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Dr Eberhard W Lisse

2022-Jul-13 16:14 UTC

[R] How to parse a really silly date with lubridate

Hi,

while all of the below work in a character vector, none works in
the tibble.

The following

	DDATA %>%
		add_column(as.tibble(lubridate::dmy_hm(DDATA$Date)),
			.before = "Period") %>%
		rename(NewDate=value) %>%
		select(Date,NewDate) %>%
		filter(between(as.Date(NewDate),as.Date('2022-07-09'),
			as.Date('2022-07-10')))

does work

	# A tibble: 3 ? 2
	  Date                  NewDate
	  <chr>                 <dttm>
	1 9. Jul 2022 at 11:39  2022-07-09 11:39:00
	2 10. Jul 2022 at 01:58 2022-07-10 01:58:00
	3 10. Jul 2022 at 11:26 2022-07-10 11:26:00

but I wonder if that can not be done more elegantly, ie by direct
replacements in the column.

greetings, el

On 2022-07-13 16:48 , Rui Barradas wrote:
[...]
 > d <- c('9. Jul 2022 at 11:39', '10. Jul 2022 at 01:58')
 > lubridate::dmy_hm(d)
 > #> [1] "2022-07-09 11:39:00 UTC" "2022-07-10 01:58:00
UTC"
[...]

On 2022-07-13 16:03 , Ben Tupper wrote:
[...]
 > s = c("9. Jul 2022 at 11:39", "10. Jul 2022 at 01:58")
 > as.POSIXct(s, format = "%d. %b %Y at %H:%M")
 > as.POSIXct(s, format = "%d. %b %Y at %H:%M", tz =
"UTC")
[...]

On 2022-07-13 15:52 , Ivan Krylov wrote:
[...]
 > Sys.setlocale('LC_TIME', 'C')
 > strptime(
 >   c('9. Jul 2022 at 11:39', '10. Jul 2022 at 01:58'),
 >   '%d. %b %Y at %H:%M'
 > )
[...]


 > ?s 14:40 de 13/07/2022, Dr Eberhard Lisse escreveu:
 >>
 >> Hi,
 >>
 >> I have data file which generated by an otherwise very nice (diabetes
 >> log) app, but exports dates really silly.
 >>
 >> After reading the enclosed mwe.csv into R like so
 >>
 >>       MWE <- read_delim('mwe.csv', delim = ';')
%>%
 >>          select(Date) %>%
 >>          print()
 >>
 >>
 >> this comes out as:
 >>
 >>       # A tibble: 2 ? 1
 >>      Date
 >>      <chr>
 >>       1 9. Jul 2022 at 11:39
 >>       2 10. Jul 2022 at 01:58
 >>
 >>
 >> No matter what I try I am not able to parse this inside R to get at
 >> proper dates (I have loaded tidyverse and lubridate).
 >>
 >> I can easily do somethig
 >>
 >>       csvq  -d ';' -t '%e. %b %Y at %H:%i' \
 >>          'SELECT Date as oridate,
 >>              DATETIME_FORMAT(Date, "%Y-%m-%d %H:%m") AS date
 >>          FROM mwe'
 >>
 >>       +-----------------------+------------------+
 >>       |        oridate        |       date       |
 >>       +-----------------------+------------------+
 >>       | 9. Jul 2022 at 11:39  | 2022-07-09 11:07 |
 >>       | 10. Jul 2022 at 01:58 | 2022-07-10 01:07 |
 >>       +-----------------------+------------------+
 >>
 >> and hence could easily do something like
 >>
 >>       csvq  -d ';' -t '%e. %b %Y at %H:%i' \
 >>        'ALTER mwe
 >>        SET Date = DATETIME_FORMAT(Date, "%Y-%m-%d
%H:%m")'
 >>
 >> but would rather like to be able to do it inside R and would therefor
 >> appreciate any advice in this regard.
 >>
 >>
 >> greetings, el
 >>
 >

Rui Barradas

2022-Jul-13 21:40 UTC

head link

[R] How to parse a really silly date with lubridate

Hello,

Are you looking for mutate? In the example below I haven't included the 
filter, since the tibble only has 2 rows. But the date column is coerced 
to an actual datetime class in place, without the need for NewDate.

suppressPackageStartupMessages({
   library(tibble)
   library(dplyr)
})

DDATA <- tibble(Date = c('9. Jul 2022 at 11:39', '10. Jul 2022 at
01:58'))

DDATA %>%
   mutate(Date = lubridate::dmy_hm(Date))
#> # A tibble: 2 ? 1
#>   Date
#>   <dttm>
#> 1 2022-07-09 11:39:00
#> 2 2022-07-10 01:58:00


