Question

I’m trying to replace elements of a data.frame containing “#N/A” with “NULL”, and I’m running into problems:

foo <- data.frame("day"= c(1, 3, 5, 7), "od" = c(0.1, "#N/A", 0.4, 0.8))

indices_of_NAs <- which(foo == "#N/A") 

replace(foo, indices_of_NAs, "NULL")

Error in [<-.data.frame(*tmp*, list, value = “NULL”) :
new columns would leave holes after existing columns

I think that the problem is that my index is treating the data.frame as a vector, but that the replace function is treating it differently somehow, but I’m not sure what the issue is?

Answer 1

NULL really means “nothing”, not “missing” so it cannot take the place of an actual value – for missing R uses NA.

You can use the replacement method of is.na to directly update the selected elements, this will work with a logical result. (Using which for indices will only work with is.na, direct use of [ invokes list access, which is the cause of your error).

foo <- data.frame("day"= c(1, 3, 5, 7), "od" = c(0.1, "#N/A", 0.4, 0.8)) 
NAs <- foo == "#N/A"

## by replace method
is.na(foo)[NAs] <- TRUE

 ## or directly
 foo[NAs] <- NA

But, you are already dealing with strings (actually a factor by default) in your od column by forced coercion when it was created with c(), and you might need to treat columns individually. Any numeric column will never have a match on the string “#N/A”, for example.