How do I calculate the date six months from the current date using the datetime Python module?

Question

I am using the datetime Python module. I am looking to calculate the date 6 months from the current date. Could someone give me a little help doing this?

The reason I want to generate a date 6 months from the current date is to produce a review date. If the user enters data into the system it will have a review date of 6 months from the date they entered the data.

You will have to be more specific : when is six months from march 31th? And from august 30th? — kmkaplan, Feb 13 '09 at 15:21
Yes the edit helps: it means you can aproximate 6 months to 183 days with no ill effect. So adding 183 days to today will do the trick. — kmkaplan, Feb 13 '09 at 16:42
The above comments strike me as silly. The concept of "adding six months" is quite clear -- take the month component and add `6` to it, with support for rolling over the year (and cycling the month back to `1`) if we go past December. This happens to be exactly what `relativedelta` does and is in fact what every programming language with support for this sort of concept does. — Kirk Woll, Jul 18 '12 at 18:30
@KirkWoll: I am sure it is quite clear. But still different for whoever speaks. Python: `date(2015, 3, 31) + relativedelta(months = 6)` gives `datetime.date(2015, 9, 30)`. Perl: `DateTime->new(year=>2000, month=>3, day=>31)->add(months=>6)` gives `2000-10-01T00:00:00`. Php: `date_create('2000-03-31', new DateTimeZone('UTC'))->add(new DateInterval('P6M'))` gives 2000-10-01. Pick your poison. — kmkaplan, Feb 10 '15 at 10:12
@kmkaplan The problem with adding 183 days or similar approaches is that for certain dates they produce an answer which doesn't match ANYONE's definition of what 'adding six months' should be. — jwg, Mar 04 '15 at 09:43
related: [How to convert years to seconds](http://stackoverflow.com/a/32658742/4279) — jfs, Jan 29 '16 at 08:30
... adding 182 seems more pragmatic for generating a *review date*: it keeps the day of the week intact. — Wolf, Mar 27 '19 at 11:35
@Wolf Maybe… It turns out `6*365/12` does give `182` because of integer division ;-) — kmkaplan, Apr 26 '20 at 09:16

score 1340 · Answer 1 · edited Mar 25 '16 at 19:17

1340

I found this solution to be good. (This uses the python-dateutil extension)

from datetime import date
from dateutil.relativedelta import relativedelta

six_months = date.today() + relativedelta(months=+6)

The advantage of this approach is that it takes care of issues with 28, 30, 31 days etc. This becomes very useful in handling business rules and scenarios (say invoice generation etc.)

$ date(2010,12,31)+relativedelta(months=+1)
  datetime.date(2011, 1, 31)

$ date(2010,12,31)+relativedelta(months=+2)
  datetime.date(2011, 2, 28)

edited Mar 25 '16 at 19:17

Paul

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answered Dec 10 '10 at 06:22

Mahendra

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the +6 is an indicator that it could be -6, same thing applies to days and years as well :) – securecurve Feb 20 '13 at 11:04
that's weird, I'm using python 2.7.1 and from dateutil.relativedelta import relativedelta won't work – sliders_alpha Mar 22 '13 at 09:05
6

@sliders_alpha You need to install the python-dateutil package (pip install python-dateutil) – poiuytrez Nov 19 '13 at 15:29
23

This is the correct solution. If the OP would have asked 6 years instead of 6 months, then the approved answer would fail miserably. So one should keep in mind that the answers are more valuable the more generalized they are. – Daniel F May 24 '14 at 20:34
12

Kindly note that the `relativedelta` function also takes in `month` as an argument, which basically replaces / sets / fixes month in the passed date, which is very different than adding more months. Just a forewarning for folks wondering if the function is broken because they forgot the extra *s* in *months*. – shad0w_wa1k3r Jul 30 '19 at 09:19
@shad0w_wa1k3r this comment should be part of the answer- I just spent a frustrating 15 minutes wondering why it wasn't working – user2682863 Aug 17 '19 at 21:34
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I'm curious: does anyone know why this isn't included by default? Timedelta seems packaged with `datetime` by default. I actually assumed I could pass "months" into timedelta. – dTanMan Nov 12 '19 at 03:14
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@dTanMan Since it is not clear what should be the result of adding one month to date `(2020, 1, 30)`. Any attempt to solve it is not perfect and will raise a debate. All these corner cases and additional features are left to be covered by the external `dateutil` package. – Jeyekomon Jan 21 '20 at 10:12
2

wow can't believe i have to scroll down this much – JChao Apr 13 '21 at 08:53
@dTanMan one challenge with timedelta is that the biggest unit is days, so adding months and years is not straightforward but the .replace() function could work: dateTimeObj = dateTimeObj.replace(year = dateTimeObj.year - 1), though this becomes challenging when adding days months goes past the edge of a month / year – Peej1226 Dec 21 '22 at 13:38

score 73 · Answer 2 · edited Jun 24 '22 at 07:43

73

Well, that depends what you mean by 6 months from the current date.

Using natural months:

inc = 6
year = year + (month + inc - 1) // 12
month = (month + inc - 1) % 12 + 1

Using a banker's definition, 6*30:
```
date += datetime.timedelta(6 * 30)
```

edited Jun 24 '22 at 07:43

Nico Schlömer

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answered Feb 13 '09 at 15:29

vartec

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Could you throw in the half-year definition (183 days) plus the 26 weeks definition, too? It helps to have them all in one place. – S.Lott Feb 13 '09 at 18:02
12

just a quick remark: I think, for month, the formula would be instead (month + 5) % 12 + 1 b/c for june, your formula gives 0 whereas the expected result is 12... despite this little error, to my mind, your answer is the one that best answers the question – pierroz Mar 15 '10 at 16:02
4

and same for year: it should be year + (month + 5)/12 – pierroz Mar 15 '10 at 16:15
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What if the date is 31, and the month six month later can't have 31 days (which is the case for most months with 31 days)? – akv Apr 06 '11 at 11:37
3

