[R] Problem with filling dataframe's column
javad bayat
j@b@y@t194 @end|ng |rom gm@||@com
Tue Jun 13 21:46:51 CEST 2023
Dear all;
I used these codes and I get what I wanted.
Sincerely
pat = c("Level 12","Level 22","0")
data3 = data2[-which(data2$Layer == pat),]
dim(data2)
[1] 281549 9
dim(data3)
[1] 244075 9
On Tue, Jun 13, 2023 at 11:36 AM Eric Berger <ericjberger using gmail.com> wrote:
> Hi Javed,
> grep returns the positions of the matches. See an example below.
>
> > v <- c("abc", "bcd", "def")
> > v
> [1] "abc" "bcd" "def"
> > grep("cd",v)
> [1] 2
> > w <- v[-grep("cd",v)]
> > w
> [1] "abc" "def"
> >
>
>
> On Tue, Jun 13, 2023 at 8:50 AM javad bayat <j.bayat194 using gmail.com> wrote:
> >
> > Dear Rui;
> > Hi. I used your codes, but it seems it didn't work for me.
> >
> > > pat <- c("_esmdes|_Des Section|0")
> > > dim(data2)
> > [1] 281549 9
> > > grep(pat, data2$Layer)
> > > dim(data2)
> > [1] 281549 9
> >
> > What does grep function do? I expected the function to remove 3 rows of
> the
> > dataframe.
> > I do not know the reason.
> >
> >
> >
> >
> >
> >
> > On Mon, Jun 12, 2023 at 5:16 PM Rui Barradas <ruipbarradas using sapo.pt>
> wrote:
> >
> > > Às 23:13 de 12/06/2023, javad bayat escreveu:
> > > > Dear Rui;
> > > > Many thanks for the email. I tried your codes and found that the
> length
> > > of
> > > > the "Values" and "Names" vectors must be equal, otherwise the results
> > > will
> > > > not be useful.
> > > > For some of the characters in the Layer column that I do not need to
> be
> > > > filled in the LU column, I used "NA".
> > > > But I need to delete some of the rows from the table as they are
> useless
> > > > for me. I tried this code to delete entire rows of the dataframe
> which
> > > > contained these three value in the Layer column: It gave me the
> following
> > > > error.
> > > >
> > > >> data3 = data2[-grep(c("_esmdes","_Des Section","0"), data2$Layer),]
> > > > Warning message:
> > > > In grep(c("_esmdes", "_Des Section", "0"), data2$Layer) :
> > > > argument 'pattern' has length > 1 and only the first element
> will
> > > be
> > > > used
> > > >
> > > >> data3 = data2[!grepl(c("_esmdes","_Des Section","0"), data2$Layer),]
> > > > Warning message:
> > > > In grepl(c("_esmdes", "_Des Section", "0"), data2$Layer) :
> > > > argument 'pattern' has length > 1 and only the first element
> will be
> > > > used
> > > >
> > > > How can I do this?
> > > > Sincerely
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > On Sun, Jun 11, 2023 at 5:03 PM Rui Barradas <ruipbarradas using sapo.pt>
> > > wrote:
> > > >
> > > >> Às 13:18 de 11/06/2023, Rui Barradas escreveu:
> > > >>> Às 22:54 de 11/06/2023, javad bayat escreveu:
> > > >>>> Dear Rui;
> > > >>>> Many thanks for your email. I used one of your codes,
> > > >>>> "data2$LU[which(data2$Layer == "Level 12")] <- "Park"", and it
> works
> > > >>>> correctly for me.
> > > >>>> Actually I need to expand the codes so as to consider all
> "Levels" in
> > > >> the
> > > >>>> "Layer" column. There are more than hundred levels in the Layer
> > > column.
> > > >>>> If I use your provided code, I have to write it hundred of time as
> > > >> below:
> > > >>>> data2$LU[which(data2$Layer == "Level 1")] <- "Park";
> > > >>>> data2$LU[which(data2$Layer == "Level 2")] <- "Agri";
> > > >>>> ...
> > > >>>> ...
> > > >>>> ...
> > > >>>> .
> > > >>>> Is there any other way to expand the code in order to consider
> all of
> > > >> the
> > > >>>> levels simultaneously? Like the below code:
> > > >>>> data2$LU[which(data2$Layer == c("Level 1","Level 2", "Level 3",
> ...))]
> > > >> <-
> > > >>>> c("Park", "Agri", "GS", ...)
> > > >>>>
> > > >>>>
> > > >>>> Sincerely
> > > >>>>
> > > >>>>
> > > >>>>
> > > >>>>
> > > >>>> On Sun, Jun 11, 2023 at 1:43 PM Rui Barradas <
> ruipbarradas using sapo.pt>
> > > >>>> wrote:
> > > >>>>
> > > >>>>> Às 21:05 de 11/06/2023, javad bayat escreveu:
> > > >>>>>> Dear R users;
> > > >>>>>> I am trying to fill a column based on a specific value in
> another
> > > >>>>>> column
> > > >>>>> of
> > > >>>>>> a dataframe, but it seems there is a problem with the codes!
