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.
