Get (year,month) for the last X months

Question

I got a very simple thing to to in python: I need a list of tuples (year,month) for the last x months starting (and including) from today. So, for x=10 and today(July 2011), the command should output:

[(2011, 7), (2011, 6), (2011, 5), (2011, 4), (2011, 3), 
(2011, 2), (2011, 1), (2010, 12), (2010, 11), (2010, 10)]

Only the default datetime implementation of python should be used. I came up with the following solution:

import datetime
[(d.year, d.month) for d in [datetime.date.today()-datetime.timedelta(weeks=4*i) for i in range(0,10)]]

This solution outputs the correct solution for my test cases but I'm not comfortable with this solution: It assumes that a month has four weeks and this is simply not true. I could replace the weeks=4 with days=30 which would make a better solution but it is still not correct.

The other solution which came to my mind is to use simple maths and subtract 1 from a months counter and if the month-counter is 0, subtract 1 from a year counter. The problem with this solution: It requires more code and isn't very readable either.

So how can this be done correctly?

Answer 1

Using relativedelta ...

import datetime
from dateutil.relativedelta import relativedelta

def get_last_months(start_date, months):
    for i in range(months):
        yield (start_date.year,start_date.month)
        start_date += relativedelta(months = -1)

>>> X = 10       
>>> [i for i in get_last_months(datetime.datetime.today(), X)]
>>> [(2013, 2), (2013, 1), (2012, 12), (2012, 11), (2012, 10), (2012, 9), (2012, 8), (2012, 7), (2012, 6), (2012, 5)]

Answer 2

I don't see it documented anywhere, but time.mktime will "roll over" into the correct year when given out-of-range, including negative, month values:

x = 10
now = time.localtime()
print([time.localtime(time.mktime((now.tm_year, now.tm_mon - n, 1, 0, 0, 0, 0, 0, 0)))[:2] for n in range(x)])