> > > >>>>>> The "Layer" and the "LU" are two different columns of the
> dataframe.
> > > >>>>>> How can I fix this?
> > > >>>>>> Sincerely
> > > >>>>>>
> > > >>>>>>
> > > >>>>>> for (i in 1:nrow(data2$Layer)){
> > > >>>>>> if (data2$Layer == "Level 12") {
> > > >>>>>> data2$LU == "Park"
> > > >>>>>> }
> > > >>>>>> }
> > > >>>>>>
> > > >>>>>>
> > > >>>>>>
> > > >>>>>>
> > > >>>>> Hello,
> > > >>>>>
> > > >>>>> There are two bugs in your code,
> > > >>>>>
> > > >>>>> 1) the index i is not used in the loop
> > > >>>>> 2) the assignment operator is `<-`, not `==`
> > > >>>>>
> > > >>>>>
> > > >>>>> Here is the loop corrected.
> > > >>>>>
> > > >>>>> for (i in 1:nrow(data2$Layer)){
> > > >>>>> if (data2$Layer[i] == "Level 12") {
> > > >>>>> data2$LU[i] <- "Park"
> > > >>>>> }
> > > >>>>> }
> > > >>>>>
> > > >>>>>
> > > >>>>>
> > > >>>>> But R is a vectorized language, the following two ways are the
> > > idiomac
> > > >>>>> ways of doing what you want to do.
> > > >>>>>
> > > >>>>>
> > > >>>>>
> > > >>>>> i <- data2$Layer == "Level 12"
> > > >>>>> data2$LU[i] <- "Park"
> > > >>>>>
> > > >>>>> # equivalent one-liner
> > > >>>>> data2$LU[data2$Layer == "Level 12"] <- "Park"
> > > >>>>>
> > > >>>>>
> > > >>>>>
> > > >>>>> If there are NA's in data2$Layer it's probably safer to use
> ?which()
> > > in
> > > >>>>> the logical index, to have a numeric one.
> > > >>>>>
> > > >>>>>
> > > >>>>>
> > > >>>>> i <- which(data2$Layer == "Level 12")
> > > >>>>> data2$LU[i] <- "Park"
> > > >>>>>
> > > >>>>> # equivalent one-liner
> > > >>>>> data2$LU[which(data2$Layer == "Level 12")] <- "Park"
> > > >>>>>
> > > >>>>>
> > > >>>>> Hope this helps,
> > > >>>>>
> > > >>>>> Rui Barradas
> > > >>>>>
> > > >>>>
> > > >>>>
> > > >>> Hello,
> > > >>>
> > > >>> You don't need to repeat the same instruction 100+ times, there is
> a
> > > way
> > > >>> of assigning all new LU values at the same time with match().
> > > >>> This assumes that you have the new values in a vector.
> > > >>
> > > >> Sorry, this is not clear. I mean
> > > >>
> > > >>
> > > >> This assumes that you have the new values in a vector, the vector
> Names
> > > >> below. The vector of values to be matched is created from the data.
> > > >>
> > > >>
> > > >> Rui Barradas
> > > >>
> > > >>>
> > > >>>
> > > >>> Values <- sort(unique(data2$Layer))
> > > >>> Names <- c("Park", "Agri", "GS")
> > > >>>
> > > >>> i <- match(data2$Layer, Values)
> > > >>> data2$LU <- Names[i]
> > > >>>
> > > >>>
> > > >>> Hope this helps,
> > > >>>
> > > >>> Rui Barradas
> > > >>>
> > > >>> ______________________________________________
> > > >>> 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.
> > > >>
> > > >>
> > > >
> > > Hello,
> > >
> > > Please cc the r-help list, R-Help is threaded and this can in the
> future
> > > be helpful to others.
> > >
> > > You can combine several patters like this:
> > >
> > >
> > > pat <- c("_esmdes|_Des Section|0")
> > > grep(pat, data2$Layer)
> > >
> > > or, programatically,
> > >
> > >
> > > pat <- paste(c("_esmdes","_Des Section","0"), collapse = "|")
> > >
> > >
> > > Hope this helps,
> > >
> > > Rui Barradas
> > >
> > >
> >
> > --
> > Best Regards
> > Javad Bayat
> > M.Sc. Environment Engineering
> > Alternative Mail: bayat194 using yahoo.com
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > 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.
>
--
Best Regards
Javad Bayat
M.Sc. Environment Engineering
Alternative Mail: bayat194 using yahoo.com
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