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

[Bert Gunter]

"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
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> >> 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
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.