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Philip

2020-Aug-02 16:24 UTC

[R] Parsing a Date

Below is some Weather Service data.  I would like to parse the forecast date
field into four different columns:

    Year
    Month
    Day
    Hour

I would like to drop the final four zeros.  Any suggestions?

forecast.date                 levels      lon           lat         HGT      RH 
TMP       UGRD    VGRD
1 2020-08-01 12:00:00 1000 mb -113.130 33.6335 75.5519 49.6484 305.495 1.40155
2.23264
2 2020-08-01 12:00:00 1000 mb -113.111 33.5142 75.9582 51.0234 305.245 1.65155
2.23264
3 2020-08-01 12:00:00 1000 mb -113.092 33.3948 76.3957 52.7734 305.057 1.90155
2.23264
4 2020-08-01 12:00:00 1000 mb -112.987 33.6495 75.9269 49.1484 305.745 1.90155
2.04514
5 2020-08-01 12:00:00 1000 mb -112.968 33.5301 76.3019 50.2734 305.495 2.08905
1.98264

Philip Heinrich



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Eric Berger

2020-Aug-02 16:33 UTC

head link

[R] Parsing a Date

If the forecast.date column is of type character you can use lubridate to
do this:
> library(lubridate)
> a <- "2020-08-01 12:00:00"
> year(a)
# [1] 2020> month(a)# [1] 8

etc


On Sun, Aug 2, 2020 at 7:24 PM Philip <herd_dog at cox.net> wrote:
> Below is some Weather Service data.  I would like to parse the forecast
> date field into four different columns:
>
>     Year
>     Month
>     Day
>     Hour
>
> I would like to drop the final four zeros.  Any suggestions?
>
> forecast.date                 levels      lon           lat         HGT
>   RH          TMP       UGRD    VGRD
> 1 2020-08-01 12:00:00 1000 mb -113.130 33.6335 75.5519 49.6484 305.495
> 1.40155 2.23264
> 2 2020-08-01 12:00:00 1000 mb -113.111 33.5142 75.9582 51.0234 305.245
> 1.65155 2.23264
> 3 2020-08-01 12:00:00 1000 mb -113.092 33.3948 76.3957 52.7734 305.057
> 1.90155 2.23264
> 4 2020-08-01 12:00:00 1000 mb -112.987 33.6495 75.9269 49.1484 305.745
> 1.90155 2.04514
> 5 2020-08-01 12:00:00 1000 mb -112.968 33.5301 76.3019 50.2734 305.495
> 2.08905 1.98264
>
> Philip Heinrich
>
>
>
>         [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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Jeff Newmiller

2020-Aug-02 16:56 UTC

head link

[R] Parsing a Date

Learn to post plain text and use dput to include data:

dta <- 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" ), levels =
c("1000 mb", "1000 mb", "1000 mb", "1000
mb", "1000 mb" ), lon = c(-113.13, -113.111, -113.092, -112.987,
-112.968), lat = c(33.6335, 33.5142, 33.3948, 33.6495, 33.5301), HGT =
c(75.5519, 75.9582, 76.3957, 75.9269, 76.3019), RH = c(49.6484, 51.0234,
52.7734, 49.1484, 50.2734), TMP = c(305.495, 305.245, 305.057, 305.745,
305.495), UGRD = c(1.40155, 1.65155, 1.90155, 1.90155, 2.08905), VGRD =
c(2.23264, 2.23264, 2.23264, 2.04514, 1.98264 )), .Names =
c("forecast.date", "levels", "lon",
"lat", "HGT", "RH", "TMP",
"UGRD", "VGRD"), class = "data.frame", row.names =
c(NA, -5L))

dta$Year <- as.integer( substr( dta$forecast.date, 1, 4 ) )
dta$Month <- as.integer( substr( dta$forecast.date, 6, 7 ) )
dta$Day <- as.integer( substr( dta$forecast.date, 9, 10 ) )
dta$Hour <- as.integer( substr( dta$forecast.date, 12, 13 ) )
dta


On August 2, 2020 9:24:24 AM PDT, Philip <herd_dog at cox.net>
wrote:>Below is some Weather Service data.  I would like to parse the forecast
>date field into four different columns:
>
>    Year
>    Month
>    Day
>    Hour
>
>I would like to drop the final four zeros.  Any suggestions?
>
>forecast.date                 levels      lon           lat         HGT
>     RH          TMP       UGRD    VGRD
>1 2020-08-01 12:00:00 1000 mb -113.130 33.6335 75.5519 49.6484 305.495
>1.40155 2.23264
>2 2020-08-01 12:00:00 1000 mb -113.111 33.5142 75.9582 51.0234 305.245
>1.65155 2.23264
>3 2020-08-01 12:00:00 1000 mb -113.092 33.3948 76.3957 52.7734 305.057
>1.90155 2.23264
>4 2020-08-01 12:00:00 1000 mb -112.987 33.6495 75.9269 49.1484 305.745
>1.90155 2.04514
>5 2020-08-01 12:00:00 1000 mb -112.968 33.5301 76.3019 50.2734 305.495
>2.08905 1.98264
>
>Philip Heinrich
>
>
>
>	[[alternative HTML version deleted]]
>
>______________________________________________
>R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
-- 
Sent from my phone. Please excuse my brevity.

Rasmus Liland

2020-Aug-02 17:26 UTC

head link

[R] Parsing a Date

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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Rui Barradas

2020-Aug-02 21:54 UTC

head link

[R] Parsing a Date

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
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.


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R help - Aug 2020 - Parsing a Date

[R] Parsing a Date

[R] Parsing a Date

[R] Parsing a Date

[R] Parsing a Date

[R] Parsing a Date