Python date time, get date 6 months from now

Question

Hey,

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

Edit: The reason I am wanting 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. Does this help?

Cheers

Eef

Answer 1

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?

Answer 2

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

Answer 3

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, 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. :)

Answer 4

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

Answer 5

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.

Answer 6

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

1) natural months,

   (day,month,year) = (day,(month+6)%12,year+(month+6)/12)

2) bankers definition, 6*30

   date += datetime.timedelta(6*30)

Answer 7

import datetime
print (datetime.date.today() + datetime.timedelta(6*365/12)).isoformat()

Answer 8

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)

Answer 9

Modified Johannes Wei's answer in the case new_month = 12. 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)

Xinhui Tang

Answer 10

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)

Answer 11

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)

Answer 12

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)

Answer 13

This solution works correctly for December, which most of the answers on this page do not. You need to first shift the months from base 1 (ie Jan = 1) to base 0 (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  = 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 soln 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)>>>