Hi
if you have full set of data, so no observation is missing, you can split your
data frame values e.g. by
split(yourdata, trunc(0:nrow(yourdata))/6)
and use lapply or for cycle.
If you need to cut according to dates you can use
?cut
POSIX date.
Cheers
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Stefano
> Sofia
> Sent: Friday, September 11, 2015 11:47 AM
> To: r-help@r-project.org
> Subject: [R] how to divide a data frame of half-hourly wind
> observations into subset of three hours
>
> Dear R-users,
> from a data frame of half-hourly wind observations (direction and
> speed, three years data), I need to evaluate the first main direction
> for each period of three hours (h03, h06, h09, h12, h15, h18, h21, h24
> of each day).
>
> The command "windRose" of the package "openair" crates a data frame
> that contains the information I need.
> Can I ask you which is the most efficient way to run the windRose
> command for each subset of three hours?
>
> The initial data frame is called wind_df is like
>
> Date Direction Speed
> 2015-06-30 00:30:00 315 2.9
> 2015-06-30 01:00:00 338 3.5
> 2015-06-30 01:30:00 23 4.8
> 2015-06-30 02:00:00 338 4.1
> 2015-06-30 02:30:00 135 1.1
> 2015-06-30 03:00:00 45 2.7
> 2015-06-30 03:30:00 1.7
> 2015-06-30 04:00:00 293 1
> 2015-06-30 04:30:00 45 1.5
> 2015-06-30 05:00:00 45 2.3
> ...
>
> The command for the wind rose creation is
> output1_df <- windRose(wind_df, ws="Speed", wd="Direction",
> layout=c(1,1), ws.int=4, angle=22.5, type="default", bias.corr=TRUE,
> cols="default", grid.line=NULL, width=2, seg=NULL, auto.text=TRUE,
> breaks=c(0,0.3,1.5,3.4,5.4,7.9,10.7,13.8,17.1,20.7,24.4,28.4,32.6,40),
> offset=5, normalise=TRUE, max.freq=NULL, paddle=TRUE, key.header=NULL,
> key.footer="(km/hr)", key.position="bottom", key=TRUE, dig.lab=2,
> statistic="prop.count", pollutant=NULL, annotate=TRUE, border=NA,
> par.settings=list(fontsize=list(text=10)))
>
> and I evaluate the first main direction through
> output2_df <- output1_df$data
> output2_df$wd[output2_df$freqs == max(output2_df$freqs)])
>
>
> Thank you for your attention and your help
> Stefano Sofia
>
>
>
>
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