Python: get all months in range?

Question

I want to get all months between now and August 2010, as a list formatted like this:

[\'2010-08-01\', \'2010-09-01\', .... , \'2016-02-01\']

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Answer 1

I had a look at the dateutil documentation. Turns out it provides an even more convenient way than using dateutil.relativedelta: recurrence rules (examples)

For the task at hand, it's as easy as

from dateutil.rrule import *
from datetime import date

months = map(
    date.isoformat,
    rrule(MONTHLY, dtstart=date(2010, 8, 1), until=date.today())
)

The fine print

Note that we're cheating a little bit, here. The elements dateutil.rrule.rrule produces are of type datetime.datetime, even if we pass dtstart and until of type datetime.date, as we do above. I let map feed them to date's isoformat function, which just turns out to convert them to strings as if it were just dates without any time-of-day information.

Therefore, the seemingly equivalent list comprehension

[day.isoformat()
    for day in rrule(MONTHLY, dtstart=date(2010, 8, 1), until=date.today())]

would return a list like

['2010-08-01T00:00:00',
 '2010-09-01T00:00:00',
 '2010-10-01T00:00:00',
 '2010-11-01T00:00:00',
 ⋮
 '2015-12-01T00:00:00',
 '2016-01-01T00:00:00',
 '2016-02-01T00:00:00']

Thus, if we want to use a list comprehension instead of map, we have to do something like

[dt.date().isoformat()
    for dt in rrule(MONTHLY, dtstart=date(2010, 8, 1), until=date.today())]