[R] adding date/time stamp

[Gabor Grothendieck]

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 <sheck at 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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