Replace <NA> in a factor column
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) 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).