Replace <NA> in a factor column
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"
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).