Since all components of cutYield have the same levels, one could do this to ensure that all levels are represented:
factor(unlist(lapply(cutYield, as.character)), levels = levels(cutYield[[1]])) On 3/15/06, Sebastian Luque <[EMAIL PROTECTED]> wrote: > "Gabor Grothendieck" <[EMAIL PROTECTED]> wrote: > > > Is this ok or is it what you are trying to avoid: > > > factor(unlist(lapply(cutYield, as.character))) > > Thank you Gabor. The problem with that is what if some levels do not > appear in any member of cutYield? In that case, the factor created above > would contain fewer levels than those present in every member of cutYield. > > Cheers, > > -- > Sebastian P. Luque > > ______________________________________________ > 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