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

Jeff Newmiller jdnewm|| @end|ng |rom dcn@d@v|@@c@@u@
Sat Nov 20 19:40:56 CET 2021


Beat me to it! But it is also worth noting that once converted to Date or POSIXct, timestamps should be treated as data without regard to how that data is displayed. When you choose to output that data you will have options as to the display format associated with the function you are using for output.

My take:

dta <- read.table( text=
"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
", header=TRUE, check.names=FALSE, as.is=TRUE)

dta

library(dplyr)
library(tidyr)

dt_fmt <- "%d/%m/%Y"

dta_long <- (   dta
            %>% pivot_longer( cols = -Buffer
                            , names_to = "dt_chr"
                            , values_to = "LST"
                            )
            %>% mutate( dt_date = as.Date( dt_chr, format = dt_fmt )
                      , dt_POSIXct = as.POSIXct( dt_chr, format = dt_fmt, tz = "Etc/GMT+8" )
                      )
            )

dta_long

On November 20, 2021 10:01:56 AM PST, 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.
>>
>
>	[[alternative HTML version deleted]]
>
>______________________________________________
>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.

-- 
Sent from my phone. Please excuse my brevity.



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