Assume the following datatable:
DT <- data.table(a=c(1,2,3,4,5,6),b=c(NaN,NaN,NaN,4,5,6),c=c(NaN,3,3,3,NaN,NaN))
How can I replace all NaN values by NA, i.e. in each column? Looking for an elegant (short) way to do this, but I am also wondering if it is possible using lapply, like I tried below.
My approach so far:
DT[,lapply(SD,function(x){x[is.nan(x)] := NA}),.SDcols=c("a","b","c")]
Actual result of my code is:
Error in
:=(x[is.nan(x)], NA) : Check that is.data.table(DT) == TRUE. Otherwise, := and:=(...) are defined for use in j, once only and in particular ways. See help(":=").
2 Answers
You can do it easily using dplyr's mutate_all function.
DT <- DT %>% mutate_all(~ifelse(is.nan(.), NA, .))
print(DT)
#a b c
# 1 NA NA
# 2 NA 3
# 3 NA 3
# 4 4 3
# 5 5 NA
# 6 6 NA
The code above is equivalent to
DT %>% mutate_all(function(x) ifelse(is.nan(x), NA, x))
Here's one way to do it using data.table syntax. I'm not sure if there is a simpler way.
DT[, names(DT) := lapply(.SD, function(x) ifelse(is.nan(x), NA, x))]
EDIT: An alternative way to do this, as posted by markus in the comments beow.
DT[, lapply(.SD, function(x) replace(x, is.nan(x), NA))]
You want something like this:
DT[DT == "NaN"] <- NA
This should ideally find all the NaN that you have and replace it with NA. If anyone knows a better way please let me know!
Hope this helps!
