Hi how many objects do you have in a list? Loop is ineffective if you use several nested loops and/or there is some unnecessary mimic of simple function.
e.g. you can use this set.seed(666) x <- runif(10) vysled <- 0 for(i in 1:length(x)) { if(vysled > x[i]) vysled <- vysled else vysled <- x[i] } vysled but I would prefer max(x) If you do not perceive performance issues there is usually no need to elaborate *apply. Regards Petr r-help-boun...@r-project.org napsal dne 10.02.2010 11:59:21: > After reading the R news, I've tried this code and it works: > > rapply(list(names(test),test), write.csv, file="filename.csv", > append=T, row.names=F) > > However, the output is structured like this: > names(test)[[1]] > names(test)[[2]] > etc... > test[[1]] > test[[2]] > etc... > > I would like to alternate names(test) and test in the output. The > desired output would be structured like this: > names(test)[[1]] > test[[1]] > names(test)[[2]] > test[[2]] > etc... > > Can someone guide me for that step? Better solutions for the whole thing > are of course welcomed! > > Thanks a lot > Ivan > > Le 2/10/2010 10:50, Ivan Calandra a écrit : > > Hi everybody! > > > > I'm still quite new in R and I don't really understand this whole > > "vectorization" thing. > > For now, I use a for loop (and it works fine), but I think it would be > > useful to replace it. > > > > I want to export the result of a test statistic, which is stored as a > > list, into a csv file. I therefore have to export each element of the > > list separately. > > Here is the code: > > ---- > > str(test) > > List of 3 > > $ output : num [1:15, 1:6] 1 2 3 4 5 6 7 8 9 10 ... > > ..- attr(*, "dimnames")=List of 2 > > .. ..$ : NULL > > .. ..$ : chr [1:6] "con.num" "psihat" "p.value" "p.crit" ... > > $ con : num [1:6, 1:15] 1 -1 0 0 0 0 1 0 -1 0 ... > > $ num.sig: int 0 > > > > for (i in 1:3){ > > write.csv(test[[i]], file="filename.csv", append=T, quote=F, > > row.names=T) > > } > > ---- > > > > As I said, I don't completely understand, but I think one of these > > "apply" function might do what I need. > > > > Thanks in advance > > Ivan > > > > ______________________________________________ > > 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. ______________________________________________ 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.