Are you the oswi who just asked a very similar question? Regardless, as Josh said, the high-performance way to do this is to use the specialty C code available through the xts package and the to.period() functions, specifically to.minutes5
Michael On Tue, May 8, 2012 at 8:48 AM, Milan Bouchet-Valat <nalimi...@club.fr> wrote: > Le mardi 08 mai 2012 à 10:44 +0200, osvald wiklander a écrit : >> >> >> >> Hi everybody, I am sorry that I am kind of spamming this forum, but I >> have searched for some input everywhere and cant really find a nice >> solution for my problem. >> >> Data looks like: >> >> price >> 2011-11-01 08:00:00 0.000000000 >> 2011-11-01 08:00:00 0.000000000 >> 2011-11-01 08:02:00 0.000000000 >> 2011-11-01 08:03:00 -0.017033339 >> 2011-11-01 08:13:00 0.000690001 >> 2011-11-01 08:24:00 0.000658411 >> 2011-11-01 08:29:00 0.000000000 >> 2011-11-01 08:29:00 0.000000000 >> 2011-11-01 08:29:00 0.000000000 >> 2011-11-01 08:29:00 0.000000000 >> 2011-11-01 08:29:00 0.002166062 >> 2011-11-01 08:44:00 0.000000000 >> 2011-11-01 08:44:00 -0.002166062 >> 2011-11-01 08:44:00 0.004321374 >> 2011-11-01 10:36:00 0.010618976 >> 2011-11-01 15:59:00 0.002092990 >> >> So the price column in fact shows the difference in the price of the >> equity. It is about 100 days. >> >> I want to be able to convert this data to regurly spaced intervals, >> like every fifth minute the previous tick is returned. >> >> So, for example I want that the new time serie should look like: >> >> price >> 2011-11-01 08:00:00 0.000000000 >> 2011-11-01 08:05:00 -0.017033339 >> 2011-11-01 08:10:00 -0.017033339 >> 2011-11-01 08:15:00 0.000690001 >> 2011-11-01 08:20:00 0.000000000 >> 2011-11-01 08:25:00 0.000658411 >> 2011-11-01 08:30:00 0.002166062 >> >> And so on for the 100 days. And the I want to do this for ten minutes, >> 30 minutes etc. >> >> Earlier I have tried with aggregatePrice() (in the RTAQ package) >> combined with a for() loop. However this didnt work so well. >> >> I assume it exist some easy way to do this? However, as I said, I have >> searched for a solution without any luck. >> >> So, what approach should I have? Have any of you done something like >> this before? > I think you should create a zoo object (from package zoo) from your > data, and use aggregate() on it (see ?aggregate.zoo). > > > My two cents > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
