How to recode a set of variables in a dataframe in R

Question

I have a dataframe with different variables containing values from 1 to 5. I want to recode some variables in the way that 5 becomes 1 and vice versa (x=6-x). I want to define a list of variables, that will be recoded like this in my dataframe.

Here is my approach using lapply. I haven't really understood it yet.

  #generate example-dataset
    var1<-sample(1:5,100,rep=TRUE)
    var2<-sample(1:5,100,rep=TRUE)
    var3<-sample(1:5,100,rep=TRUE)
    dat<-as.data.frame(cbind(var1,var2,var3))

    recode.list<-c("var1","var3")  
    recode.function<- function(x){          
    x=6-x
     }
    lapply(recode.list,recode.function,data=dat)

calculating 6-x in this case is pretty simple but if you have more specific recoding patterns you might want to check the custom "recode" function from the "car" package. — agenis - code en bois
– agenis - code en bois, Commented Nov 20, 2014 at 13:52

A5C1D2H2I1M1N2O1R2T1 · Accepted Answer · 2014-11-20 13:48:49Z

6

There's no need for an external function or for a package for this. Just use an anonymous function in lapply, like this:

df[recode.list] <- lapply(df[recode.list], function(x) 6-x)

Using [] lets us replace just those columns directly in the original dataset. This is needed since just using lapply would result in the data as a named list.

As noted in the comments, you can actually even skip lapply:

df[recode.list] <- 6 - df[recode.list]

edited Nov 20, 2014 at 13:48

answered Nov 20, 2014 at 13:43

A5C1D2H2I1M1N2O1R2T1

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2 Comments

SEMson Over a year ago

Thank you for your help. Somehow, i get following message when applying your code to my data: "Error in [.data.frame(df, recode.list) : undefined columns selected". What could this possibly mean?

A5C1D2H2I1M1N2O1R2T1 Over a year ago

@user2982730, sounds like it means that your "recode.list" includes columns that are not found in the source data.frame.

shadow · Accepted Answer · 2014-11-20 13:47:05Z

1

You can use mapvalues from plyr.

require(plyr)
# if you just want to replace 5 with 1 and vice versa
df[, recode.list] <- sapply(df[, recode.list], mapvalues, c(1, 5), c(5,1))
# if you want to apply to x=6-x to all values (in this case you don't need mapvalues)
df[, recode.list] <- sapply(df[, recode.list], mapvalues, 1:5, 5:1)

answered Nov 20, 2014 at 13:47

shadow

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1 Comment

Simone Over a year ago

mapvalues didn't work for me. instead I used recode from the car package. That said, recode is only useful when one knows how many different values the variable has and there aren't too many values that need to be recoded.

talat · Accepted Answer · 2014-11-20 13:53:35Z

1

Here's an option to do this with dplyr:

recode.function<- function(x){          
  x <- 6-x 
}

recode.list <- c("var1","var3") 

require(dplyr)
df %>% mutate_each_(funs(recode.function), recode.list)

#    var1 var2 var3
#1      2    2    4
#2      3    3    3
#3      3    5    2
#4      3    3    2
#5      4    3    3
#6      5    4    1
#...

edited Nov 20, 2014 at 13:53

answered Nov 20, 2014 at 13:40

talat

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