Hope this helps,

Rui Barradas


?s 17:14 de 13/07/2022, Dr Eberhard W Lisse escreveu:> 
> Hi,
> 
> while all of the below work in a character vector, none works in
> the tibble.
> 
> The following
> 
>  ????DDATA %>%
>  ??????? add_column(as.tibble(lubridate::dmy_hm(DDATA$Date)),
>  ??????????? .before = "Period") %>%
>  ??????? rename(NewDate=value) %>%
>  ??????? select(Date,NewDate) %>%
>  ??????? filter(between(as.Date(NewDate),as.Date('2022-07-09'),
>  ??????????? as.Date('2022-07-10')))
> 
> does work
> 
>  ????# A tibble: 3 ? 2
>  ????? Date????????????????? NewDate
>  ????? <chr>???????????????? <dttm>
>  ????1 9. Jul 2022 at 11:39? 2022-07-09 11:39:00
>  ????2 10. Jul 2022 at 01:58 2022-07-10 01:58:00
>  ????3 10. Jul 2022 at 11:26 2022-07-10 11:26:00
> 
> but I wonder if that can not be done more elegantly, ie by direct
> replacements in the column.
> 
> greetings, el
> 
> On 2022-07-13 16:48 , Rui Barradas wrote:
> [...]
>  > d <- c('9. Jul 2022 at 11:39', '10. Jul 2022 at
01:58')
>  > lubridate::dmy_hm(d)
>  > #> [1] "2022-07-09 11:39:00 UTC" "2022-07-10
01:58:00 UTC"
> [...]
> 
> On 2022-07-13 16:03 , Ben Tupper wrote:
> [...]
>  > s = c("9. Jul 2022 at 11:39", "10. Jul 2022 at
01:58")
>  > as.POSIXct(s, format = "%d. %b %Y at %H:%M")
>  > as.POSIXct(s, format = "%d. %b %Y at %H:%M", tz =
"UTC")
> [...]
> 
> On 2022-07-13 15:52 , Ivan Krylov wrote:
> [...]
>  > Sys.setlocale('LC_TIME', 'C')
>  > strptime(
>  >?? c('9. Jul 2022 at 11:39', '10. Jul 2022 at 01:58'),
>  >?? '%d. %b %Y at %H:%M'
>  > )
> [...]
> 
> 
>  > ?s 14:40 de 13/07/2022, Dr Eberhard Lisse escreveu:
>  >>
>  >> Hi,
>  >>
>  >> I have data file which generated by an otherwise very nice
(diabetes
>  >> log) app, but exports dates really silly.
>  >>
>  >> After reading the enclosed mwe.csv into R like so
>  >>
>  >>?????? MWE <- read_delim('mwe.csv', delim =
';') %>%
>  >>????????? select(Date) %>%
>  >>????????? print()
>  >>
>  >>
>  >> this comes out as:
>  >>
>  >>?????? # A tibble: 2 ? 1
>  >>????? Date
>  >>????? <chr>
>  >>?????? 1 9. Jul 2022 at 11:39
>  >>?????? 2 10. Jul 2022 at 01:58
>  >>
>  >>
>  >> No matter what I try I am not able to parse this inside R to get
at
>  >> proper dates (I have loaded tidyverse and lubridate).
>  >>
>  >> I can easily do somethig
>  >>
>  >>?????? csvq? -d ';' -t '%e. %b %Y at %H:%i' \
>  >>????????? 'SELECT Date as oridate,
>  >>????????????? DATETIME_FORMAT(Date, "%Y-%m-%d %H:%m") AS
date
>  >>????????? FROM mwe'
>  >>
>  >>?????? +-----------------------+------------------+
>  >>?????? |??????? oridate??????? |?????? date?????? |
>  >>?????? +-----------------------+------------------+
>  >>?????? | 9. Jul 2022 at 11:39? | 2022-07-09 11:07 |
>  >>?????? | 10. Jul 2022 at 01:58 | 2022-07-10 01:07 |
>  >>?????? +-----------------------+------------------+
>  >>
>  >> and hence could easily do something like
>  >>
>  >>?????? csvq? -d ';' -t '%e. %b %Y at %H:%i' \
>  >>??????? 'ALTER mwe
>  >>??????? SET Date = DATETIME_FORMAT(Date, "%Y-%m-%d
%H:%m")'
>  >>
>  >> but would rather like to be able to do it inside R and would
therefor
>  >> appreciate any advice in this regard.
>  >>
>  >>
>  >> greetings, el
>  >>
>  >
> 
> ______________________________________________
> 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.

R help - Jul 2022 - How to parse a really silly date with lubridate

[R] How to parse a really silly date with lubridate

[R] How to parse a really silly date with lubridate