Note:
> levels(factor(c(0,0,1))) ## just gives you the levels attribute
[1] "0" "1"
> as.character(factor(c(0,0,1))) ## gives you the level of each value in
the vector
[1] "0" "0" "1"
Does that answer your question or have I misunderstood.
Cheers,
Bert
On Tue, Apr 2, 2024 at 12:00 AM Kimmo Elo <kimmo....@utu.fi> wrote:
> Hi,
>
> why would this simple procedure not work?
>
> --- snip ---
> mydf <- data.frame(id_station = 1234, string_data = c(2024, 12, 1, 0, 0),
> rainfall_value= 55)
>
> mydf$string_data <- as.factor(mydf$string_data)
>
> values<-as.integer(levels(mydf$string_data))
>
> for (i in 1:length(values)) {
> assign(paste("VAR_", i, sep=""), values[i])
> }
>
> --- snip ---
>
> Best,
>
> Kimmo
>
> to, 2024-03-28 kello 14:17 +0000, Ebert,Timothy Aaron kirjoitti:
> > Here are some pieces of working code. I assume you want the second one or
> > the third one that is functionally the same but all in one statement. I
> > do not understand why it is a factor, but I will assume that there is a
> > current and future reason for that. This means I cannot alter the
> > string_data variable, or you can simplify by not making the variable a
> > factor only to turn it back into character.
> >
> > mydf <- data.frame(id_station = 1234, string_data = c(2024, 12, 1, 0, 0),
> > rainfall_value= 55)
> > mydf$string_data <- as.factor(mydf$string_data)
> >
> > mydf <- data.frame(id_station = 1234, string_data = "2024, 12, 1, 0, 0",
> > rainfall_value= 55)
> > mydf$string_data <- as.factor(mydf$string_data)
> >
> > mydf <- data.frame(id_station = 1234, string_data = as.factor("2024, 12,
> > 1, 0, 0"), rainfall_value= 55)
> >
> > mydf <- data.frame(id_station = 1234, string_data = as.factor("2024, 12,
> > 1, 0, 0"), rainfall_value= 55)
> > mydf$string_data2 <- as.character(mydf$string_data)
> >
> > #I assume there are many records in the data frame and your example is
> > for demonstration only.
> > #I cannot assume that all records are the same, though you may be able to
> > simplify if that is true.
> > #Split the string based on commas.
> > split_values <- strsplit(mydf$string_data2, ",")
> >
> > # find the maximum string length
> > max_length <- max(lengths(split_values))
> >
> > # Add new variables to the data frame
> > for (i in 1:max_length) {
> > new_var_name <- paste0("VAR_", i)
> > mydf[[new_var_name]] <- sapply(split_values, function(x)
> > ifelse(length(x) >= i, x[i], NA))
> > }
> >
> > # Convert to numeric
> > for (i in 1:max_length) {
> > new_var_name <- paste0("VAR_", i)
> > mydf[[new_var_name]] <- as.numeric(mydf[[new_var_name]])
> > }
> > # remove trash
> > mydf <- mydf[,-4]
> > # Provide more useful names
> > colnames(mydf) <- c("id_station", "string_data", "rainfall_mm", "Year",
> > "Month", "Day", "hour", "minute")
> >
> > Regards,
> > Tim
> >
> > -----Original Message-----
> > From: R-help <r-help-boun...@r-project.org> On Behalf Of Stefano Sofia
> > Sent: Thursday, March 28, 2024 7:48 AM
> > To: Fabio D'Agostino <dagostinof...@gmail.com>; r-help@R-project.org
> > Subject: Re: [R] split a factor into single elements
> >
> > [External Email]
> >
> > Sorry for my hurry.
> >
> > The correct reproducible code is different from the initial one. The
> > correct example is
> >
> >
> > mydf <- data.frame(id_station = 1234, string_data = as.factor(2024, 12,
> > 1, 0, 0), rainfall_value= 55)
> >
> >
> > In this case mydf$string_data is a factor, but of length 1 (and not 5
> > like in the initial example).
> >
> > Therefore the suggestion offered by Fabio does not work.
> >
> >
> > Any suggestion?
> >
> > Sorry again for my mistake
> >
> > Stefano
> >
> >
> >
> > (oo)
> > --oOO--( )--OOo--------------------------------------
> > Stefano Sofia PhD
> > Civil Protection - Marche Region - Italy Meteo Section Snow Section Via
> > del Colle Ameno 5
> > 60126 Torrette di Ancona, Ancona (AN)
> > Uff: +39 071 806 7743
> > E-mail: stefano.so...@regione.marche.it
> > ---Oo---------oO----------------------------------------
> >
> >
> > ________________________________
> > Da: Fabio D'Agostino <dagostinof...@gmail.com>
> > Inviato: gioved 28 marzo 2024 12:20
> > A: Stefano Sofia; r-help@R-project.org
> > Oggetto: Re: [R] split a factor into single elements
> >
> >
> > Non si ricevono spesso messaggi di posta elettronica da
> > dagostinof...@gmail.com. Informazioni sul perch
> > importante<https://aka.ms/LearnAboutSenderIdentification>
> >
> > Hi Stefano,
> > maybe something like this can help you?
> >
> > myfactor <- as.factor(c(2024, 2, 1, 0, 0))
> >
> > # Convert factor values to integers
> > first_element <- as.integer(as.character(myfactor)[1])
> > second_element <- as.integer(as.character(myfactor)[2])
> > third_element <- as.integer(as.character(myfactor)[3])
> >
> > # Print the results
> > first_element
> > [1] 2024
> > second_element
> > [1] 2
> > third_element
> > [1] 1
> >
> > # Check the type of the object
> > typeof(first_element)
> > [1] "integer"
> >
> > Fabio
> >
> > Il giorno gio 28 mar 2024 alle ore 11:29 Stefano Sofia
> > <stefano.so...@regione.marche.it<mailto:stefano.so...@regione.marche.it
> >>
> > ha scritto:
> > Dear R-list users,
> >
> > forgive me for this silly question, I did my best to find a solution with
> > no success.
> >
> > Suppose I have a factor type like
> >
> >
> > myfactor <- as.factor(2024, 2, 1, 0, 0)
> >
> >
> > There are no characters (and therefore strsplit for eample does not
> > work).
> >
> > I need to store separately the 1st, 2nd and 3rd elements as integers. How
> > can I do?
> >
> >
> > Thank you for your help
> >
> > Stefano
> >
> >
> > (oo)
> > --oOO--( )--OOo--------------------------------------
> > Stefano Sofia PhD
> > Civil Protection - Marche Region - Italy Meteo Section Snow Section Via
> > del Colle Ameno 5
> > 60126 Torrette di Ancona, Ancona (AN)
> > Uff: +39 071 806 7743
> > E-mail:
> > stefano.so...@regione.marche.it<mailto:stefano.so...@regione.marche.it>
> > ---Oo---------oO----------------------------------------
> >
> > ________________________________
> >
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