[R] General function to substitute values in a data frame

Gerrit Eichner Gerrit.Eichner at math.uni-giessen.de
Fri Nov 9 11:44:34 CET 2012


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]]
>
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