Try this: p <- 0:10/10 p * log2(p + !p)
On 7/11/06, Taka Matzmoto <[EMAIL PROTECTED]> wrote: > Dear R-users > > >prob <- c(0.5,0.4,0.3,0.1,0.0) > >cal <- prob * log(prob,base=2) > >cal > [1] -0.5000000 -0.5287712 -0.5210897 -0.3321928 NaN > > Is there any way to change NaN to zero ? > > I did come up with this by applying Ripley's relpy to my previous question > > cal <-prob*log(pmax(prob,0.00000001),base=2) > > Any suggestion ? > > Thank you > > Taka > > ______________________________________________ > [email protected] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html > ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
