[R] Date read correctly from CSV, then reformatted incorrectly by R

[Philip Monk]

I am.  Long day, poorly small children!

P

On Sat, 20 Nov 2021, 21:08 Bert Gunter, <bgunter.4567 using gmail.com> wrote:

> "I also know that '/' is a special character in R (if that's the right
> term) "
>
> That is false. I think you are confusing "/" with "\", which is R's
> *escape* character.
>
> > cat("a/nb")
> a/nb
> > cat("a\nb")
> a
> b
>
> It gets confusing especially in regex's, because "\" is used in regex
> syntax also.
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
> On Sat, Nov 20, 2021 at 12:23 PM Philip Monk <prmonk using gmail.com> wrote:
> >
> > Thanks, Andrew.  I didn't realise as.Date *only* read two formats, I
> > think I was tripped up by using %y instead of %Y, though I also know
> > that '/' is a special character in R (if that's the right term) and as
> > such know there is special syntax to use (which I don't know).
> >
> > On Sat, 20 Nov 2021 at 18:02, Andrew Simmons <akwsimmo using gmail.com> wrote:
> > >
> > > The as.Date function for a character class argument will try reading
> in two formats (%Y-%m-%d and %Y/%m/%d).
> > >
> > >
> > > This does not look like the format you have provided, which is why it
> doesn't work. Try something like:
> > >
> > >
> > > x <- c("28/10/2016", "19/11/2016", "31/12/2016", "16/01/2016",
> "05/03/2017")
> > > as.Date(x, format = "%d/%m/%Y")
> > >
> > >
> > > which produces this output:
> > >
> > >
> > > > x <- c("28/10/2016", "19/11/2016", "31/12/2016", "16/01/2016",
> "05/03/2017")
> > > > as.Date(x, format = "%d/%m/%Y")
> > > [1] "2016-10-28" "2016-11-19" "2016-12-31" "2016-01-16" "2017-03-05"
> > > >
> > >
> > >
> > > much better than before! I hope this helps
> > >
> > > On Sat, Nov 20, 2021 at 12:49 PM Philip Monk <prmonk using gmail.com> wrote:
> > >>
> > >> Thanks Eric & Jeff.
> > >>
> > >> I'll certainly read up on lubridate, and the posting guide (again)
> > >> (this should be in plain text).
> > >>
> > >> CSV extract below...
> > >>
> > >> Philip
> > >>
> > >> Buffer    28/10/2016    19/11/2016    31/12/2016    16/01/2017
> 05/03/2017
> > >> 100    2.437110889    -8.69674895    3.239299816    2.443183304
> 2.346743827
> > >> 200    2.524329899    -7.688862068    3.386811734    2.680347706
> 2.253885237
> > >> 300    2.100784256    -8.059855835    3.143786507    2.615152896
> 2.015645973
> > >> 400    1.985608385    -10.6707206    2.894572791    2.591925038
> 2.057913137
> > >> 500    1.824982163    -9.122519736    2.560350727    2.372226799
> 1.995863839
> > >>
> > >>
> > >> On Sat, 20 Nov 2021 at 17:08, Philip Monk <prmonk using gmail.com> wrote:
> > >> >
> > >> > Hello,
> > >> >
> > >> > Simple but infuriating problem.
> > >> >
> > >> > Reading in CSV of data using :
> > >> >
> > >> > ```
> > >> > # CSV file has column headers with date of scene capture in format
> dd/mm/yyyy
> > >> > # check.names = FALSE averts R incorrectly processing dates due to
> '/'
> > >> > data <- read.csv("C:/R_data/Bungala (b2000) julian.csv",
> check.names = FALSE)
> > >> >
> > >> > # Converts data table from wide (many columns) to long (many rows)
> and creates the new object 'data_long'
> > >> > # Column 1 is the 'Buffer' number (100-2000), Columns 2-25 contain
> monthly data covering 2 years (the header row being the date, and rows 2-21
> being a value for each buffer).
> > >> > # Column headers for columns 2:25 are mutated into a column called
> 'Date', values for each buffer and each date into the column 'LST'
> > >> > data_long <- data %>% pivot_longer(cols = 2:25, names_to = "Date",
> values_to = "LST")
> > >> >
> > >> > # Instructs R to treat the 'Date' column data as a date
> > >> > data_long$Date <- as.Date(data_long$Date)
> > >> > ```
> > >> >
> > >> > Using str(data), I can see that R has correctly read the dates in
> the format %d/%m/%y (e.g. 15/12/2015) though has the data type as chr.
> > >> >
> > >> > Once changing the type to 'Date', however, the date is
> reconfigured.  For instance, 15/01/2010 (15 January 2010), becomes
> 0015-01-20.
> > >> >
> > >> > I've tried ```data_long$Date <- as.Date(data_long$Date, format =
> "%d/%m.%y")```, and also ```tryformat c("%d/%m%y")```, but either the error
> persists or I get ```NA```.
> > >> >
> > >> > How do I make R change Date from 'chr' to 'date' without it going
> wrong?
> > >> >
> > >> > Suggestions/hints/solutions would be most welcome.  :)
> > >> >
> > >> > Thanks for your time,
> > >> >
> > >> > Philip
> > >> >
> > >> > Part-time PhD Student (Environmental Science)
> > >> > Lancaster University, UK.
> > >> >
> > >> > ~~~~~
> > >> >
> > >> > I asked a question a few weeks ago and put together the answer I
> needed from the responses but didn't know how to say thanks on this list.
> So, thanks Andrew Simmons, Bert Gunter, Jeff Newmiller and Daniel Nordlund!
> > >>
> > >> ______________________________________________
> > >> R-help using 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 using 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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