30 days isn't 'the bankers definition'. There are LOTS of ways of counting days for interest payments etc., and the several methods based on 30/360 :) are just a few of them. http://en.wikipedia.org/wiki/Day_count_convention#30.2F360_methods – jwg Mar 04 '15 at 09:46
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Downvote because the first solution ``(day, month, year) = (day, (month+6)%12, year+(month+6)/12)`` can buggy, as it generates invalid dates such as ``(31, 8, 2015)`` -> ``(31, 2, 2016)`` – ohw Jan 16 '16 at 23:40
does not work with python 3. There you get a float year (e.g. 2020.6666666666667) – stackunderflow Jul 21 '20 at 19:24
@dev-null, just wrap year in `int()` – ruslan_krivoshein Aug 11 '20 at 14:26
3

@ruslan_krivoshein, ok that works. But I think it would be even better to use the quotient operator // in the last expression: `year + (month + 5)//12`. Then there's no type conversion needed and the purpose is clearer. – stackunderflow Aug 12 '20 at 19:40
For method 1, the month should be the "old" month variable. Otherwise it's broken. Edit: never mind, it WAS correct but someone changed and broke it. – fireattack Jun 24 '22 at 06:53
@nico-schlömer would you mind to modify it one more time to fix the issue mentioned above? Thanks (I can't edit this answer due to "queue being full". – fireattack Jun 24 '22 at 06:56
@nico-schlömer Sorry I wasn't clear. They were not swapped. The issue is that with the old way, which is `(month, year) = ((month + 5) % 12 + 1, year + (month + 5)/12)` both variables are changed together, so the new variable of `month` won't be used when evaluating `year`. Your change, which split it into two lines, would have that problem and gives incorrect result. See: https://stackoverflow.com/questions/11502268/how-does-pythons-comma-operator-work-during-assignment – fireattack Jun 25 '22 at 09:34
I finally be able to modified it myself, so it should be solved now. Thanks for your time – fireattack Jun 25 '22 at 09:44

score 65 · Answer 3 · answered Sep 24 '18 at 11:25

65

With Python 3.x you can do it like this:

from datetime import datetime, timedelta
from dateutil.relativedelta import *

date = datetime.now()
print(date)
# 2018-09-24 13:24:04.007620

date = date + relativedelta(months=+6)
print(date)
# 2019-03-24 13:24:04.007620

but you will need to install python-dateutil module:

pip install python-dateutil

answered Sep 24 '18 at 11:25

Luka Lopusina

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I'd recommend `from dateutil.relativedelta import relativedelta`. Using `import *` isn't explicit. – Sam Morgan Apr 29 '20 at 13:55
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`pd.DateOffset` exactly the same thing in newer `pandas`, `date = date + pd.DateOffset(months=6)` without need to install `dateutil`, see help – gregV May 04 '21 at 19:40

score 24 · Answer 4 · answered Jul 07 '10 at 17:59

So, here is an example of the dateutil.relativedelta which I found useful for iterating through the past year, skipping a month each time to the present date:

>>> import datetime
>>> from dateutil.relativedelta import relativedelta
>>> today = datetime.datetime.today()
>>> month_count = 0
>>> while month_count < 12:
...  day = today - relativedelta(months=month_count)
...  print day
...  month_count += 1
... 
2010-07-07 10:51:45.187968
2010-06-07 10:51:45.187968
2010-05-07 10:51:45.187968
2010-04-07 10:51:45.187968
2010-03-07 10:51:45.187968
2010-02-07 10:51:45.187968
2010-01-07 10:51:45.187968
2009-12-07 10:51:45.187968
2009-11-07 10:51:45.187968
2009-10-07 10:51:45.187968
2009-09-07 10:51:45.187968
2009-08-07 10:51:45.187968

As with the other answers, you have to figure out what you actually mean by "6 months from now." If you mean "today's day of the month in the month six years in the future" then this would do:

datetime.datetime.now() + relativedelta(months=6)

score 20 · Answer 5 · edited Feb 02 '17 at 10:53

20

For beginning of month to month calculation:

from datetime import timedelta
from dateutil.relativedelta import relativedelta

end_date = start_date + relativedelta(months=delta_period) + timedelta(days=-delta_period)

edited Feb 02 '17 at 10:53

JRodDynamite

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answered Feb 02 '17 at 10:49

breddi

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This isn't the relativedelta solution mentioned above in the comments. Keep scrolling for the solution with 600+ upvotes. – Nostalg.io Feb 12 '18 at 20:16

Arpan Saini · Answer 6 · 2021-08-12T17:05:06.977

Python can use datautil package for that, Please see the example below

It's not Just limited to that, you can pass combination of days, Months and Years at the same time also.

import datetime
from dateutil.relativedelta import relativedelta

# subtract months
proc_dt = datetime.date(2021,8,31)
proc_dt_minus_3_months = proc_dt + relativedelta(months=-3)
print(proc_dt_minus_3_months)

# add months
proc_dt = datetime.date(2021,8,31)
proc_dt_plus_3_months = proc_dt + relativedelta(months=+3)
print(proc_dt_plus_3_months)

# subtract days:
proc_dt = datetime.date(2021,8,31)
proc_dt_minus_3_days = proc_dt + relativedelta(days=-3)
print(proc_dt_minus_3_days)

# add days days:
proc_dt = datetime.date(2021,8,31)
proc_dt_plus_3_days = proc_dt + relativedelta(days=+3)
print(proc_dt_plus_3_days)

# subtract years:
proc_dt = datetime.date(2021,8,31)
proc_dt_minus_3_years = proc_dt + relativedelta(years=-3)
print(proc_dt_minus_3_years)

# add years:
proc_dt = datetime.date(2021,8,31)
proc_dt_plus_3_years = proc_dt + relativedelta(years=+3)
print(proc_dt_plus_3_years)

Results:

2021-05-31

2021-11-30

2021-08-28

2021-09-03

2018-08-31

2024-08-31

"Python has inbuild libraries for that" -> no, dateutil is not part of the python standard library. — Doug, Aug 09 '21 at 07:27
@Doug thank for correcting me, I didn't notice that. i got it installed long time back. — Arpan Saini, Aug 09 '21 at 23:17

score 19 · Answer 7 · edited Jul 15 '22 at 20:33

This solution works correctly for December, which most of the answers on this page do not. You need to first shift the months from a 1-based index (ie Jan = 1) to a 0-based index (ie Jan = 0) before using modulus ( % ) or integer division ( // ), otherwise November (11) plus 1 month gives you 12, which when finding the remainder ( 12 % 12 ) gives 0.

