A belated note/warning regarding class membership in R objects. The correct method for identification of "labelled" is not to test for with an is function or equality {==) but rather with inherits. Methods that test for a specific location will not pick up cases where the order of existing classes are not the ones assumed.

You can avoid creating "labelled" variables in spss.get with the argument: , use.value.labels=FALSE.

w <- spss.get( /tmp/my.sav , use.value.labels=FALSE, datevars=c( birthdate , deathdate ))

The code from Bhattacharya could fail if the class of the labelled vector were simply "labelled" rather than c("labelled", "factor") in which case it should have been:

class(x[[i]]) <- NULL  # no error from assignment of empty vector

The error you report can be reproduced with this code:

> b <- 4:6
> label(b) <-  B Label 
> str(b)
Class  labelled   atomic [1:3] 4 5 6
  ..- attr(*, "label")= chr "B Label"
> class(b) <- class(b)[-1]
Error in class(b) <- class(b)[-1] : 
  invalid replacement object to be a class string

Here s how I get rid of the labels altogether. Similar to Jyotirmoy s solution but works for a vector as well as a data.frame. (Partial credits to Frank Harrell)

clear.labels <- function(x) {
  if(is.list(x)) {
    for(i in 1 : length(x)) class(x[[i]]) <- setdiff(class(x[[i]]),  labelled ) 
    for(i in 1 : length(x)) attr(x[[i]],"label") <- NULL
  }
  else {
    class(x) <- setdiff(class(x), "labelled")
    attr(x, "label") <- NULL
  }
  return(x)
}

Use as follows:

my.unlabelled.df <- clear.labels(my.labelled.df)

EDIT

Here s a bit of a cleaner version of the function, same results:

clear.labels <- function(x) {
  if(is.list(x)) {
    for(i in seq_along(x)) {
      class(x[[i]]) <- setdiff(class(x[[i]]),  labelled ) 
      attr(x[[i]],"label") <- NULL
    } 
  } else {
    class(x) <- setdiff(class(x), "labelled")
    attr(x, "label") <- NULL
  }
  return(x)
}

You can try out the read.spss function from the foreign package.

A rough and ready way to get rid of the labelled class created by spss.get

for (i in 1:ncol(x)) {
    z<-class(x[[i]])
    if (z[[1]]== labelled ){
       class(x[[i]])<-z[-1]
       attr(x[[i]], label )<-NULL
    }
}

But can you please give an example where labelled causes problems?

If I have a variable MAED in a data frame x created by spss.get, I have:

> class(x$MAED)
[1] "labelled" "factor"  
> is.factor(x$MAED)
[1] TRUE

So well-written code that expects a factor (say) should not have any problems.

Suppose:

library(Hmisc)
w <- spss.get( ... )

You could remove the labels of a variable called "var1" by using:

attributes(w$var1)$label <- NULL

If you also want to remove the class "labbled", you could do:

class(w$var1) <- NULL 

or if the variable has more than one class:

class(w$var1) <- class(w$var1)[-which(class(w$var1)=="labelled")]

Hope this helps!

Well, I figured out that unclass function can be utilized to remove classes (who would tell, aye?!):

library(Hmisc)
# let s presuppose that variable x is gathered through spss.get() function
# and that x is factor
> class(x)
[1] "labelled" "factor"
> foo <- unclass(x)
> class(foo)
[1] "integer"

It s not the luckiest solution, just imagine back-converting bunch of vectors... If anyone tops this, I ll check it as an answer...