=28) + timedelta(days=10)).replace(day=today.day)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: day is out of range for month

So this is a valid solution if you need to move to the first day of the next month, as you always know that the next month has day 1 (.replace(day=1)). Otherwise, to move to the last available day, you might want to use:

>>> today
datetime.datetime(2016, 1, 31, 11, 47, 27, 811253)
>>> next_month = (today.replace(day=28) + timedelta(days=10))
>>> import calendar
>>> next_month.replace(day=min(today.day, 
                               calendar.monthrange(next_month.year, next_month.month)[1]))
datetime.datetime(2016, 2, 29, 11, 47, 27, 811253)
Respondent: dcolish

Solution #14:

I was looking to solve the related problem of finding the date for the first of the following month, regardless of the day in the given date. This does not find the same day 1 month later.

So, if all you want is to put in December 12, 2014 (or any day in December) and get back January 1, 2015, try this:

import datetime

def get_next_month(date):
    month = (date.month % 12) + 1
    year = date.year + (date.month + 1 > 12)
    return datetime.datetime(year, month, 1)
Respondent: amol

Solution #15:

def add_month(d,n=1): return type(d)(d.year+(d.month+n-1)/12, (d.month+n-1)%12+1, 1)
Respondent: Ari Lacenski

Solution #16:

A solution without the use of calendar:

def add_month_year(date, years=0, months=0):
    year, month = date.year + years, date.month + months + 1
    dyear, month = divmod(month - 1, 12)
    rdate = datetime.date(year + dyear, month + 1, 1) - datetime.timedelta(1)
    return rdate.replace(day = min(rdate.day, date.day))
Respondent: Neil C. Obremski

Solution #17:

Just Use This:

import datetime
today = datetime.datetime.today()
nextMonthDatetime = today + datetime.timedelta(days=(today.max.day - today.day)+1)
Respondent: Oliver Wienand

Solution #18:

This is what I came up with

from calendar  import monthrange

def same_day_months_after(start_date, months=1):
    target_year = start_date.year + ((start_date.month + months) / 12)
    target_month = (start_date.month + months) % 12
    num_days_target_month = monthrange(target_year, target_month)[1]
    return start_date.replace(year=target_year, month=target_month, 
        day=min(start_date.day, num_days_target_month))
Respondent: ayged

Solution #19:

def month_sub(year, month, sub_month):
    result_month = 0
    result_year = 0
    if month > (sub_month % 12):
        result_month = month - (sub_month % 12)
        result_year = year - (sub_month / 12)
    else:
        result_month = 12 - (sub_month % 12) + month
        result_year = year - (sub_month / 12 + 1)
    return (result_year, result_month)

def month_add(year, month, add_month):
    return month_sub(year, month, -add_month)

>>> month_add(2015, 7, 1)                        
(2015, 8)
>>> month_add(2015, 7, 20)
(2017, 3)
>>> month_add(2015, 7, 12)
(2016, 7)
>>> month_add(2015, 7, 24)
(2017, 7)
>>> month_add(2015, 7, -2)
(2015, 5)
>>> month_add(2015, 7, -12)
(2014, 7)
>>> month_add(2015, 7, -13)
(2014, 6)
Respondent: jaywhy13

Solution #20:

example using the time object:

start_time = time.gmtime(time.time())    # start now

#increment one month
start_time = time.gmtime(time.mktime([start_time.tm_year, start_time.tm_mon+1, start_time.tm_mday, start_time.tm_hour, start_time.tm_min, start_time.tm_sec, 0, 0, 0]))
Respondent: damn_c

Solution #21:

My very simple solution, which doesn’t require any additional modules:

def addmonth(date):
    if date.day < 20:
        date2 = date+timedelta(32)
    else :
        date2 = date+timedelta(25)
    date2.replace(date2.year, date2.month, day)
    return date2
Respondent: karl

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

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