Hi I'll study your answers.
I could also try gsub("01", "00", x) N times but it could be very slow if N is large In fact when I wrote 111110011 I mean a vector 1 1 1 1 1 0 0 1 1 not a string, but I wrote it more compactly. I also could by shifting the elements of the vector one position and ANDing the result with the original. And again shifting 2 postions and so on up to N. But it's very slow. -- View this message in context: http://r.789695.n4.nabble.com/adding-zeroes-after-old-zeroes-in-a-vector-tp2534824p2534982.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ R-help@r-project.org 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.