One more variation on the solution, no idea how it compares in speed. Using your x ...
> ifelse(x, unlist(mapply(seq, to = rle(x)$lengths, from = 1)), 0) [1] 1 2 3 0 0 1 2 0 1 Feng, Ken wrote: > Hi, > > I start with an array of booleans: > > x <- c( TRUE, TRUE, TRUE, FALSE, FALSE, TRUE, TRUE, FALSE, TRUE ); > > I want to define an y <- f(x) such that: > > y <- c( 1, 2, 3, 0, 0, 1, 2, 0, 1 ); > > In other words, do a cumsum when I see a TRUE, but reset to 0 if I see a > FALSE. > > I know I can do this with a very slow and ugly loop or maybe use apply, > but I was hoping there are some R experts out there who can show me > a cleaner/more elegant solution? > > Thanks in advance. > > - Ken > > ______________________________________________ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.