(And dont suggest "(month % 12) + 1" or Oct + 1 = december!)

def AddMonths(d,x):
    newmonth = ((( d.month - 1) + x ) % 12 ) + 1
    newyear  = int(d.year + ((( d.month - 1) + x ) / 12 ))
    return datetime.date( newyear, newmonth, d.day)

However ... This doesnt account for problem like Jan 31 + one month. So we go back to the OP - what do you mean by adding a month? One solution is to backtrack until you get to a valid day, given that most people would presume the last day of jan, plus one month, equals the last day of Feb. This will work on negative numbers of months too. Proof:

>>> import datetime
>>> AddMonths(datetime.datetime(2010,8,25),1)
datetime.date(2010, 9, 25)
>>> AddMonths(datetime.datetime(2010,8,25),4)
datetime.date(2010, 12, 25)
>>> AddMonths(datetime.datetime(2010,8,25),5)
datetime.date(2011, 1, 25)
>>> AddMonths(datetime.datetime(2010,8,25),13)
datetime.date(2011, 9, 25)
>>> AddMonths(datetime.datetime(2010,8,25),24)
datetime.date(2012, 8, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-1)
datetime.date(2010, 7, 25)
>>> AddMonths(datetime.datetime(2010,8,25),0)
datetime.date(2010, 8, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-12)
datetime.date(2009, 8, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-8)
datetime.date(2009, 12, 25)
>>> AddMonths(datetime.datetime(2010,8,25),-7)
datetime.date(2010, 1, 25)>>>

Fixes the "Jan 31 + one month" problem: days_in_month = calendar.monthrange(newyear, newmonth)[1]; newday = min(d.day, days_in_month); (bumps the day down if too big eg Feb 31 to Feb 28/29) — Curtis Yallop, Nov 30 '15 at 18:33
This worked for me only by converting the newyear to an integer data type def AddMonths(d,x): newmonth = ((( d.month - 1) + x ) % 12 ) + 1 newyear = d.year + int((( d.month - 1) + x ) / 12 ) return datetime.date( newyear, newmonth, d.day) — Manmeet Singh, Nov 29 '19 at 10:28
Why fiddle around with all these magic numbers when the correct solution has already been implemented in the `dateutil` package? — Roland Illig, Aug 26 '22 at 20:31
@RolandIllig works great for my situation because the tool I'm using does not allow package installation — Jak, Oct 19 '22 at 06:09
how about if the day is 2020-3-1, and i need to go back a full month to get 2020-1-30 or 2020-1-31 — dataviews, Apr 13 '23 at 01:23

score 18 · Answer 8 · edited Apr 22 '20 at 09:29

18

What do you mean by "6 months"?

Is 2009-02-13 + 6 months == 2009-08-13? Or is it 2009-02-13 + 6*30 days?

import mx.DateTime as dt

#6 Months
dt.now()+dt.RelativeDateTime(months=6)
#result is '2009-08-13 16:28:00.84'

#6*30 days
dt.now()+dt.RelativeDateTime(days=30*6)
#result is '2009-08-12 16:30:03.35'

More info about mx.DateTime

edited Apr 22 '20 at 09:29

Lii

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answered Feb 13 '09 at 15:21

Johannes Weiss

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rpanai · Answer 9 · 2019-11-07T11:45:49.100

This doesn't answer the specific question (using datetime only) but, given that others suggested the use of different modules, here there is a solution using pandas.

import datetime as dt
import pandas as pd

date = dt.date.today() - \
       pd.offsets.DateOffset(months=6)

print(date)

2019-05-04 00:00:00

Which works as expected in leap years

date = dt.datetime(2019,8,29) - \
       pd.offsets.DateOffset(months=6)
print(date)

2019-02-28 00:00:00

score 16 · Answer 10 · answered Feb 13 '09 at 15:29

16

There's no direct way to do it with Python's datetime.

Check out the relativedelta type at python-dateutil. It allows you to specify a time delta in months.

answered Feb 13 '09 at 15:29

Akbar ibrahim

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score 14 · Answer 11 · edited Mar 12 '19 at 21:35

14

Just use the timetuple method to extract the months, add your months and build a new dateobject. If there is a already existing method for this I do not know it.

import datetime

def in_the_future(months=1):
    year, month, day = datetime.date.today().timetuple()[:3]
    new_month = month + months
    return datetime.date(year + (new_month / 12), (new_month % 12) or 12, day)

The API is a bit clumsy, but works as an example. Will also obviously not work on corner-cases like 2008-01-31 + 1 month. :)

edited Mar 12 '19 at 21:35

kolypto

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answered Feb 13 '09 at 15:23

pi.

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Error in your code: `new_month % 12` should be `(new_month % 12) or 12`. Otherwise if you try this in November you will get an error. :) – Jordan Reiter May 02 '11 at 21:45
This is actually much cleaner than the accepted solution. This doesn't require any new import, just basic maths – Simon PA Apr 02 '19 at 16:26
return datetime.date(year + (new_month/ / 12), (new_month % 12) or 12, day) – paytam Apr 19 '20 at 05:57
This didn't handle new_month % 12 == 0, it just skip a whole year – Asoul Jun 24 '22 at 08:53
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Doing `new_month / 12` and `new_month % 12` is wrong since month uses a 1-based index. You need to convert to 0-based indexing, perform division/modulus, and then convert back to 1-based indexing. – jamesdlin Jul 15 '22 at 20:51

Mike Davies · Answer 12 · 2012-01-12T22:39:29.587

I know this was for 6 months, however the answer shows in google for "adding months in python" if you are adding one month:

import calendar

date = datetime.date.today()    //Or your date

datetime.timedelta(days=calendar.monthrange(date.year,date.month)[1])

this would count the days in the current month and add them to the current date, using 365/12 would ad 1/12 of a year can causes issues for short / long months if your iterating over the date.

