To be clear, I take no credit for the rather extraordinary function cll shown
below:
mutate(Date = lubridate::dmy_hm(Date))
I would pretty much never have constructed such an interesting and rather
unnecessary line of code.
ALL the work is done within the parentheses:
Date = lubridate::dmy_hm(Date))
The above creates a tibble and assigns it to the name of Date. but first it
takes a vector called Date containing text and uses a function to parse it and
return a form more suitable to use as a date.
So what does mutate() do when it is called with a tibble and asked to do
nothing? I mean it sees something more like this:
Mutate(.data=Date)
What is missing are the usual assignment statements to modify existing columns
or make new ones. So it does nothing and exists without complaint.
My suggestion was more like the following which uses mutate. For illustration,
I will not use the name "Date" repeatedly and make new columns and uses an
original_date as the source but note I am not saying the lubridate used will
work as I do not see how it handles the initial part of the string containing
an index number.
old_tibble <- tibble(useless=original_date)
new_tibble <- mutate(old_tibble, useful = lubridate::dmy_hm(Date))
The above can be more compact by making the tibble directly in the first
argument, and it can be done using old or new pipelines.
The reason the suggested way worked is because it used the vectorized methods
of base-R and I mentioned there was no reason you must use dplyr for this or
many other things especially when it is simple.
Now if you wanted to make multiple new columns containing character or integer
versions of the month and other parts of the date or even calculate what day of
the week that full date was in that year, then mutate can be very useful as you
can keep adding requests to make a new column using all old and new columns
already specified.
Sometimes when code works, we don't look to see if it works inadvertently, LOL!
-----Original Message-----
From: R-help <r-help-boun...@r-project.org> On Behalf Of Dr Eberhard W Lisse
Sent: Wednesday, July 13, 2022 5:49 PM
To: r-help@r-project.org
Subject: Re: [R] How to parse a really silly date with lubridate
Bui,
thanks, this what Avi suggested in an email to me as well and works.
It's so easy if you know it :-)-O
el
On 2022-07-13 23:40 , Rui Barradas wrote:
> 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
[...]
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