How to increment datetime by custom months in python without using library
Question:
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?
Answers:
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()
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)
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)
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.
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)
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]))
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))
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
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
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'))
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 require python-dateutil module. If you are on Linux you need to run this command in the terminal in order to install it.
sudo apt-get update && sudo apt-get install python-dateutil
Explanation : Add month value in python
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 🙂
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)
def add_month(d,n=1): return type(d)(d.year+(d.month+n-1)/12, (d.month+n-1)%12+1, 1)
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)
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)
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)
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
Just Use This:
import datetime
today = datetime.datetime.today()
nextMonthDatetime = today + datetime.timedelta(days=(today.max.day - today.day)+1)
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
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))
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)
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?
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()
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)
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)
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.
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)
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]))
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))
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
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
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'))
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 require python-dateutil module. If you are on Linux you need to run this command in the terminal in order to install it.
sudo apt-get update && sudo apt-get install python-dateutil
Explanation : Add month value in python
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 🙂
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)
def add_month(d,n=1): return type(d)(d.year+(d.month+n-1)/12, (d.month+n-1)%12+1, 1)
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)
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)
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)
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
Just Use This:
import datetime
today = datetime.datetime.today()
nextMonthDatetime = today + datetime.timedelta(days=(today.max.day - today.day)+1)
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
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))
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)