David Ragazzi · Answer 13 · 2018-08-17T14:25:06.360

Using Python standard libraries, i.e. without dateutil or others, and solving the 'February 31st' problem:

import datetime
import calendar

def add_months(date, months):
    months_count = date.month + months

    # Calculate the year
    year = date.year + int(months_count / 12)

    # Calculate the month
    month = (months_count % 12)
    if month == 0:
        month = 12

    # Calculate the day
    day = date.day
    last_day_of_month = calendar.monthrange(year, month)[1]
    if day > last_day_of_month:
        day = last_day_of_month

    new_date = datetime.date(year, month, day)
    return new_date

Testing:

>>>date = datetime.date(2018, 11, 30)

>>>print(date, add_months(date, 3))
(datetime.date(2018, 11, 30), datetime.date(2019, 2, 28))

>>>print(date, add_months(date, 14))
(datetime.date(2018, 12, 31), datetime.date(2020, 2, 29))

This has a bug: If the target month is December, it will return a date one year later than needed. Example: `add_months(datetime.date(2018, 11, 30), 1)` returns `datetime.date(2019, 12, 30)` (the year should be 2018, not 2019). So, better use a dedicated, well tested library for this! If you really need to only use standard library modules, see [my answer to a similar question](https://stackoverflow.com/a/35067015/40076). — taleinat, May 06 '20 at 15:40

score 10 · Answer 14 · answered Feb 13 '09 at 15:26

10

Dateutil package has implementation of such functionality. But be aware, that this will be naive, as others pointed already.

answered Feb 13 '09 at 15:26

zgoda

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dateutil is awesome. It can be installed with easy_install too. – Soviut Feb 13 '09 at 15:42
Excellent. Thanks for suggesting that. That seems to be a god-sent. – ayaz Apr 21 '10 at 07:38
What do you mean by _naive_? – Roland Illig Aug 26 '22 at 20:36

Diego Magalhães · Answer 15 · 2019-08-01T19:52:56.377

6

A quick suggestion is Arrow

pip install arrow

>>> import arrow

>>> arrow.now().date()
datetime.date(2019, 6, 28)
>>> arrow.now().shift(months=6).date()
datetime.date(2019, 12, 28)

edited Aug 01 '19 at 19:52

answered Jun 28 '19 at 12:57

Diego Magalhães

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Use .datetime to avoid timezone (aware / naive) warning: arrow.now().shift(months=6).datetime – Flo Mar 10 '21 at 19:40

score 5 · Answer 16 · edited May 25 '11 at 12:44

5

I have a better way to solve the 'February 31st' problem:

def add_months(start_date, months):
    import calendar

    year = start_date.year + (months / 12)
    month = start_date.month + (months % 12)
    day = start_date.day

    if month > 12:
        month = month % 12
        year = year + 1

    days_next = calendar.monthrange(year, month)[1]
    if day > days_next:
        day = days_next

    return start_date.replace(year, month, day)

I think that it also works with negative numbers (to subtract months), but I haven't tested this very much.

edited May 25 '11 at 12:44

bstpierre

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answered May 25 '11 at 11:00

amoyafoss

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This answer has the merit of correctly solving the problem using only built-in Python libraries. Check stackoverflow.com/a/4131114/302264 for an equivalent but slightly more concise answer. – Eduardo Dobay Apr 04 '18 at 12:23
What type is `start_date` supposed to have? – Roland Illig Aug 26 '22 at 20:40
Only works in December, due to the floating-point division `months / 12`. – Roland Illig Aug 26 '22 at 20:48

score 4 · Answer 17 · answered Nov 17 '09 at 16:37

4

The QDate class of PyQt4 has an addmonths function.

>>>from PyQt4.QtCore import QDate  
>>>dt = QDate(2009,12,31)  
>>>required = dt.addMonths(6) 

>>>required
PyQt4.QtCore.QDate(2010, 6, 30)

>>>required.toPyDate()
datetime.date(2010, 6, 30)

answered Nov 17 '09 at 16:37

user213043

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score 3 · Answer 18 · edited Feb 08 '12 at 16:02

3

Modified Johannes Wei's answer in the case 1new_month = 121. This works perfectly for me. The months could be positive or negative.

def addMonth(d,months=1):
    year, month, day = d.timetuple()[:3]
    new_month = month + months
    return datetime.date(year + ((new_month-1) / 12), (new_month-1) % 12 +1, day)

edited Feb 08 '12 at 16:02

stealthyninja

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answered Jan 16 '10 at 02:02

Xinhui Tang

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It DOESN'T "work perfectly" when the day in the start date is greater than the number of days in the target month. Example: 2001-01-31 plus one month tries to create a date 2001-02-31. – John Machin Jan 16 '10 at 09:25

score 3 · Answer 19 · answered Jun 06 '11 at 12:25

3

Modified the AddMonths() for use in Zope and handling invalid day numbers:

def AddMonths(d,x):
    days_of_month = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    newmonth = ((( d.month() - 1) + x ) % 12 ) + 1
    newyear  = d.year() + ((( d.month() - 1) + x ) // 12 ) 
    if d.day() > days_of_month[newmonth-1]:
      newday = days_of_month[newmonth-1]
    else:
      newday = d.day() 
    return DateTime( newyear, newmonth, newday)

answered Jun 06 '11 at 12:25

Jost

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This does not seem to consider leap years (29 days in feb). – handle Apr 04 '18 at 12:19
Instead get the `days_of_month` with calendar.monthrange(year, month)[-1] – SurpriseDog Oct 31 '22 at 18:16

score 3 · Answer 20 · edited Dec 15 '11 at 20:21

3

import time

def add_month(start_time, months):  

        ret = time.strptime(start_time, '%Y-%m-%d')
        t = list(ret)

        t[1] += months

        if t[1] > 12:
            t[0] += 1 + int(months / 12)

            t[1] %= 12

        return int(time.mktime(tuple(t)))

edited Dec 15 '11 at 20:21

answered Aug 11 '11 at 02:28

ares

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Robert Lujo · Answer 21 · 2014-01-20T23:58:04.530

Yet another solution - hope someone will like it:

def add_months(d, months):
    return d.replace(year=d.year+months//12).replace(month=(d.month+months)%12)

This solution doesn't work for days 29,30,31 for all cases, so more robust solution is needed (which is not so nice anymore :) ):

def add_months(d, months):
    for i in range(4):
        day = d.day - i
        try:
            return d.replace(day=day).replace(year=d.year+int(months)//12).replace(month=(d.month+int(months))%12)
        except:
            pass
    raise Exception("should not happen")

score 2 · Answer 22 · edited May 23 '17 at 10:31

From this answer, see parsedatetime. Code example follows. More details: unit test with many natural-language -> YYYY-MM-DD conversion examples, and apparent parsedatetime conversion challenges/bugs.

