Here's a toy example which you can apply the logic of: dfr <- expand.grid(1:3,1:2) results <- apply(dfr, 1, sum) dfr[results==4,]
On 25 January 2013 22:19, Andras Farkas <motyoc...@yahoo.com> wrote: > > Dear All > > I have the following data (somewhat simplyfied): > > TINF <-1 > a <-c(500,750,1000,1250,1500,1750,2000) > b <-c(8,12,18,24,36,48,60,72,96) > > following function: > > infcprodessa <-function (D, tin, tau, ts) > (D * (1 - exp(-0.048 * tin))/(tin * (0.048*79) * (1 - exp(-0.048 * tau)))) > * exp(-0.048 * (ts - tin)) > > z <-sapply(1:1, function(n) infcprodessa(1000,TINF,12,12-TINF)) > > is there a way to select the combination of respective a and b values that > would result in a calculated z that is between 15 and 20? In this case the a > would be 1000 and the b would be 12 (other combinations are also possible), > but how could I automatically find them? perhaps a loop? > > Apreciate the help, > > Sincerely, > > Andras > [[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. > ______________________________________________ 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.
