Hello,
And another solution, taking advantage of Rasmus' one:
simplify2array(parallel::mclapply(c(
� "%Y",
� "%m",
� "%d",
� "%H"), function(fmt, x) {
��� as.integer(format(as.POSIXct(x), format = fmt))
}, x = dta$forecast.date))
#���� [,1] [,2] [,3] [,4]
#[1,] 2020��� 8��� 1�� 12
#[2,] 2020��� 8��� 1�� 12
#[3,] 2020��� 8��� 1�� 12
#[4,] 2020��� 8��� 1�� 12
#[5,] 2020��� 8��� 1�� 12
The data set dta is Jeff's, it's in dput format.
Hope this helps,
Rui Barradas
�s 18:26 de 02/08/2020, Rasmus Liland escreveu:
> On 2020-08-02 09:24 -0700, Philip wrote:
> | Below is some Weather Service data. I
> | would like to parse the forecast date
> | field into four different columns:
> | Year, Month, Day, Hour
>
> Dear Philip,
>
> I'm largely re-iterating Eric and Jeff's
> excellent solutions:
>
> > dat <- structure(list(forecast.date =
> + c("2020-08-01 12:00:00",
> + "2020-08-01 12:00:00",
> + "2020-08-01 12:00:00",
> + "2020-08-01 12:00:00",
> + "2020-08-01 12:00:00"
> + ), TMP = c("305.495", "305.245",
> + "305.057", "305.745", "305.495"
> + )), row.names = c(NA, 5L),
> + class = "data.frame")
> > t(apply(simplify2array(
> + strsplit(dat$forecast.date, "-| |:")),
> + 2, as.numeric))
> [,1] [,2] [,3] [,4] [,5] [,6]
> [1,] 2020 8 1 12 0 0
> [2,] 2020 8 1 12 0 0
> [3,] 2020 8 1 12 0 0
> [4,] 2020 8 1 12 0 0
> [5,] 2020 8 1 12 0 0
> > simplify2array(parallel::mclapply(c(
> + lubridate::year,
> + lubridate::month,
> + lubridate::day,
> + lubridate::hour), function(FUN, x) {
> + FUN(x)
> + }, x=dat$forecast.date))
> [,1] [,2] [,3] [,4]
> [1,] 2020 8 1 12
> [2,] 2020 8 1 12
> [3,] 2020 8 1 12
> [4,] 2020 8 1 12
> [5,] 2020 8 1 12
>
> V
>
> r
>
> ______________________________________________
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