Dear Olivella,
Assuming that x is your data, here is one possible way to do what you want:

t(
   apply(x, 1, function(x){
                       index <- x %in% c(1,2)
                       x[index] <- cls[index]
                       x[!index] <- NA
                       x
                       }
           )
  )

How big is the data you are dealing with?

HTH,

Jorge


On Wed, Jul 8, 2009 at 4:25 PM, Olivella <olivella@wustl.edu> wrote:

>
> Hello,
>
> I wish to perform a substitution of certain numerical values in a data
> matrix with the corresponding column name. For instance, if I have a data
> matrix
> V1  V2  V3
> 2    0    1
> 0    1    2
> 1    5    0
> 5    0    0
>
> I want to substitute the `1' and the `2' for the corresponding column name,
> and make everything else `NA' like this
> V1    V2    V3
> V1    NA    V3
> NA    V2    V3
> V1    NA    NA
> NA    NA    NA
>
> I have done this using an explicit `for' loop, but it takes a really long
> time to finish. Is there any way I can do this using `apply' or some form
> of
> implicit looping?
>
> Thank you for your help,
>
> SO
> --
> View this message in context:
> http://www.nabble.com/Substituting-numerical-values-using-%60apply%27-tp24398687p24398687.html
> Sent from the R help mailing list archive at Nabble.com.
>
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