New function to evaluate any in i grouped by keyby or by

[tsp]

Hi
Would it be possible to add a function that evaluates any() in i by each group given by keyby or by. So instead of

DT<-data.table(Id=rep(1:3,each=3),Trt=c("A","A","B","C","B","B","A","A","A"),Dur=1:9)
DT[Id %in% DT[Trt=="A",Id],sum(Dur),by=Id]

the final line would be replaced by
DT(anyby(Trt=="A"),sum(Dur),by=Id)

[rikivillalba]

These forms are equivalent

DT[Id %in% Id[Trt=="A"],sum(Dur),by=Id]
DT[, .SD[any(Trt=="A"), .(sum(Dur))], by = Id]

The last one is particularly close to your use case. Note the .(), if you omit it , you'll obtain zero for the sum of id's with no A's in Trt

[MichaelChirico]

IIUC #788 solves this:

DT[, by=Id, having=any(Trt=='A'), sum(Dur)]

In current code, slightly more canonical (readable?) than the above suggestion with .SD is

DT[, if(any(Trt=='A')) sum(Dur), by=Id]

[Kamgang-B]

Another way, likely more useful when Trt contains NAs (parhaps more readable), is:

DT[, if("A" %in% Trt) sum(Dur), by=Id]

The (current) approach suggested by @MichaelChirico would fail if there is a group (in Id) that contains only NAs, or only NAs and values different from "A"; See data below (modified):

DT <- data.table(Id=rep(1:3,each=3),Trt=c("A","A","B","C","B","B","B",NA,NA),Dur=1:9)