Is there a simple way to increment a datetime object one month in Python?
Question:
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?
Answers:
>>> 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.
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:
- keeps the time information if given a datetime.datetime object
- 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)
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.
Check out from dateutil.relativedelta import *
for adding a specific amount of time to a date, you can continue to use timedelta for the simple stuff i.e.
import datetime
from dateutil.relativedelta import *
use_date = datetime.datetime.now()
use_date = use_date + datetime.timedelta(minutes=+10)
use_date = use_date + datetime.timedelta(hours=+1)
use_date = use_date + datetime.timedelta(days=+1)
use_date = use_date + datetime.timedelta(weeks=+1)
or you can start using relativedelta
use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(years=+1)
for the last day of next month:
use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(day=31)
Right now this will provide 29/02/2016
for the penultimate day of next month:
use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(day=31)
use_date = use_date+relativedelta(days=-1)
last Friday of the next month:
use_date = use_date+relativedelta(months=+1, day=31, weekday=FR(-1))
2nd Tuesday of next month:
new_date = use_date+relativedelta(months=+1, day=1, weekday=TU(2))
As @mrroot5 points out dateutil’s rrule functions can be applied, giving you an extra bang for your buck, if you require date occurences.
for example:
Calculating the last day of the month for 9 months from the last day of last month.
Then, calculate the 2nd Tuesday for each of those months.
from dateutil.relativedelta import *
from dateutil.rrule import *
from datetime import datetime
use_date = datetime(2020,11,21)
#Calculate the last day of last month
use_date = use_date+relativedelta(months=-1)
use_date = use_date+relativedelta(day=31)
#Generate a list of the last day for 9 months from the calculated date
x = list(rrule(freq=MONTHLY, count=9, dtstart=use_date, bymonthday=(-1,)))
print("Last day")
for ld in x:
print(ld)
#Generate a list of the 2nd Tuesday in each of the next 9 months from the calculated date
print("n2nd Tuesday")
x = list(rrule(freq=MONTHLY, count=9, dtstart=use_date, byweekday=TU(2)))
for tuesday in x:
print(tuesday)
Last day
2020-10-31 00:00:00
2020-11-30 00:00:00
2020-12-31 00:00:00
2021-01-31 00:00:00
2021-02-28 00:00:00
2021-03-31 00:00:00
2021-04-30 00:00:00
2021-05-31 00:00:00
2021-06-30 00:00:00
2nd Tuesday
2020-11-10 00:00:00
2020-12-08 00:00:00
2021-01-12 00:00:00
2021-02-09 00:00:00
2021-03-09 00:00:00
2021-04-13 00:00:00
2021-05-11 00:00:00
2021-06-08 00:00:00
2021-07-13 00:00:00
rrule could be used to find the next date occurring on a particular day.
e.g. the next 1st of January occurring on a Monday (Given today is the 4th November 2021)
from dateutil.relativedelta import *
from dateutil.rrule import *
from datetime import *
year = rrule(YEARLY,dtstart=datetime.now(),bymonth=1,bymonthday=1,byweekday=MO)[0].year
year
2024
or the next 5 x 1st of January’s occurring on a Monday
years = rrule(YEARLY,dtstart=datetime.now(),bymonth=1,bymonthday=1,byweekday=MO)[0:5]
for i in years:print(i.year)
...
2024
2029
2035
2046
2052
The first Month next Year that starts on a Monday:
>>> month = rrule(YEARLY,dtstart=datetime.date(2023, 1, 1),bymonthday=1,byweekday=MO)[0]
>>> month.strftime('%Y-%m-%d : %B')
'2023-05-01 : May'
If you need the months that start on a Monday between 2 dates:
months = rrule(YEARLY,dtstart=datetime.date(2025, 1, 1),until=datetime.date(2030, 1, 1),bymonthday=1,byweekday=MO)
>>> for m in months:
... print(m.strftime('%Y-%m-%d : %B'))
...
2025-09-01 : September
2025-12-01 : December
2026-06-01 : June
2027-02-01 : February
2027-03-01 : March
2027-11-01 : November
2028-05-01 : May
2029-01-01 : January
2029-10-01 : October
This is by no means an exhaustive list of what is available.
Documentation is available here: https://dateutil.readthedocs.org/en/latest/
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?
