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Several factors same levels

3 messages · dereksloan, Dimitris Rizopoulos, Peter Dalgaard

dereksloan
# 
This is probably very simple but I'm new to R so apologies for being stupid.

I have some data with No coded as 0 and yes coded as 1.

e.g. 

id sex alcohol smoker
1   M     0        1
2   F      1        0 
3   M     0        0 

I realise I can covert the numerical variable back to a factor by

falcohol<-factor(alcohol,levels=0:1)
levels<-c("No","Yes") 
fsmoker<-factor(smoker,levels=0:1)
levels<-c("No","Yes") 

but can I do this for all factors using the same levels in a single command
to make it easier (i.e. switch all 0/1, no/yes variables without having to
recode the same factor levels for them all)?

Derek

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Dimitris Rizopoulos
# 
one way is the following:

DF <- data.frame(id = 1:6,
     sex = gl(2, 3, labels = c("M", "F")),
     x = sample(0:1, 6, TRUE), y = sample(0:1, 6, TRUE),
     z = rnorm(6))

f <- function (x) {
     if (all(unique(x) %in% 0:1))
         factor(x, levels = 0:1, labels = c("No", "Yes"))
     else
         x
}

DF[] <- lapply(DF, f)


I hope it helps.

Best,
Dimitris
On 4/19/2011 10:57 AM, dereksloan wrote:

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Peter Dalgaard
    

    

      
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On Apr 19, 2011, at 10:57 , dereksloan wrote:

        

            

      
      
      
    

        
Yes, if they are in the same data frame. The idiom goes roughly like this

ix <- c("alcohol", "smoker")
new <- paste("f", ix, sep="")
dd[new] <- lapply(dd[ix], factor, levels=0:1, labels=c("No","Yes"))

(You might store the transformed variables back where they came from, just make sure to do it exactly once! That would avoid creating new variable names and give you a bit more flexibility in the choice of index type. There are also various possibilities to actually compute the indices, e.g.

isBin <- function(x) all(na.omit(x) %in% 0:1)
ixl <- sapply(dd, isBin)
ix <- names(dd)[ixl]
...
)