Roger Bivand <[EMAIL PROTECTED]> writes: > > rlex$lengths[rlex$values] > [1] 1 3 2 5 1 4 1 1 1 3 1 1 2 > > cetnost > [1] 1 3 2 5 1 4 1 1 1 3 1 1 2 > > rle() is interpreted too, like your solution, so I'm not sure how it will > scale.
Not spectacularly better, but I don't think Peter is doing what he thinks he's doing... > > > > Example 2 > > x<-sample(c(T,F),40321*51, replace=T) > > dd<-matrix(x,40321,51) > > system.time(cetnost <- lapply(dd,function(x) as.numeric(table(which(x)- > > cumsum(x[which(x)]))))) > > Timing stopped at: 750.63 1 775.6 NA NA dd is not a list or data frame, so lapply is doing something for each of the 2 million cells. Was this intended instead: > system.time(cetnost <- apply(dd,2,function(x) as.numeric(table(which(x)- + cumsum(x[which(x)]))))) [1] 8.45 0.10 13.84 0.00 0.00 rle() helps a bit but not orders of magnitude: > system.time(cetnost <- apply(dd,2,function(x) ((z <- rle(x))$lengths)[z$values])) [1] 2.88 0.03 5.32 0.00 0.00 (This problem has a memory foot print of more than 200MB, so total timings vary wildly depending on whether swapping occurs.) -- O__ ---- Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