#!/usr/bin/env python
# -*- coding: utf-8 -*-
import time, calendar
from datetime import date

# from https://github.com/bear/parsedatetime
import parsedatetime as pdt

def print_todays_date():
    todays_day_of_week = calendar.day_name[date.today().weekday()]
    print "today's date = " + todays_day_of_week + ', ' + \
                              time.strftime('%Y-%m-%d')

def convert_date(natural_language_date):
    cal = pdt.Calendar()
    (struct_time_date, success) = cal.parse(natural_language_date)
    if success:
        formal_date = time.strftime('%Y-%m-%d', struct_time_date)
    else:
        formal_date = '(conversion failed)'
    print '{0:12s} -> {1:10s}'.format(natural_language_date, formal_date)

print_todays_date()
convert_date('6 months')

The above code generates the following from a MacOSX machine:

$ ./parsedatetime_simple.py 
today's date = Wednesday, 2015-05-13
6 months     -> 2015-11-13
$

score 2 · Answer 23 · edited May 23 '17 at 12:10

2

How about this? Not using another library (dateutil) or timedelta? building on vartec's answer I did this and I believe it works:

import datetime

today = datetime.date.today()
six_months_from_today = datetime.date(today.year + (today.month + 6)/12, (today.month + 6) % 12, today.day)

I tried using timedelta, but because it is counting the days, 365/2 or 6*356/12 does not always translate to 6 months, but rather 182 days. e.g.

day = datetime.date(2015, 3, 10)
print day
>>> 2015-03-10

print (day + datetime.timedelta(6*365/12))
>>> 2015-09-08

I believe that we usually assume that 6 month's from a certain day will land on the same day of the month but 6 months later (i.e. 2015-03-10 --> 2015-09-10, Not 2015-09-08)

I hope you find this helpful.

edited May 23 '17 at 12:10

Community

1
1

answered Sep 10 '15 at 04:44

Babak K

469
5
13

3

`day + datetime.timedelta(6*365/12)` won't work always, as some years have 365 days and others have 366 days. – 3kstc Jan 08 '18 at 07:12
`day + datetime.timedelta(6*365/12)` is the wrong way – xinnjie Oct 27 '22 at 08:33

score 2 · Answer 24 · answered Feb 18 '16 at 06:32

Here's a example which allows the user to decide how to return a date where the day is greater than the number of days in the month.

def add_months(date, months, endOfMonthBehaviour='RoundUp'):
    assert endOfMonthBehaviour in ['RoundDown', 'RoundIn', 'RoundOut', 'RoundUp'], \
        'Unknown end of month behaviour'
    year = date.year + (date.month + months - 1) / 12
    month = (date.month + months - 1) % 12 + 1
    day = date.day
    last = monthrange(year, month)[1]
    if day > last:
        if endOfMonthBehaviour == 'RoundDown' or \
            endOfMonthBehaviour == 'RoundOut' and months < 0 or \
            endOfMonthBehaviour == 'RoundIn' and months > 0:
            day = last
        elif endOfMonthBehaviour == 'RoundUp' or \
            endOfMonthBehaviour == 'RoundOut' and months > 0 or \
            endOfMonthBehaviour == 'RoundIn' and months < 0:
            # we don't need to worry about incrementing the year
            # because there will never be a day in December > 31
            month += 1
            day = 1
    return datetime.date(year, month, day)


>>> from calendar import monthrange
>>> import datetime
>>> add_months(datetime.datetime(2016, 1, 31), 1)
datetime.date(2016, 3, 1)
>>> add_months(datetime.datetime(2016, 1, 31), -2)
datetime.date(2015, 12, 1)
>>> add_months(datetime.datetime(2016, 1, 31), -2, 'RoundDown')
datetime.date(2015, 11, 30)

blacknight12321 · Answer 25 · 2017-03-30T06:56:38.263

2

given that your datetime variable is called date:

date=datetime.datetime(year=date.year+int((date.month+6)/12),
                       month=(date.month+6)%13 + (1 if (date.month + 
                       months>12) else 0), day=date.day)

edited Mar 30 '17 at 06:56

answered Mar 30 '17 at 05:57

blacknight12321

91
1
5

Nandlal Yadav · Answer 26 · 2017-12-07T01:13:57.260

2

General function to get next date after/before x months.

from datetime import date

def after_month(given_date, month):
    yyyy = int(((given_date.year * 12 + given_date.month) + month)/12)
    mm = int(((given_date.year * 12 + given_date.month) + month)%12)

    if mm == 0:
        yyyy -= 1
        mm = 12
    return given_date.replace(year=yyyy, month=mm)


if __name__ == "__main__":
    today = date.today()
    print(today)

    for mm in [-12, -1, 0, 1, 2, 12, 20 ]:
        next_date = after_month(today, mm)
        print(next_date)

edited Dec 07 '17 at 01:13

answered Dec 06 '17 at 23:59

Nandlal Yadav

21
2

This is clearly the best answer. Simple and effective, without rounding issues. – jprobichaud Apr 03 '18 at 15:08

tony diep · Answer 27 · 2011-03-10T03:06:59.383

import datetime


'''
Created on 2011-03-09

@author: tonydiep
'''

def add_business_months(start_date, months_to_add):
    """
    Add months in the way business people think of months. 
    Jan 31, 2011 + 1 month = Feb 28, 2011 to business people
    Method: Add the number of months, roll back the date until it becomes a valid date
    """
    # determine year
    years_change = months_to_add / 12

