Try this: do.call(sprintf, c("%9.2f\t%d\t%d\t%8.3f", as.list(v[iv])))
On 5/3/06, Paul Roebuck <[EMAIL PROTECTED]> wrote: > How would one go about getting sprintf to use the > values of a vector without having to specify each > argument individually? > > > v <- c(1, 2, -1.197114, 0.1596687) > > iv <- c(3, 1, 2, 4) > > sprintf("%9.2f\t%d\t%d\t%8.3f", v[3], v[1], v[2], v[4]) > [1] " -1.20\t1\t2\t 0.160" > > Essentially, desired effect would be something like: > > sprintf("%9.2f\t%d\t%d\t%8.3f", v[iv]) # wish it worked > > ---------------------------------------------------------- > SIGSIG -- signature too long (core dumped) > > ______________________________________________ > 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