[R] Dividing tick-data into intervalls

[Milan Bouchet-Valat]

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