>>> 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.
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:
- keeps the time information if given a datetime.datetime object
- doesn’t use try/catch, instead using
calendar.monthrangefrom thecalendarmodule 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)
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.
Check out from dateutil.relativedelta import *
for adding a specific amount of time to a date, you can continue to use timedelta for the simple stuff i.e.
import datetime
from dateutil.relativedelta import *
use_date = datetime.datetime.now()
use_date = use_date + datetime.timedelta(minutes=+10)
use_date = use_date + datetime.timedelta(hours=+1)
use_date = use_date + datetime.timedelta(days=+1)
use_date = use_date + datetime.timedelta(weeks=+1)
or you can start using relativedelta
use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(years=+1)
for the last day of next month:
use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(day=31)
Right now this will provide 29/02/2016
for the penultimate day of next month:
use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(day=31)
use_date = use_date+relativedelta(days=-1)
last Friday of the next month:
use_date = use_date+relativedelta(months=+1, day=31, weekday=FR(-1))
2nd Tuesday of next month:
new_date = use_date+relativedelta(months=+1, day=1, weekday=TU(2))
As @mrroot5 points out dateutil’s rrule functions can be applied, giving you an extra bang for your buck, if you require date occurences.
for example:
Calculating the last day of the month for 9 months from the last day of last month.
Then, calculate the 2nd Tuesday for each of those months.
from dateutil.relativedelta import *
from dateutil.rrule import *
from datetime import datetime
use_date = datetime(2020,11,21)
#Calculate the last day of last month
use_date = use_date+relativedelta(months=-1)
use_date = use_date+relativedelta(day=31)
#Generate a list of the last day for 9 months from the calculated date
x = list(rrule(freq=MONTHLY, count=9, dtstart=use_date, bymonthday=(-1,)))
print("Last day")
for ld in x:
print(ld)
#Generate a list of the 2nd Tuesday in each of the next 9 months from the calculated date
print("n2nd Tuesday")
x = list(rrule(freq=MONTHLY, count=9, dtstart=use_date, byweekday=TU(2)))
for tuesday in x:
print(tuesday)
Last day
2020-10-31 00:00:00
2020-11-30 00:00:00
2020-12-31 00:00:00
2021-01-31 00:00:00
2021-02-28 00:00:00
2021-03-31 00:00:00
2021-04-30 00:00:00
2021-05-31 00:00:00
2021-06-30 00:00:00
2nd Tuesday
2020-11-10 00:00:00
2020-12-08 00:00:00
2021-01-12 00:00:00
2021-02-09 00:00:00
2021-03-09 00:00:00
2021-04-13 00:00:00
2021-05-11 00:00:00
2021-06-08 00:00:00
2021-07-13 00:00:00
rrule could be used to find the next date occurring on a particular day.
e.g. the next 1st of January occurring on a Monday (Given today is the 4th November 2021)
from dateutil.relativedelta import *
from dateutil.rrule import *
from datetime import *
year = rrule(YEARLY,dtstart=datetime.now(),bymonth=1,bymonthday=1,byweekday=MO)[0].year
year
2024
or the next 5 x 1st of January’s occurring on a Monday
years = rrule(YEARLY,dtstart=datetime.now(),bymonth=1,bymonthday=1,byweekday=MO)[0:5]
for i in years:print(i.year)
...
2024
2029
2035
2046
2052
The first Month next Year that starts on a Monday:
>>> month = rrule(YEARLY,dtstart=datetime.date(2023, 1, 1),bymonthday=1,byweekday=MO)[0]
>>> month.strftime('%Y-%m-%d : %B')
'2023-05-01 : May'
If you need the months that start on a Monday between 2 dates:
months = rrule(YEARLY,dtstart=datetime.date(2025, 1, 1),until=datetime.date(2030, 1, 1),bymonthday=1,byweekday=MO)
>>> for m in months:
... print(m.strftime('%Y-%m-%d : %B'))
...
2025-09-01 : September
2025-12-01 : December
2026-06-01 : June
2027-02-01 : February
2027-03-01 : March
2027-11-01 : November
2028-05-01 : May
2029-01-01 : January
2029-10-01 : October
This is by no means an exhaustive list of what is available.
Documentation is available here: https://dateutil.readthedocs.org/en/latest/