Replace <NA> in a factor column

Question

I want to replace <NA> 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 comes from a foreign csv file I have to deal with.

df <- data.frame(a=sample(0:10, size=10, replace=TRUE),                  b=sample(20:30, size=10, replace=TRUE)) df[df$a==0,'a'] <- NA df$a <- as.factor(df$a) 

Could look like this

      a  b 1     1 29 2     2 23 3     3 23 4     3 22 5     4 28 6  <NA> 24 7     2 21 8     4 25 9  <NA> 29 10    3 24 

Now I want to replace the <NA> values with a number.

df[is.na(df$a), 'a'] <- 88 In `[<-.factor`(`*tmp*`, iseq, value = c(88, 88)) :   invalid factor level, NA generated 

I think I missed a fundamental R concept about factors. Am I? I can not understand why it doesn't work. I think invalid factor level means that 88 is not a valid level in that factor, right? So I have to tell the factor column that there is another level?

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"