Hi Well, I am not sure if this is what you want but same result can be achieved by
vec1 <- (probs>=p00)*(probs>=p10) Petr > -----Original Message----- > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- > project.org] On Behalf Of Gopi Goteti > Sent: Friday, August 24, 2012 6:50 AM > To: r-help@r-project.org > Subject: [R] updating elements of a vector sequentially - is there a > faster way? > > I would like to know whether there is a faster way to do the below > operation (updating vec1). > > My objective is to update the elements of a vector (vec1), where a > particular element i is dependent on the previous one. I need to do > this on vectors that are 1 million or longer and need to repeat that > process several hundred times. The for loop works but is slow. If there > is a faster way, please let me know. > > probs <- c(.1, .3, .2, .4, .7, .9, .3, .4, .5, .6) p10 <- 0.6 p00 <- > 0.4 > vec1 <- rep(0, 10) > for (i in 2:10) { > vec1[i] <- ifelse(vec1[i-1] == 0, > ifelse(probs[i]<p10, 0, 1), > ifelse(probs[i]<p00, 0, 1)) } > > Thanks > GG > > [[alternative HTML version deleted]] > > ______________________________________________ > 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. ______________________________________________ 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.
