Add extra level to factors in dataframe
The levels
function accept the levels(x) <- value
call. Therefore, it's very easy to add different levels:
f1 <- factor(c("a", "a", NA, NA, "b", NA, "a", "c", "a", "c", "b"))
str(f1)
Factor w/ 3 levels "a","b","c": 1 1 NA NA 2 NA 1 3 1 3 ...
levels(f1) <- c(levels(f1),"No Answer")
f1[is.na(f1)] <- "No Answer"
str(f1)
Factor w/ 4 levels "a","b","c","No Answer": 1 1 4 4 2 4 1 3 1 3 ...
You can then loop it around all variables in a data.frame:
f1 <- factor(c("a", "a", NA, NA, "b", NA, "a", "c", "a", "c", "b"))
f2 <- factor(c("c", NA, "b", NA, "b", NA, "c" ,"a", "d", "a", "b"))
f3 <- factor(c(NA, "b", NA, "b", NA, NA, "c", NA, "d" , "e", "a"))
df1 <- data.frame(f1,n1=1:11,f2,f3)
str(df1)
'data.frame': 11 obs. of 4 variables:
$ f1: Factor w/ 3 levels "a","b","c": 1 1 NA NA 2 NA 1 3 1 3 ...
$ n1: int 1 2 3 4 5 6 7 8 9 10 ...
$ f2: Factor w/ 4 levels "a","b","c","d": 3 NA 2 NA 2 NA 3 1 4 1 ...
$ f3: Factor w/ 5 levels "a","b","c","d",..: NA 2 NA 2 NA NA 3 NA 4 5 ...
for(i in 1:ncol(df1)) if(is.factor(df1[,i])) levels(df1[,i]) <- c(levels(df1[,i]),"No Answer")
df1[is.na(df1)] <- "No Answer"
str(df1)
'data.frame': 11 obs. of 4 variables:
$ f1: Factor w/ 4 levels "a","b","c","No Answer": 1 1 4 4 2 4 1 3 1 3 ...
$ n1: int 1 2 3 4 5 6 7 8 9 10 ...
$ f2: Factor w/ 5 levels "a","b","c","d",..: 3 5 2 5 2 5 3 1 4 1 ...
$ f3: Factor w/ 6 levels "a","b","c","d",..: 6 2 6 2 6 6 3 6 4 5 ...
You could define a function that adds the levels to a factor, but just returns anything else:
addNoAnswer <- function(x){
if(is.factor(x)) return(factor(x, levels=c(levels(x), "No Answer")))
return(x)
}
Then you just lapply
this function to your columns
df <- as.data.frame(lapply(df, addNoAnswer))
That should return what you want.