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
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