Try posting your data using 'dput' so it is easily read for testing. On Mon, Oct 18, 2010 at 11:17 AM, Gregory Ryslik <rsa...@comcast.net> wrote: > Unfortunately, that gives me null everywhere. Here's the data I have for > all.predicted.values and max.growth. Perhaps this will help. Thus I want > all.predicted.values[[1]][[4]] then all.predicted.values[[2]][3]] and then > all.predicted.values[[3]][[4]]. > > I've attached what your statement outputs at the end. > > Thanks again! > > Browse[2]> max.growth > [[1]] > [1] 4 > > [[2]] > [1] 3 > > [[3]] > [1] 4 > > Browse[2]> all.predicted.values > [[1]] > [[1]][[1]] > [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 > 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 > [55] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 > 0 0 0 0 0 0 0 0 0 0 > Levels: 0 1 2 > > [[1]][[2]] > [1] 2 2 2 0 2 0 2 2 2 2 2 2 2 2 0 2 2 2 2 2 2 2 2 2 2 0 0 2 2 2 2 0 0 0 2 2 > 0 0 2 2 0 2 2 2 2 2 0 2 2 2 0 2 2 0 > [55] 0 0 2 0 2 0 0 0 0 2 2 2 2 0 2 2 2 0 2 2 0 0 2 2 2 2 2 2 2 0 0 0 2 0 2 2 > 2 2 0 2 2 2 0 2 0 0 > Levels: 0 1 2 > > [[1]][[3]] > [1] 0 0 0 0 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 2 0 2 2 2 0 0 0 2 0 0 2 0 0 0 0 0 > 0 0 2 0 0 0 0 0 2 2 0 0 0 2 0 0 0 0 > [55] 0 0 2 0 2 0 0 0 0 2 2 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 > 2 0 0 0 0 0 0 2 0 0 > Levels: 0 1 2 > > [[1]][[4]] > [1] 0 0 0 0 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 2 0 2 2 2 0 0 0 2 0 0 2 0 0 0 0 0 > 0 0 2 0 0 0 0 0 2 2 0 0 0 2 0 0 0 0 > [55] 0 0 2 0 2 0 0 0 0 2 2 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 > 2 0 0 0 0 0 0 2 0 0 > Levels: 0 1 2 > > > [[2]] > [[2]][[1]] > [1] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 > 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 > [55] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 > 2 2 2 2 2 2 2 2 2 2 > Levels: 0 1 2 > > [[2]][[2]] > [1] 2 2 2 2 1 2 2 2 2 2 1 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 1 2 > 2 2 1 2 2 1 1 2 2 2 2 2 2 2 2 1 2 2 > [55] 2 2 2 2 1 2 2 2 2 1 2 2 1 1 1 2 2 2 1 2 1 2 1 2 1 2 2 2 1 1 2 2 1 2 2 1 > 1 2 1 1 1 2 2 1 2 2 > Levels: 0 1 2 > > [[2]][[3]] > [1] 2 2 2 0 1 2 2 2 2 2 1 2 2 2 0 1 2 1 2 2 2 2 2 2 2 0 0 2 1 2 2 2 0 0 1 2 > 0 0 1 2 0 1 1 2 2 2 0 2 2 2 0 1 2 2 > [55] 0 2 2 2 1 0 0 0 0 1 2 2 1 1 1 2 2 0 1 2 1 0 1 2 1 2 2 2 1 1 2 2 1 2 2 1 > 1 2 1 1 1 2 2 1 0 2 > Levels: 0 1 2 > > > [[3]] > [[3]][[1]] > [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 > 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 > [55] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 > 0 0 0 0 0 0 0 0 0 0 > Levels: 0 1 2 > > [[3]][[2]] > [1] 2 2 2 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 0 0 2 2 2 2 2 0 0 2 2 > 2 0 2 2 0 2 2 2 2 2 0 2 2 2 0 2 2 2 > [55] 0 2 2 2 2 2 0 0 2 2 2 2 2 2 2 2 2 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 > 2 2 2 2 2 2 2 2 2 2 > Levels: 0 1 2 > > [[3]][[3]] > [1] 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 > 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 > [55] 0 0 0 0 1 0 0 0 0 1 0 0 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 1 1 0 0 1 0 0 1 > 1 0 1 1 1 0 0 1 1 0 > Levels: 0 1 2 > > [[3]][[4]] > [1] 2 2 2 0 1 0 2 2 0 2 1 2 2 0 0 1 1 1 1 0 2 0 0 0 2 0 0 0 1 2 0 0 0 0 1 2 > 0 0 1 2 0 1 1 2 0 0 0 2 2 0 0 1 2 0 > [55] 0 0 0 0 1 0 0 0 0 1 0 2 1 1 1 2 0 0 1 2 1 1 1 2 1 2 2 2 1 1 0 0 1 0 2 1 > 1 2 1 1 1 2 0 1 1 0 > Levels: 0 1 2 > > > Browse[2]> > predicted.values.for.max.growth<-diag(sapply(all.predicted.values,'[[','max.growth')) > Browse[2]> predicted.values.for.max.growth > [[1]] > NULL > > [[2]] > [1] 0 > > [[3]] > [1] 0 > > [[4]] > [1] 0 > > [[5]] > NULL > > [[6]] > [1] 0 > > [[7]] > [1] 0 > > [[8]] > [1] 0 > > [[9]] > NULL > > > > On Oct 18, 2010, at 11:08 AM, Henrique Dallazuanna wrote: > >> Try this: >> >> diag(sapply(all.predicted.values, '[[', 'max.growth')) >> >> >> On Mon, Oct 18, 2010 at 12:59 PM, Gregory Ryslik <rsa...@comcast.net> wrote: >> Hi, >> >> I have a list of n items and the ith element has m_i elements within it. >> >> I want to do something like: >> >> predicted.values<- lapply(all.predicted.values,'[[',max.growth[[i]]) >> >> Where max.growth[[i]] is the element I want to extract from each of the ith >> predicted elements. Thus, for example, I want to extract the max.growth[[1]] >> element from all.predicted.values[[1]] (which is itself a list). Then I >> want to extract max.growth[[2]] element from all.predicted.values[[2]]. >> >> I realize I can do this with a for loop but then if I can do this as one >> line that would be preferable. >> >> Thanks! >> >> Greg >> [[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. >> >> >> >> -- >> Henrique Dallazuanna >> Curitiba-Paraná-Brasil >> 25° 25' 40" S 49° 16' 22" O > > > [[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. > >
-- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? ______________________________________________ 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.