    # determine if there is carryover from adding months
    if (start_date.month + (months_to_add % 12) > 12 ):
        years_change = years_change + 1

    new_year = start_date.year + years_change

    # determine month
    work = months_to_add % 12
    if 0 == work:
        new_month = start_date.month
    else:
        new_month = (start_date.month + (work % 12)) % 12

    if 0 == new_month:
        new_month = 12 

    # determine day of the month
    new_day = start_date.day
    if(new_day in [31, 30, 29, 28]):
        #user means end of the month
        new_day = 31


    new_date = None
    while (None == new_date and 27 < new_day):
        try:
            new_date = start_date.replace(year=new_year, month=new_month, day=new_day)
        except:
            new_day = new_day - 1   #wind down until we get to a valid date

    return new_date


if __name__ == '__main__':
    #tests
    dates = [datetime.date(2011, 1, 31),
             datetime.date(2011, 2, 28),
             datetime.date(2011, 3, 28),
             datetime.date(2011, 4, 28),
             datetime.date(2011, 5, 28),
             datetime.date(2011, 6, 28),
             datetime.date(2011, 7, 28),
             datetime.date(2011, 8, 28),
             datetime.date(2011, 9, 28),
             datetime.date(2011, 10, 28),
             datetime.date(2011, 11, 28),
             datetime.date(2011, 12, 28),
             ]
    months = range(1, 24)
    for start_date in dates:
        for m in months:
            end_date = add_business_months(start_date, m)
            print("%s\t%s\t%s" %(start_date, end_date, m))

score 2 · Answer 28 · answered Oct 19 '18 at 15:17

2

Im chiming in late, but

check out Ken Reitz Maya module,

https://github.com/kennethreitz/maya

something like this may help you, just change hours=1 to days=1 or years=1

>>> from maya import MayaInterval

# Create an event that is one hour long, starting now.
>>> event_start = maya.now()
>>> event_end = event_start.add(hours=1)

>>> event = MayaInterval(start=event_start, end=event_end)

answered Oct 19 '18 at 15:17

perfecto25

772
9
13

2

can also use Pendulum module (Maya is based on Pendulum) https://github.com/sdispater/pendulum – perfecto25 Oct 19 '18 at 15:21

Breno Monteiro · Answer 29 · 2020-06-08T09:46:34.453

The "python-dateutil" (external extension) is a good solution, but you can do it with build-in Python modules (datetime and datetime)

I made a short and simple code, to solve it (dealing with year, month and day)

(running: Python 3.8.2)

from datetime import datetime
from calendar import monthrange

# Time to increase (in months)
inc = 12

# Returns mod of the division for 12 (months)
month = ((datetime.now().month + inc) % 12) or 1

# Increase the division by 12 (months), if necessary (+ 12 months increase)
year = datetime.now().year + int((month + inc) / 12)

# (IF YOU DON'T NEED DAYS,CAN REMOVE THE BELOW CODE)
# Returns the same day in new month, or the maximum day of new month
day = min(datetime.now().day,monthrange(year, month)[1])

print("Year: {}, Month: {}, Day: {}".format(year, month, day))

by using "math.floor" instead of "int", decrease functionality would also be supported — Hamed, Nov 13 '20 at 08:15

score 2 · Answer 30 · answered Nov 15 '20 at 16:02

I often need last day of month to remain last day of month. To solve that I add one day before calculation and then subtract it again before return.

from datetime import date, timedelta

# it's a lot faster with a constant day
DAY = timedelta(1)

def add_month(a_date, months):
    "Add months to date and retain last day in month."
    next_day = a_date + DAY
    # calculate new year and month
    m_sum = next_day.month + months - 1
    y = next_day.year + m_sum // 12
    m = m_sum % 12 + 1
    try:
        return date(y, m, next_day.day) - DAY
    except ValueError:
        # on fail return last day in month
        # can't fail on december so I don't bother changing the year
        return date(y, m + 1, 1) - DAY

score 2 · Answer 31 · answered Dec 31 '11 at 18:10

Rework of an earlier answer by user417751. Maybe not so pythonic way, but it takes care of different month lengths and leap years. In this case 31 January 2012 + 1 month = 29 February 2012.

import datetime
import calendar

def add_mths(d, x):
    newday = d.day
    newmonth = (((d.month - 1) + x) % 12) + 1
    newyear  = d.year + (((d.month - 1) + x) // 12)
    if newday > calendar.mdays[newmonth]:
        newday = calendar.mdays[newmonth]
        if newyear % 4 == 0 and newmonth == 2:
            newday += 1
    return datetime.date(newyear, newmonth, newday)

marshallpenguin · Answer 32 · 2010-03-13T19:10:28.400

0

This is what I came up with. It moves the correct number of months and years but ignores days (which was what I needed in my situation).

import datetime

month_dt = 4
today = datetime.date.today()
y,m = today.year, today.month
m += month_dt-1
year_dt = m//12
new_month = m%12
new_date = datetime.date(y+year_dt, new_month+1, 1)

edited Mar 13 '10 at 19:10

answered Mar 02 '10 at 21:06

marshallpenguin

167
2
8

score 0 · Answer 33 · answered Apr 01 '10 at 13:55

I use this function to change year and month but keep day:

def replace_month_year(date1, year2, month2):
    try:
        date2 = date1.replace(month = month2, year = year2)
    except:
        date2 = datetime.date(year2, month2 + 1, 1) - datetime.timedelta(days=1)
    return date2

You should write:

new_year = my_date.year + (my_date.month + 6) / 12
new_month = (my_date.month + 6) % 12
new_date = replace_month_year(my_date, new_year, new_month)

score 0 · Answer 34 · answered Jan 01 '17 at 15:38

Another solution: calculate sum of days in month for next n month and add result to current date.

import calendar
import datetime

def date_from_now(months):
    today = datetime.datetime.today()

    month = today.month
    year = today.year
    sum_days = 0

    for i in range(int(months)):
        month += 1

        if month == 13:
            month = 1
            year += 1

        sum_days += calendar.monthrange(year, month)[1]

    return datetime.date.today() + datetime.timedelta(sum_days)

print(date_from_now(12)) # if to day is 2017-01-01, output: 2019-01-01

score 0 · Answer 35 · answered Feb 02 '11 at 10:45

I think it would be safer to do something like this instead of manually adding days:

import datetime
today = datetime.date.today()

def addMonths(dt, months = 0):
    new_month = months + dt.month
    year_inc = 0
    if new_month>12:
        year_inc +=1
        new_month -=12
    return dt.replace(month = new_month, year = dt.year+year_inc)

newdate = addMonths(today, 6)

score 0 · Answer 36 · answered Feb 02 '11 at 14:44

I solved this problem like this:

import calendar
from datetime import datetime
moths2add = 6
now = datetime.now()
current_year = now.year
current_month = now.month
#count days in months you want to add using calendar module
days = sum(
  [calendar.monthrange(current_year, elem)[1] for elem in range(current_month, current_month + moths)]
    )
print now + days

