R help - Nov 2012 - General function to substitute values in a data frame

Hi R users,

I need a way to substitute the values 10:31 to the letters A:V (i.e 10=A,
11=B, ..., 31=V) in a data frame.

For example:
> y<-c(10,11,12,13)
> z<-c(28,29,30,31)
> df<-data.frame(y,z)
> df
   y  z
1 10 28
2 11 29
3 12 30
4 13 31

Then I would substitute it and obtain a data frame like this as a result of
the function:
> w<-c("A","B","C","D")     #
without actually writing this part down, of
course.
> x<-c("S","T","U","V")     #
without actually writing this part down, of
course.
> df2<-data.frame(w,x)
> df2
  w x
1 A S
2 B T
3 C U
4 D V

Apparently the function "replace" can do the
job:
> attach(df)
> replace(y, y==10,"A")
[1] "A"  "11" "12" "13"

But then I would have to do it letter by letter and build the data frame
again. I would not mind doing this for one small data frame but I do have
several large ones, so I was wondering if that's a way that I can write
only one function to perform the action?

I found another way, but it looks kind of silly:
> ifelse(y==10,"A", ifelse(y==11,"B",
ifelse(y==12,"C", ... )))
Anyway, I would have to rewrite this for every column as well. So what I
really want is something that I could use for the whole data frame (or at
least lapply it), like:
> change.to.letters<-function (x) {if x==7 replace(x, x==7, "A")
                      if (x==6) replace(x,x==6,"B")
                        ...................}

                      # and so on... but of course this one does not work,
I just wrote down what I suppose it should looks like. Then I could
use:
> change.to.letters(y)  # or
> lapply(df, FUN=change.to.letters)
Any help would be much appreciated!

Thanks!

    Fabricius

	[[alternative HTML version deleted]]

Hello, Fabricius,

does
> as.data.frame( lapply( df, function( x) LETTERS[ x-9]))
what you want?

Other (here maybe less flexible) ways:
> transform( df, y = LETTERS[y - 9], z = LETTERS[ z - 9])
> within( df, {y <- LETTERS[y - 9]; z <- LETTERS[ z - 9]})
Hth  --  Gerrit


On Fri, 9 Nov 2012, Fabricius Domingos wrote:
> Hi R users,
>
> I need a way to substitute the values 10:31 to the letters A:V (i.e 10=A,
> 11=B, ..., 31=V) in a data frame.
>
> For example:
>> y<-c(10,11,12,13)
>> z<-c(28,29,30,31)
>> df<-data.frame(y,z)
>> df
>   y  z
> 1 10 28
> 2 11 29
> 3 12 30
> 4 13 31
>
> Then I would substitute it and obtain a data frame like this as a result of
> the function:
>> w<-c("A","B","C","D")     #
without actually writing this part down, of
> course.
>> x<-c("S","T","U","V")     #
without actually writing this part down, of
> course.
>> df2<-data.frame(w,x)
>> df2
>  w x
> 1 A S
> 2 B T
> 3 C U
> 4 D V
>
> Apparently the function "replace" can do the job:
>> attach(df)
>> replace(y, y==10,"A")
> [1] "A"  "11" "12" "13"
>
> But then I would have to do it letter by letter and build the data frame
> again. I would not mind doing this for one small data frame but I do have
> several large ones, so I was wondering if that's a way that I can write
> only one function to perform the action?
>
> I found another way, but it looks kind of silly:
>> ifelse(y==10,"A", ifelse(y==11,"B",
ifelse(y==12,"C", ... )))
>
> Anyway, I would have to rewrite this for every column as well. So what I
> really want is something that I could use for the whole data frame (or at
> least lapply it), like:
>> change.to.letters<-function (x) {if x==7 replace(x, x==7,
"A")
>                      if (x==6) replace(x,x==6,"B")
>                        ...................}
>
>                      # and so on... but of course this one does not work,
> I just wrote down what I suppose it should looks like. Then I could use:
>> change.to.letters(y)  # or
>> lapply(df, FUN=change.to.letters)
>
> Any help would be much appreciated!
>
> Thanks!
>
>    Fabricius
>
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> 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.

---------------------------------------------------------------------
Dr. Gerrit Eichner                   Mathematical Institute, Room 212
gerrit.eichner at math.uni-giessen.de   Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104          Arndtstr. 2, 35392 Giessen, Germany
Fax: +49-(0)641-99-32109        http://www.uni-giessen.de/cms/eichner

Hello,

Try the following. I've changed the name of your data.frame to
'dat',
'df' is an R function.


replace.letter <- function(x, first = 1, upper = TRUE){
     if(upper)
         LETTERS[x - first + 1]
     else
         letters[x - first + 1]
}

replace.letter(10:31, first = 10)

y <- c(10,11,12,13)
z <- c(28,29,30,31)
dat <- data.frame(y,z)

apply(dat, 2, replace.letter, first = 10)


Also, the use of attach() is disadvised, it can be confusing.

