Replace in a factor column

Question

I want to replace values in a factors column with a valid value. But I can not find a way. This example is only for demonstration. The original data

Answer 1

1) addNA If fac is a factor addNA(fac) is the same factor but with NA added as a level. See ?addNA

To force the NA level to be 88:

facna <- addNA(fac)
levels(facna) <- c(levels(fac), 88)

giving:

> facna
 [1] 1  2  3  3  4  88 2  4  88 3 
Levels: 1 2 3 4 88

1a) This can be written in a single line as follows:

`levels<-`(addNA(fac), c(levels(fac), 88))

2) factor It can also be done in one line using the various arguments of factor like this:

factor(fac, levels = levels(addNA(fac)), labels = c(levels(fac), 88), exclude = NULL)

2a) or equivalently:

factor(fac, levels = c(levels(fac), NA), labels = c(levels(fac), 88), exclude = NULL)

3) ifelse Another approach is:

factor(ifelse(is.na(fac), 88, paste(fac)), levels = c(levels(fac), 88))

4) forcats The forcats package has a function for this:

library(forcats)

fct_explicit_na(fac, "88")
## [1] 1  2  3  3  4  88 2  4  88 3 
## Levels: 1 2 3 4 88

Note: We used the following for input fac

fac <- structure(c(1L, 2L, 3L, 3L, 4L, NA, 2L, 4L, NA, 3L), .Label = c("1", 
"2", "3", "4"), class = "factor")

Update: Have improved (1) and added (1a). Later added (4).

Answer 2

other way to do is:

#check levels
levels(df$a)
#[1] "3"  "4"  "7"  "9"  "10"

#add new factor level. i.e 88 in our example
df$a = factor(df$a, levels=c(levels(df$a), 88))

#convert all NA's to 88
df$a[is.na(df$a)] = 88

#check levels again
levels(df$a)
#[1] "3"  "4"  "7"  "9"  "10" "88"

Answer 3

My way would be a little bit traditional by using factor function:

a <- factor(a, 
            exclude = NULL, 
            levels = c(levels(a), NA),
            labels = c(levels(a), "None"))

You can replace "None" with appropriate replacement that you want (0L for example)

Answer 4

The problem is that NA is not a level of that factor:

> levels(df$a)
[1] "2"  "4"  "5"  "9"  "10"

You can't change it straight away, but the following will do the trick:

df$a <- as.numeric(as.character(df$a))
df[is.na(df$a),1] <- 88
df$a <- as.factor(df$a)
> df$a
 [1] 9  88 3  9  5  9  88 8  3  9 
Levels: 3 5 8 9 88
> levels(df$a)
[1] "3"  "5"  "8"  "9"  "88"

Answer 5

I had similar issues and I want to add what I consider the most pragmatic (and also tidy) solution:

Convert the column to a character column, use mutate and a simple ifelse-statement to change the NA values to what you want the factor level to be (I have chosen "None"), convert it back to a factor column:

df %>% mutate(
a = as.character(a),
a = ifelse(is.na(a), "None", a),
a = as.factor(a)
)

Clean and painless because you do not actually have to dabble with NA values when they occur in a factor column. You bypass the weirdness and end up with a clean factor variable.

Answer 6

The basic concept of a factor variable is that it can only take specific values, i.e., the levels. A value not in the levels is invalid.

You have two possibilities:

If you have a variable that follows this concept, make sure to define all levels when you create it, even those without corresponding values.

Or make the variable a character variable and work with that.

PS: Often these problems result from data import. For instance, what you show there looks like it should be a numeric variable and not a factor variable.