Thanks. This helps. How do I generate P? Will this work? p1<-pnorm(mean=0, std=1) p2<-pnorm(mean=0, std=1)
x <- cbind(x, v1=ifelse(x[,'p'] > 0.4, 1, 0), v2=ifelse(x[,'2'] > 0.6, 0, 1)) If the 'data set' is a dataframe, the following will work: x$v1 <- ifelse(x$p > 0.4, 1, 0) x$v2 <- ifelse(x$p > 0.6, 1, 0) If it is matrix, try x <- cbind(x, v1=ifelse(x[,'p'] > 0.4, 1, 0), v2=ifelse(x[,'p'] > 0.6, 1, 2)) On Sat, Oct 10, 2009 at 6:32 PM, jim holtman <jholt...@gmail.com> wrote: > If the 'data set' is a dataframe, the following will work: > > x$v1 <- ifelse(x$p > 0.4, 1, 0) > x$v2 <- ifelse(x$p > 0.6, 1, 0) > > If it is matrix, try > > x <- cbind(x, v1=ifelse(x[,'p'] > 0.4, 1, 0), v2=ifelse(x[,'p'] > 0.6, 1, > 2)) > > If helps a lot if you follow the posting rules and provide commented, > minimal, self-contained, reproducible code. > > On Sat, Oct 10, 2009 at 6:04 PM, Ashta <sewa...@gmail.com> wrote: > > Hi all, > > > > I have a data set called x with 200 rows and 12 columns. I want > > create two more columns based on probability. ie > > if p >0 .4 then v1 =1 else v1=0; > > if p >0 .6 then v2 =1 else v2=0; > > > > Finally x will have 14 variables. > > > > Can any one show me how to do that? > > > > Thanks > > Ashta > > > > > > . > > > > [[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. > > > > > > -- > Jim Holtman > Cincinnati, OH > +1 513 646 9390 > > What is the problem that you are trying to solve? > [[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.
