Add extra level to factors in dataframe
Solution 1
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 ...
Solution 2
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.
Solution 3
I have a very simple answer that may not directly address your specific scenario, but is a simple way to do this generally
levels(df$column) <- c(levels(df$column), newFactorLevel)
Solution 4
Since this question was last answered this has become possible using fct_explicit_na()
from the forcats
package. I add here the example given in the documentation.
f1 <- factor(c("a", "a", NA, NA, "a", "b", NA, "c", "a", "c", "b"))
table(f1)
# f1
# a b c
# 4 2 2
f2 <- forcats::fct_explicit_na(f1)
table(f2)
# f2
# a b c (Missing)
# 4 2 2 3
Default value is (Missing)
but this can be changed via the na_level
argument.
Solution 5
Expanding on ilir's answer and its comment, you can check if a column is a factor and that it does not already contain the new level, then add the level and thus make the function re-runable:
addLevel <- function(x, newlevel=NULL) {
if(is.factor(x)) {
if (is.na(match(newlevel, levels(x))))
return(factor(x, levels=c(levels(x), newlevel)))
}
return(x)
}
You can then apply it like so:
dataFrame$column <- addLevel(dataFrame$column, "newLevel")
enedene
Updated on February 23, 2020Comments
-
enedene about 4 years
I have a data frame with numeric and ordered factor columns. I have lot of NA values, so no level is assigned to them. I changed NA to "No Answer", but levels of the factor columns don't contain that level, so here is how I started, but I don't know how to finish it in an elegant way:
addNoAnswer = function(df) { factorOrNot = sapply(df, is.factor) levelsList = lapply(df[, factorOrNot], levels) levelsList = lapply(levelsList, function(x) c(x, "No Answer")) ...
Is there a way to directly apply new levels to factor columns, for example, something like this:
df[, factorOrNot] = lapply(df[, factorOrNot], factor, levelsList)
Of course, this doesn't work correctly.
I want the order of levels preserved and "No Answer" level added to last place.
-
msoftrain over 9 yearsJust a little suggestion to make this function more generic. I've encountered the need to add a new level to a given factor a number of times (e.g., when merging datasets), so others might be in that case too: addLevel <- function(x, newlevel=NULL){ if(is.factor(x)) return(factor(x, levels=c(levels(x), newlevel))) return(x) }
-
David Arenburg almost 7 yearsIt's probably better to do something like
df[] <- lapply(df, addNoAnswer)
instead (haven't tested it with your function though). -
Uwe almost 7 yearsGood suggestion. Hadley's
forcats
package has turned out to be a great help to me when I had to solve tricky as well as trivial situations with factors. -
EngrStudent over 2 yearspart of the value of this answer is that it cleanly and easily generalizes to cases beyond the original, while answering the original question well.