[R] Parsing a Date

[Rui Barradas]

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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