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Home/ Questions/Q 92481
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Asked: May 10, 20262026-05-10T23:11:37+00:00 2026-05-10T23:11:37+00:00

I have an rrule instance e.g. r = rrule(WEEKLY, byweekday=SA, count=10, dtstart=parse(‘20081001’)) where dtstart

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I have an rrule instance e.g.

     r = rrule(WEEKLY, byweekday=SA, count=10, dtstart=parse('20081001')) 

where dtstart and byweekday may change.

If I then want to generate the ten dates that follow on from this rrule, what’s the best way of doing it? Can I assign a new value to the _dtstart member of r? That seems to work but I’m not sure.

e.g.

     r._dtstart = list(r)[-1] or something like that 

Otherwise I guess I’ll create a new rrule and access _dtstart, _count, _byweekday etc. of the original instance.

EDIT:

I’ve thought about it more, and I think what I should be doing is omitting the ‘count’ argument when I create the first rrule instance. I can still get ten occurrences the first time I use the rrule

instances = list(r[0:10])  

and then afterwards I can get more

more = list(r[10:20])  

I think that solves my problem without any ugliness

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  1. 2026-05-10T23:11:38+00:00Added an answer on May 10, 2026 at 11:11 pm

    Firstly, r._dtstart = list(r)[-1] will give you the last date in the original sequence of dates. If you use that, without modification, for the beginning of a new sequence, you will end up with a duplicate date, i.e. the last date of the first sequence will be the same as the first date of the new sequence, which is probably not what you want:

    >>> from dateutil.rrule import * >>> import datetime  >>> r = rrule(WEEKLY, byweekday=SA, count=10, dtstart=datetime.datetime(2008,10,01)) >>> print list(r) [datetime.datetime(2008, 10, 4, 0, 0), datetime.datetime(2008, 10, 11, 0, 0), datetime.datetime(2008, 10, 18, 0, 0), datetime.datetime(2008, 10, 25, 0, 0), datetime.datetime(2008, 11, 1, 0, 0), datetime.datetime(2008, 11, 8, 0, 0), datetime.datetime(2008, 11, 15, 0, 0), datetime.datetime(2008, 11, 22, 0, 0), datetime.datetime(2008, 11, 29, 0, 0), datetime.datetime(2008, 12, 6, 0, 0)] >>> r._dtstart = r[-1] >>> print list(r) [datetime.datetime(2008, 12, 6, 0, 0), datetime.datetime(2008, 12, 13, 0, 0), datetime.datetime(2008, 12, 20, 0, 0), datetime.datetime(2008, 12, 27, 0, 0), datetime.datetime(2009, 1, 3, 0, 0), datetime.datetime(2009, 1, 10, 0, 0), datetime.datetime(2009, 1, 17, 0, 0), datetime.datetime(2009, 1, 24, 0, 0), datetime.datetime(2009, 1, 31, 0, 0), datetime.datetime(2009, 2, 7, 0, 0)] 

    Furthermore, it is considered poor form to manipulate r._dtstart as it is clearly intended to be a private attribute.

    Instead, do something like this:

    >>> r = rrule(WEEKLY, byweekday=SA, count=10, dtstart=datetime.datetime(2008,10,01)) >>> r2 = rrule(WEEKLY, byweekday=SA, count=r.count(), dtstart=r[-1] + datetime.timedelta(days=1)) >>> print list(r) [datetime.datetime(2008, 10, 4, 0, 0), datetime.datetime(2008, 10, 11, 0, 0), datetime.datetime(2008, 10, 18, 0, 0), datetime.datetime(2008, 10, 25, 0, 0), datetime.datetime(2008, 11, 1, 0, 0), datetime.datetime(2008, 11, 8, 0, 0), datetime.datetime(2008, 11, 15, 0, 0), datetime.datetime(2008, 11, 22, 0, 0), datetime.datetime(2008, 11, 29, 0, 0), datetime.datetime(2008, 12, 6, 0, 0)] >>> print list(r2) [datetime.datetime(2008, 12, 13, 0, 0), datetime.datetime(2008, 12, 20, 0, 0), datetime.datetime(2008, 12, 27, 0, 0), datetime.datetime(2009, 1, 3, 0, 0), datetime.datetime(2009, 1, 10, 0, 0), datetime.datetime(2009, 1, 17, 0, 0), datetime.datetime(2009, 1, 24, 0, 0), datetime.datetime(2009, 1, 31, 0, 0), datetime.datetime(2009, 2, 7, 0, 0), datetime.datetime(2009, 2, 14, 0, 0)] 

    This codes does not access any private attributes of rrule (although you might have to look at _byweekday).

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