your specification was to have the data centered at 15 minute intervals and
extend 7.5 minutes on either side. on my output you see the time of the lower
value of the interval. 11:52:30 is the lower value of the 12:00:00 time. so
you can just add back the 7.5 minutes to the time value to get it centered on
the time you want.
Sent from my iPad
On Aug 28, 2012, at 9:14, Jason Gilmore <w...@wjgilmore.com> wrote:
> Hi Jim,
>
> Oh wow thanks so much. One question though, did you notice that the 12:00:00
> slot is missing entirely from the output? It shows 11:52:30 and 12:07:30, but
> not 12:00:00. Actually after another review I see that it's missing the 12,
> 12:15, and 12:30?
>
> >2012-07-22 11:52:30 2012-07-22 12:07:30 2012-07-22 12:22:30 2012-07-22
> >12:37:30
> > NA 22.33333 NA
> > 24.00000
> >2012-07-22 12:52:30
> > 20.00000
>
> Any ideas how to fix this?
>
> Thanks again!
> Jason
>
>
> On Mon, Aug 27, 2012 at 1:40 PM, jim holtman <jholt...@gmail.com> wrote:
> try this:
>
>
> > x <- read.table(text = "2012-07-22 12:12:00, 21
> + 2012-07-22 12:15:00, 22
> + 2012-07-22 12:18:00, 24
> + 2012-07-22 12:39:00, 21
> + 2012-07-22 12:45:00, 25
> + 2012-07-22 12:49:00, 26
> + 2012-07-22 12:53:00, 20
> + 2012-07-22 13:00:00, 18
> + 2012-07-22 13:06:00, 22", colClasses = c("POSIXct", "integer"), sep = ',')
> > # get minimum at an hour granularity
> > tMin <- trunc(min(x$V1), units = 'hour')
> > # back off 7.5 minute
> > tMin <- tMin - (7.5 * 60)
> > # create sequence for 'cut'
> > cSeq <- seq(tMin, max(x$V1) + (7.5 * 60), by = '15 min')
> > # now split and average
> > cCut <- cut(x$V1, cSeq)
> > # compute means
> > tapply(x$V2, cCut, mean)
> 2012-07-22 11:52:30 2012-07-22 12:07:30 2012-07-22 12:22:30 2012-07-22
> 12:37:30
> NA 22.33333 NA
> 24.00000
> 2012-07-22 12:52:30
> 20.00000
> >
>
>
> On Mon, Aug 27, 2012 at 9:53 AM, Jason Gilmore <w...@wjgilmore.com> wrote:
> >
> > Hi,
> >
> > I'm pretty new to R and have run into a task which although I'm certain is
> > within R's capabilities, falls outside of mine. :-) Consider the following
> > data set:
> >
> > 2012-07-22 12:12:00, 21
> > 2012-07-22 12:15:00, 22
> > 2012-07-22 12:18:00, 24
> > 2012-07-22 12:39:00, 21
> > 2012-07-22 12:45:00, 25
> > 2012-07-22 12:49:00, 26
> > 2012-07-22 12:53:00, 20
> > 2012-07-22 13:00:00, 18
> > 2012-07-22 13:06:00, 22
> >
> > My task involves creating a data set which *averages* these values at a
> > resolution of 15 minutes, meaning that I need to average the values falling
> > within 7.5 minutes of a 15 minute increment. Therefore given the above data
> > set I need to treat it as three groups:
> >
> > 2012-07-22 12:12:00, 21
> > 2012-07-22 12:15:00, 22
> > 2012-07-22 12:18:00, 24
> >
> > 2012-07-22 12:39:00, 21
> > 2012-07-22 12:45:00, 25
> > 2012-07-22 12:49:00, 26
> >
> > 2012-07-22 12:53:00, 20
> > 2012-07-22 13:00:00, 18
> > 2012-07-22 13:06:00, 22
> >
> > The end result should look like this:
> >
> > 2012-07-22 12:15:00, 22.33
> > 2012-07-22 12:30:00, NA <- Because this 15 minute slot did not previously
> > exist
> > 2012-07-22 12:45:00, 24
> > 2012-07-22 1:00:00, 20
> >
> > Any help much appreciated. I've been working on this for several hours with
> > little success. I'm able to identify the missing (NA) value using zoo/xts
> > but can't seem to sort out the averaging matter.
> >
> > Thanks so much!
> > Jason
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > 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.
>
>
>
>
> --
> Jim Holtman
> Data Munger Guru
>
> What is the problem that you are trying to solve?
> Tell me what you want to do, not how you want to do it.
>
[[alternative HTML version deleted]]
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