Hi,
Perhaps this might work for you. It leverages findInterval() and a
simple look-up-table of times to do the grouping. I made it return NA
when computing the mean when there are fewer than the three
observations.
Cheers,
Ben
n <- 144
x <- data.frame(
datetime = seq(from = as.POSIXct("2018-02-01 00:00:00", tz = "UTC"),
by = "10 min",
length = n),
vmax = sample(10:50, n, replace = TRUE)
)
lut <- seq(from = x$datetime[1],
to = x$datetime[n],
by = "30 min") + 1 # add one second so that 00 sorts
with 40, 50, 00
# and the other grouping is 10, 20 30
x$interval <- findInterval(x$datetime, lut)
x
y <- aggregate(vmax ~ interval, data = x,
FUN = function(x){
if (length(x) < 3){
r <- NA
} else {
r <- mean(x)
}
r
})
y
On Sun, Dec 6, 2020 at 1:59 PM Stefano Sofia
<stefano.so...@regione.marche.it> wrote:
>
> Hi Jim.
> I studied and implemented your solution in details. The idea is great, but
> after a sharp revision I came to the conclusion that unfortunately it des not
> work correctly: for the "am" side (10, 20, 30 minutes) it works well because
> the hour is exactly the same, while for the "pm" side (40, 50, 00) the
> algorithm it doesn't because the hour related to 40 and 50 minutes is
> different from the hour related to 00 (which is the following one). Am I
> wrong?
> I tried to fix it keeping the easy structure of the algorithm, but with no
> success.
>
> Any hint for that?
> Thank you for your attention and your help
>
> Stefano
>
>
> (oo)
> --oOO--( )--OOo----------------
> Stefano Sofia PhD
> Civil Protection - Marche Region
> Meteo Section
> Snow Section
> Via del Colle Ameno 5
> 60126 Torrette di Ancona, Ancona
> Uff: 071 806 7743
> E-mail: stefano.so...@regione.marche.it
> ---Oo---------oO----------------
>
> ________________________________________
> Da: Jim Lemon [drjimle...@gmail.com]
> Inviato: giovedì 3 dicembre 2020 4.41
> A: Stefano Sofia
> Cc: r-help mailing list
> Oggetto: Re: [R] change frequency of wind data correctly
>
> Hi again,
> Didn't realize that the example didn't even span a full day.
>
> ssdf<-read.table(text="date_POSIX time_POSIX vmax
> 2018-02-01 00:00:00 27
> 2018-02-01 00:10:00 41
> 2018-02-01 00:20:00 46
> 2018-02-01 00:30:00 39
> 2018-02-01 00:40:00 34
> 2018-02-01 00:50:00 32
> 2018-02-01 01:00:00 37
> 2018-02-01 01:10:00 31
> 2018-02-01 01:20:00 26
> 2018-02-01 01:30:00 29
> 2018-02-01 01:40:00 24
> 2018-02-01 01:50:00 35",
> header=TRUE,stringsAsFactors=FALSE)
> # extract the hour
> ssdf$hour<-
> as.numeric(unlist(lapply(strsplit(ssdf$time_POSIX,":"),"[",1)))
> # get the time of day as seconds from the time field
> ssdf$mins<-
> as.numeric(unlist(lapply(strsplit(ssdf$time_POSIX,":"),"[",2)))
> # create an AM/PM variable
> ssdf$ampm<-ifelse(ssdf$mins > 0 & ssdf$mins <= 30,"am","pm")
> # drop first row
> ssdf<-ssdf[-1,]
> means<-aggregate(vmax~hour+ampm,ssdf,mean)
>
> This does a full day. To do more, add the date_POSIX field to the
> aggregate command. If you have the date and time in one field you'll
> have to split that. That will distinguish the AM/PM means in each day
> as well as hour.
>
> Jim
>
> On Thu, Dec 3, 2020 at 2:10 PM Jim Lemon <drjimle...@gmail.com> wrote:
> >
> > Hi Stefano,
> > I read in your date-time as two separate fields for convenience. You
> > can split your single field at the space to get the same result.
