buhtz buhtz - 3 months ago 17
R Question

replace <NA> in a factor column in R

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)
lv <- levels(facna)
levels(facna) <- replace(lv, is.na(lv), 88)

giving:

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

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")