score 0 · Answer 37 · answered Jul 24 '18 at 14:02

I used the replace() method and write this recursive function. the dt is a datetime.datetime object:

def month_timedelta(dt, m):
    y = m // 12
    dm = m % 12
    if y == 0:
        if dt.month + m <= 12:
            return dt.replace(month = dt.month + m)
        else:
            dy = (dt.month + m) // 12
            ndt = dt.replace(year=dt.year + dy)
            return ndt.replace(month=(ndt.month + m) % 12)
    else:
        return month_timedelta(dt.replace(year=dt.year + y),dm)

score 0 · Answer 38 · answered Mar 03 '21 at 22:01

My implementation based on taleinat's answer:

import datetime
import calendar

def add_months(orig_date, month_count = 1):
    while month_count > 12:
        month_count -= 12
        orig_date = add_months(orig_date, 12)
    new_year = orig_date.year
    new_month = orig_date.month + month_count
    # 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)

With this function you can add as many months as you'd like.

from datetime import date
dt = date(2021, 1, 31)

print(add_months(dt, 49))

returns 2025-02-28

Utkonos · Answer 39 · 2022-07-27T23:49:45.973

I know that there are many answers to this question already, but using collections.deque and then the rotate() method, a function can be made that takes a datetime object as the input and then outputs a new datetime object that is one "business month" later than the current one. If the day of the month does not exist in the next month, then it subtracts one until it arrives on a valid day of the month and then returns that object.

import collections
import datetime

def next_month(dt: datetime.datetime):
    month_list = list(range(1, 12 + 1))
    months = collections.deque(month_list)
    while True:
        this_month = list(months)[0]
        if dt.month == this_month:
            break
        months.rotate(-1)
    months.rotate(-1)
    month_plus = list(months)[0]
    for i in range(4):
        try:
            return dt.replace(month=month_plus, day=dt.day - i)
        except ValueError:
            continue

The same end result can be done using itertools.cycle.

import datetime
import itertools

def next_month(dt: datetime.datetime):
    month_list = list(range(1, 12 + 1))
    month = itertools.cycle(month_list)
    while True:
        if next(month) == dt.month:
            break
    month_plus = next(month)
    for i in range(4):
        try:
            return dt.replace(month=month_plus, day=dt.day - i)
        except ValueError:
            continue

score 0 · Answer 40 · answered Jun 11 '23 at 18:17

I actually managed to do this using datetime and MonthDelta

pip install MonthDelta

Then

from datetime import datetime
from monthdelta import monthdelta

# Date and time today, now
now        = datetime.now()

# Date next six months from now
six_months = now + monthdelta(6)

print( now.strftime("%Y-%m-%d") )
print( six_months.strftime("%Y-%m-%d") )

>>> 2023-06-11
>>> 2023-12-11

newby · Answer 41 · 2012-05-25T08:21:58.573

-1

In this function, n can be positive or negative.

def addmonth(d, n):
    n += 1
    dd = datetime.date(d.year + n/12, d.month + n%12, 1)-datetime.timedelta(1)
    return datetime.date(dd.year, dd.month, min(d.day, dd.day))

edited May 25 '12 at 08:21

answered May 25 '12 at 08:15

newby

7
2

Frosty Snowman · Answer 42 · 2013-05-28T14:15:38.107

We probably should use dateutil.relativedelta

however for academic interest I will just add that before I discovered it I was goint to use this:

try:
vexpDt = K.today.replace(K.today.year + (K.today.month+6)//12, (K.today.month+5)%12+1, K.today.day)
except:
vexpDt = K.today.replace(K.today.year + (K.today.month+6)//12, (K.today.month+6)%12+1, 1) - timedelta(days = 1)

it seems quite simple but still catches all the issues like 29,30,31

it also works for - 6 mths by doing -timedelta

nb - don't be confused by K.today its just a variable in my program

score -1 · Answer 43 · answered Oct 09 '13 at 10:17

    def addDay(date, number):
        for i in range(number)
            #try to add a day
            try:
                date = date.replace(day = date.day + 1)
            #in case it's impossible ex:january 32nd add a month and restart at day 1
            except:
                #add month part
                try:
                    date = date.replace(month = date.month +1, day = 1)
                except:
                    date = date.replace(year = date.year +1, month = 1, day = 1)

For everyone still reading this post. I think this code is way clearer, especially compared to code using modulo(%).

Sorry for any grammatical error, english is so not my main language

Daniel Margarido · Answer 44 · 2016-01-16T20:52:57.047

This is what I do when I need to add months or years and don't want to import more libraries.

import datetime
__author__ = 'Daniel Margarido'


# Check if the int given year is a leap year
# return true if leap year or false otherwise
def is_leap_year(year):
    if (year % 4) == 0:
        if (year % 100) == 0:
            if (year % 400) == 0:
                return True
            else:
                return False
        else:
            return True
    else:
        return False


THIRTY_DAYS_MONTHS = [4, 6, 9, 11]
THIRTYONE_DAYS_MONTHS = [1, 3, 5, 7, 8, 10, 12]

# Inputs -> month, year Booth integers
# Return the number of days of the given month
def get_month_days(month, year):
    if month in THIRTY_DAYS_MONTHS:   # April, June, September, November
        return 30
    elif month in THIRTYONE_DAYS_MONTHS:   # January, March, May, July, August, October, December
        return 31
    else:   # February
        if is_leap_year(year):
            return 29
        else:
            return 28

# Checks the month of the given date
# Selects the number of days it needs to add one month
# return the date with one month added
def add_month(date):
    current_month_days = get_month_days(date.month, date.year)
    next_month_days = get_month_days(date.month + 1, date.year)

    delta = datetime.timedelta(days=current_month_days)
    if date.day > next_month_days:
        delta = delta - datetime.timedelta(days=(date.day - next_month_days) - 1)

    return date + delta


def add_year(date):
    if is_leap_year(date.year):
        delta = datetime.timedelta(days=366)
    else:
        delta = datetime.timedelta(days=365)

    return date + delta


# Validates if the expected_value is equal to the given value
def test_equal(expected_value, value):
    if expected_value == value:
        print "Test Passed"
        return True

    print "Test Failed : " + str(expected_value) + " is not equal to " str(value)
    return False