Hope this helps,

Rui Barradas

Em 09-11-2012 09:33, Fabricius Domingos escreveu:
> Hi R users,
>
> I need a way to substitute the values 10:31 to the letters A:V (i.e 10=A,
> 11=B, ..., 31=V) in a data frame.
>
> For example:
>> y<-c(10,11,12,13)
>> z<-c(28,29,30,31)
>> df<-data.frame(y,z)
>> df
>     y  z
> 1 10 28
> 2 11 29
> 3 12 30
> 4 13 31
>
> Then I would substitute it and obtain a data frame like this as a result of
> the function:
>> w<-c("A","B","C","D")     #
without actually writing this part down, of
> course.
>> x<-c("S","T","U","V")     #
without actually writing this part down, of
> course.
>> df2<-data.frame(w,x)
>> df2
>    w x
> 1 A S
> 2 B T
> 3 C U
> 4 D V
>
> Apparently the function "replace" can do the job:
>> attach(df)
>> replace(y, y==10,"A")
> [1] "A"  "11" "12" "13"
>
> But then I would have to do it letter by letter and build the data frame
> again. I would not mind doing this for one small data frame but I do have
> several large ones, so I was wondering if that's a way that I can write
> only one function to perform the action?
>
> I found another way, but it looks kind of silly:
>> ifelse(y==10,"A", ifelse(y==11,"B",
ifelse(y==12,"C", ... )))
> Anyway, I would have to rewrite this for every column as well. So what I
> really want is something that I could use for the whole data frame (or at
> least lapply it), like:
>> change.to.letters<-function (x) {if x==7 replace(x, x==7,
"A")
>                        if (x==6) replace(x,x==6,"B")
>                          ...................}
>
>                        # and so on... but of course this one does not work,
> I just wrote down what I suppose it should looks like. Then I could use:
>> change.to.letters(y)  # or
>> lapply(df, FUN=change.to.letters)
> Any help would be much appreciated!
>
> Thanks!
>
>      Fabricius
>
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> 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.

HI,
May be this:
res<- data.frame(apply(df,2,function(x)
ifelse(grepl("\\d+",x),LETTERS[x-9],NA)))
res
#? y z
#1 A S
#2 B T
#3 C U
#4 D V
#or
?apply(df,2,function(x) LETTERS[x-9])
A.K.





----- Original Message -----
From: Fabricius Domingos <fabriciusm at gmail.com>
To: r-help at r-project.org
Cc: 
Sent: Friday, November 9, 2012 4:33 AM
Subject: [R] General function to substitute values in a data frame

Hi R users,

I need a way to substitute the values 10:31 to the letters A:V (i.e 10=A,
11=B, ..., 31=V) in a data frame.

For example:
> y<-c(10,11,12,13)
> z<-c(28,29,30,31)
> df<-data.frame(y,z)
> df
?  y? z
1 10 28
2 11 29
3 12 30
4 13 31

Then I would substitute it and obtain a data frame like this as a result of
the function:
> w<-c("A","B","C","D")? ?  #
without actually writing this part down, of
course.
> x<-c("S","T","U","V")? ?  #
without actually writing this part down, of
course.
> df2<-data.frame(w,x)
> df2
? w x
1 A S
2 B T
3 C U
4 D V

Apparently the function "replace" can do the
job:
> attach(df)
> replace(y, y==10,"A")
[1] "A"? "11" "12" "13"

But then I would have to do it letter by letter and build the data frame
again. I would not mind doing this for one small data frame but I do have
several large ones, so I was wondering if that's a way that I can write
only one function to perform the action?

I found another way, but it looks kind of silly:
> ifelse(y==10,"A", ifelse(y==11,"B",
ifelse(y==12,"C", ... )))
Anyway, I would have to rewrite this for every column as well. So what I
really want is something that I could use for the whole data frame (or at
least lapply it), like:
> change.to.letters<-function (x) {if x==7 replace(x, x==7, "A")
? ? ? ? ? ? ? ? ? ? ? if (x==6) replace(x,x==6,"B")
? ? ? ? ? ? ? ? ? ? ? ? ...................}

? ? ? ? ? ? ? ? ? ? ? # and so on... but of course this one does not work,
I just wrote down what I suppose it should looks like. Then I could
use:
> change.to.letters(y)? # or
> lapply(df, FUN=change.to.letters)
Any help would be much appreciated!

Thanks!

? ? Fabricius

??? [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list
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.