Unable to assign columns using (names(.SD)) :=

[skanskan]

Hello.

I want to modify the columns containing a given pattern in their names.

Say I have this toy example:
myDT <- data.table(cod1=1:20, cod2=20:1, other1=rnorm(20), other2=rnorm(20))

And I want to convert the content of the columns whose name start by "cod" to factors.

With dplyr I can use this compact notation:
myDT %>% mutate(across(contains("cod"),as.factor))

With data.table I would use this:

cols <- grep("cod", names(myDT), value = T)
myDT[,(cols):= lapply(.SD, as.factor), .SDcols = cols] 

It's much longer and it generates new variables we don't need anymore.
Is there any shorter/simpler/faster alternative?

I have tried this:
myDT[, (names(.SD)) := lapply(.SD, as.factor) , .SDcols = patterns("cod")]
but it doesn't work.

[jangorecki]

There is FR for that already, and maybe even PR as well.

[MichaelChirico]

duplicate of #795, will be fixed in #4163