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 <jholtman@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 <sewashm@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?
>

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