Hi, Thanks for the help. Both of the responses work the way I need it to.
Cheers, Peter On Thu, May 30, 2013 at 4:34 AM, Rui Barradas <ruipbarra...@sapo.pt> wrote: > Hello, > > You can write a helper function. > > are.equal <- function(x, y, eps = .Machine$double.eps^0.5) abs(x - y) < eps > > x[are.equal(x, 0.15)] > > > Hope this helps, > > Rui Barradas > > Em 30-05-2013 02:27, Peter Lomas escreveu: > >> Hello, >> >> I have a whole bunch of data to two decimal places. I've done some >> arithmetic with them, so floating point becomes an issue. >> >> x <- c(1, 0.15,(0.1+.05),0.4) >> >> I want to do something like this: >> >> x[x==0.15] >> >> But you'll notice that is troublesome with the well known floating point >> issue. So really I need to do something like this: >> >> x[all.equal(x, 0.15)] >> >> But that doesn't work because all.equal wants to compare objects and not >> each element. >> >> I could do: >> >> x[round(x,2) ==0.15] >> >> It seems to work in this case, but as I've been working with my data I'm >> concerned its unreliable. What is the most efficient way of subsetting >> data using a machine-tolerance equal numeric value? >> >> Thanks R-Helpers. >> >> Peter >> >> [[alternative HTML version deleted]] >> >> ______________________________**________________ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/**listinfo/r-help<https://stat.ethz.ch/mailman/listinfo/r-help> >> PLEASE do read the posting guide http://www.R-project.org/** >> posting-guide.html <http://www.R-project.org/posting-guide.html> >> and provide commented, minimal, self-contained, reproducible code. >> >> [[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.