# Test leap year
print "---------- Test leap year ----------"
test_equal(True, is_leap_year(2012))
test_equal(True, is_leap_year(2000))
test_equal(False, is_leap_year(1900))
test_equal(False, is_leap_year(2002))
test_equal(False, is_leap_year(2100))
test_equal(True, is_leap_year(2400))
test_equal(True, is_leap_year(2016))

# Test add month
print "---------- Test add month ----------"
test_equal(datetime.date(2016, 2, 1), add_month(datetime.date(2016, 1, 1)))
test_equal(datetime.date(2016, 6, 16), add_month(datetime.date(2016, 5, 16)))
test_equal(datetime.date(2016, 3, 15), add_month(datetime.date(2016, 2, 15)))
test_equal(datetime.date(2017, 1, 12), add_month(datetime.date(2016, 12, 12)))
test_equal(datetime.date(2016, 3, 1), add_month(datetime.date(2016, 1, 31)))
test_equal(datetime.date(2015, 3, 1), add_month(datetime.date(2015, 1, 31)))
test_equal(datetime.date(2016, 3, 1), add_month(datetime.date(2016, 1, 30)))
test_equal(datetime.date(2016, 4, 30), add_month(datetime.date(2016, 3, 30)))
test_equal(datetime.date(2016, 5, 1), add_month(datetime.date(2016, 3, 31)))

# Test add year
print "---------- Test add year ----------"
test_equal(datetime.date(2016, 2, 2), add_year(datetime.date(2015, 2, 2)))
test_equal(datetime.date(2001, 2, 2), add_year(datetime.date(2000, 2, 2)))
test_equal(datetime.date(2100, 2, 2), add_year(datetime.date(2099, 2, 2)))
test_equal(datetime.date(2101, 2, 2), add_year(datetime.date(2100, 2, 2)))
test_equal(datetime.date(2401, 2, 2), add_year(datetime.date(2400, 2, 2)))

Just create a datetime.date() object, call add_month(date) to add a month and add_year(date) to add a year.

``add_month(datetime(2015, 1, 31))`` would yield ``2015-03-03`` which is probably not what people would expect for ``adding one month``. — ohw, Jan 16 '16 at 19:58
Fixed, now 2015-03-01 would be returned instead of 2015-03-03 — Daniel Margarido, Jan 16 '16 at 20:53

Nandlal Yadav · Answer 45 · 2017-12-07T01:13:25.367

-1

Using below given function you can get date after/before x months.

from datetime import date

def next_month(given_date, month):
    yyyy = int(((given_date.year * 12 + given_date.month) + month)/12)
    mm = int(((given_date.year * 12 + given_date.month) + month)%12)

    if mm == 0:
        yyyy -= 1
        mm = 12
    return given_date.replace(year=yyyy, month=mm)


if __name__ == "__main__":
    today = date.today()
    print(today)

    for mm in [-12, -1, 0, 1, 2, 12, 20 ]:
        next_date = next_month(today, mm)
        print(next_date)

edited Dec 07 '17 at 01:13

answered Dec 07 '17 at 00:08

Nandlal Yadav

21
2

This answer is an exact duplicate of an answer you posted 8 minutes earlier to this same question. Please delete this redundant answer. – Tom Aranda Dec 07 '17 at 00:19

score -1 · Answer 46 · answered Mar 23 '18 at 13:15

I could not find an exact solution to this question so i'll post my solution in case it may be of any help using stantard Calendar and datetime libs. this works for add and substract months, and accounts for month-end rolls and cases where the final month has less days than the initial one. I also have a more generalized solution if you are looking for more complex manipulation, it adds regular intervals (days, months, years, quarters, semeters, etc) like: '1m', '-9m', '-1.5y', '-3q', '1s' etc.

from datetime import datetime
from calendar import monthrange
def date_bump_months(start_date, months):
    """
    bumps months back and forth. 
    --> if initial date is end-of-month, i will move to corresponding month-end
    --> ir inital date.day is greater than end of month of final date, it casts it to momth-end
    """
    signbit = -1 if months < 0 else 1
    d_year, d_month = divmod(abs(months),12)    
    end_year = start_date.year + d_year*signbit 
    end_month = 0
    if signbit ==-1:            
        if d_month < start_date.month:
            end_month = start_date.month - d_month
        else:
            end_year -=1
            end_month = 12 - (d_month - start_date.month)
    else:
        end_month +=start_date.month
        if end_month  > 12:
            end_year +=1
            end_month -=12
    # check if we are running end-of-month dates
    eom_run = monthrange(start_date.year, start_date.month)[1]==start_date.day
    eom_month = monthrange((end_year), (end_month))[1]
    if eom_run:
        end_day = eom_month 
    else:
        end_day = min(start_date.day, eom_month )    
    return date(end_year, end_month, end_day)

Mark · Answer 47 · 2018-12-12T13:03:02.920

My modification to Tony Diep's answer, possibly marginally more elegant (Python 2 of course, matching the date of the question & original answers, for Python 3 modify as necessary, including / to // at least):

def add_months(date, months):
    month = date.month + months - 1
    year = date.year + (month / 12)
    month = (month % 12) + 1
    day = date.day
    while (day > 0):
        try:
            new_date = date.replace(year=year, month=month, day=day)
            break
        except:
            day = day - 1    
    return new_date

adds months according to a 'business needs' interpretation that dates mapping beyond the end of the month, should map to the end of the month rather than into the following month.

score -6 · Answer 48 · answered Feb 13 '09 at 15:20

-6

Use the python datetime module to add a timedelta of six months to datetime.today() .

http://docs.python.org/library/datetime.html

You will of course have to solve the issue raised by Johannes Weiß-- what do you mean by 6 months?

answered Feb 13 '09 at 15:20

Devin Jeanpierre

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timedelta doesn't support months and so sidesteps the possible answers to the question "how many days in 6-month?" Eef's code will set a review date so I would suggest one could consider setting the timedelta using days (6*30). If the period represents clients' access to a product/service then a business definition may be required/preferred. – Carl Sep 21 '09 at 15:46

How do I calculate the date six months from the current date using the datetime Python module?

48 Answers48

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