> I think you should use a single data.table; it's much more straightforward > in that case: > > qq <- data.table(runif(18),id=rep(1:2,each=9)) > qq[,freq:=.N,by=.(id,seq(nrow(qq))%/%5)]
Thanks for the idea, unfortunately I need lists. The example is just a minimal example to show the problem. The actual code will use large data.tables in lists and I want to eventually use mclapply to parallelize the computation. Sebastian _______________________________________________ datatable-help mailing list [email protected] https://lists.r-forge.r-project.org/cgi-bin/mailman/listinfo/datatable-help
