Hi Rangesh, Perhaps I mis-understand your question, but it could be as simple as...
p <- 1-exp(-exp(s)) In R, this is vectorized such that a new vector is calculated - one value for each value of s, so p will have the same length as s. In the "Introduction to R" read up on vectors and how to avoid loops. cheers! Sean On 06/08/05, Rangesh Kunnavakkam <[EMAIL PROTECTED]> wrote: > I am new to R,so excuse me if its too basic. I have a vector of scores > S= {s1,s2,s3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,sn} > n could be very large. > I want to find p-value for each score and I have the formula > pvalue= 1-exp(-exp(S)) > so any ideas how to get this new vector P={p1,p2,p3,p4,p5....pn} > thankyou in advance > Rangesh > > ______________________________________________ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html > ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html