General function to substitute values in a data frame

as.data.frame( lapply( df, function( x) LETTERS[ x-9]))

transform( df, y = LETTERS[y - 9], z = LETTERS[ z - 9])

within( df, {y <- LETTERS[y - 9]; z <- LETTERS[ z - 9]})

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

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