Try this:
> aggregate(z$mph, trunc(time(z), "hour"), mean)
(09/01/05 00:00:00) (09/01/05 01:00:00) (09/01/05 02:00:00)
9.27500 10.08333 9.20000
On Tue, Dec 30, 2008 at 6:30 PM, Sherri Heck <sh...@ucar.edu> wrote:
> Dear All-
>
> I have a dataset that is comprised of the following (LST = yymmddhhMM):
>
>
>
> LST in mph Deg DegF DegF2 % volts Deg mph2 w/m2
> 0509010000 0.00 7.8 216.9 45.1 -999 24.4 -999 -999 10.6
> 0.2
> 0509010005 0.00 8.6 206.6 45.1 -999 25.2 -999 -999 11.7
> 0.2
> 0509010010 0.00 7.8 199.2 44.9 -999 25.4 -999 -999 12.8
> 0.2
> 0509010015 0.00 7.7 197.4 44.8 -999 25.4 -999 -999 10.4
> 0.2
> 0509010020 0.00 7.6 203.9 44.8 -999 25.3 -999 -999 10.0
> 0.2
> 0509010025 0.00 9.3 200.9 44.9 -999 25.3 -999 -999 11.8
> 0.2
> 0509010030 0.00 9.4 200.3 44.7 -999 25.5 -999 -999 12.2
> 0.2
> 0509010035 0.00 10.0 199.2 44.6 -999 25.9 -999 -999 13.0
> 0.2
> 0509010040 0.00 9.5 201.5 44.5 -999 25.9 -999 -999 13.3
> 0.2
> 0509010045 0.00 10.8 200.4 44.5 -999 26.1 -999 -999 13.0
> 0.2
> 0509010050 0.00 11.8 198.4 44.5 -999 26.1 -999 -999 13.3
> 0.2
> 0509010055 0.00 11.0 197.4 44.5 -999 25.5 -999 -999 13.3
> 0.2
> 0509010100 0.00 9.7 202.0 44.6 -999 25.1 -999 -999 13.0
> 0.2
> 0509010105 0.00 9.0 215.1 44.7 -999 24.9 -999 -999 12.2
> 0.2
> 0509010110 0.00 10.1 223.1 44.6 -999 25.1 -999 -999 13.2
> 0.2
> 0509010115 0.00 10.4 231.2 44.5 -999 25.5 -999 -999 12.0
> 0.2
> 0509010120 0.00 11.0 237.4 44.2 -999 25.9 -999 -999 11.7
> 0.2
> 0509010125 0.00 10.6 241.0 44.2 -999 26.0 -999 -999 11.8
> 0.2
> 0509010130 0.00 11.1 242.2 44.1 -999 26.2 -999 -999 12.2
> 0.2
> 0509010135 0.00 10.6 240.0 44.0 -999 26.5 -999 -999 11.5
> 0.2
> 0509010140 0.00 10.1 241.0 44.0 -999 26.4 -999 -999 11.5
> 0.2
> 0509010145 0.00 9.8 243.2 44.0 -999 26.6 -999 -999 10.7
> 0.2
> 0509010150 0.00 9.3 240.3 43.9 -999 27.0 -999 -999 10.0
> 0.2
> 0509010155 0.00 9.3 239.2 43.8 -999 26.8 -999 -999 10.0
> 0.2
> 0509010200 0.00 9.2 240.1 43.8 -999 26.6 -999 -999 9.8
> 0.2
> 0509010205 0.00 9.0 240.0 43.8 -999 26.6 -999 -999 9.4
> 0.2
> 0509010210 0.00 9.2 245.0 43.9 -999 26.3 -999 -999 9.8
> 0.2
> 0509010215 0.00 9.4 253.2 44.1 -999 26.4 -999 -999 10.6
> 0.2
>
> The data are recorded in 5 minute intervals and I would like to condense it
> into hourly means for "mph". For example, I would like the hourly avg of
> mph so that the output would be as follows:
>
> Year Month Day Hour mph
> 2005 1 1 0 12
> 2005 1 1 1 7
> 2005 1 1 2 11, etc.
>
>
> It seems I am able to get the averages but not output the corresponding
> date/time stamp. From looking at previous help questions, I think I need to
> us "ts" and "aggregate". Gabor taught me how to convert the date/time
> stamp to an easier to manage format (his help is shown below). This is what
> I have so far.
> library(zoo)
> library(chron)
>
> z <- read.zoo("SPL 2005 2008 met data 5 min wout full hdr.txt", header =
> TRUE, na.strings = -999,
> format = "%y%m%d%H%M", FUN = as.chron,
> colClasses = c("character", rep("numeric", 10)))
> z.ts <- ts(z, frequency=12) #avging 5 min intervals to get hourly avg.
> ww <- matrix(aggregate(z.ts[,2], FUN=mean))
>
> any thoughts as to how to add the time stamp is greatly welcomed!
>
> sherri heck
>
> ______________________________________________
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>
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