> >
> > ssdf<-read.table(text="date_POSIX time_POSIX vmax
> > 2018-02-01 00:00:00 27
> > 2018-02-01 00:10:00 41
> > 2018-02-01 00:20:00 46
> > 2018-02-01 00:30:00 39
> > 2018-02-01 00:40:00 34
> > 2018-02-01 00:50:00 32",
> > header=TRUE,stringsAsFactors=FALSE)
> > # get the time of day as seconds from the time field
> > ssdf$seconds<-as.numeric(strptime(ssdf$time_POSIX,"%H:%M:%S"))
> > # subtract whatever current date strptime guesses for the date
> > ssdf$seconds<-ssdf$seconds-min(ssdf$seconds)
> > # create an AM/PM variable
> > ssdf$ampm<-ifelse(ssdf$seconds > 0 & ssdf$seconds <= 1800,"am","pm")
> > means<-aggregate(vmax~ampm,ssdf,mean)
> >
> > Jim
> >
> > On Thu, Dec 3, 2020 at 4:55 AM Stefano Sofia
> > <stefano.so...@regione.marche.it> wrote:
> > >
> > > Dear list users,
> > > I have wind data with frequency of 10 minutes (three years data). For
> > > simplicity let me use only max wind speed.
> > > I need to reduce the frequency to 30 minutes, at 00 (taking the mean of
> > > data at 40, 50 and 00 minutes) and at 30 (taking the mean of data at 10,
> > > 20 and 30 minutes) of each hour.
> > >
> > > The simple code here reported works well, but the column "interval"
> > > groups data forward, not backward:
> > >
> > > init_day <- as.POSIXct("2018-02-01-00-00", format="%Y-%m-%d-%H-%M",
> > > tz="Etc/GMT-1")
> > > fin_day <- as.POSIXct("2018-02-01-02-00", format="%Y-%m-%d-%H-%M",
> > > tz="Etc/GMT-1")
> > > mydf <- data.frame(data_POSIX=seq(init_day, fin_day, by="10 mins"))
> > > mydf$vmax <- round(rnorm(13, 35, 10))
> > > mydf$interval <- cut(mydf$data_POSIX, , breaks="30 min")
> > > means <- aggregate(vmax ~ interval, mydf, mean)
> > >
> > > data_POSIX vmax interval
> > > 1 2018-02-01 00:00:00 27 2018-02-01 00:00:00
> > > 2 2018-02-01 00:10:00 41 2018-02-01 00:00:00
> > > 3 2018-02-01 00:20:00 46 2018-02-01 00:00:00
> > > 4 2018-02-01 00:30:00 39 2018-02-01 00:30:00
> > > 5 2018-02-01 00:40:00 34 2018-02-01 00:30:00
> > > 6 2018-02-01 00:50:00 32 2018-02-01 00:30:00
> > > ...
> > >
> > > I should work with
> > >
> > > data_POSIX vmax interval
> > > 1 2018-02-01 00:00:00 27 2018-02-01 00:00:00
> > > 2 2018-02-01 00:10:00 41 2018-02-01 00:30:00
> > > 3 2018-02-01 00:20:00 46 2018-02-01 00:30:00
> > > 4 2018-02-01 00:30:00 39 2018-02-01 00:30:00
> > > 5 2018-02-01 00:40:00 34 2018-02-01 00:00:00
> > > 6 2018-02-01 00:50:00 32 2018-02-01 00:00:00
> > > ...
> > >
> > >
> > > Is there a way to modify this code to groupp data correctly? (I would
> > > prefer using only the base package)
> > >
> > > Thank you for your help
> > > Stefano
> > >
> > >
> > >
> > > (oo)
> > > --oOO--( )--OOo----------------
> > > Stefano Sofia PhD
> > > Civil Protection - Marche Region
> > > Meteo Section
> > > Snow Section
> > > Via del Colle Ameno 5
> > > 60126 Torrette di Ancona, Ancona
> > > Uff: 071 806 7743
> > > E-mail: stefano.so...@regione.marche.it
> > > ---Oo---------oO----------------
> > >
> > > ________________________________
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> informazioni confidenziali, pertanto è destinato solo a persone autorizzate
> alla ricezione. I messaggi di posta elettronica per i client di Regione
> Marche possono contenere informazioni confidenziali e con privilegi legali.
> Se non si è il destinatario specificato, non leggere, copiare, inoltrare o
> archiviare questo messaggio. Se si è ricevuto questo messaggio per errore,
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> di necessità ed urgenza, la risposta al presente messaggio di posta
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--
Ben Tupper
Bigelow Laboratory for Ocean Science
East Boothbay, Maine
http://www.bigelow.org/
https://eco.bigelow.org
______________________________________________
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