Hi lapply should do it if values is a data frame with first column of names and second column numbers.
lapply(v, function(x) values[match(x, values[ ,1]), 2]) Regards Petr r-help-boun...@r-project.org napsal dne 14.04.2009 14:01:50: > Dear All, > > Here's my problem. I have two lists: > > > v > [[1]] > [1] "five" > [[2]] > [1] "four" > [[3]] > [1] "three" > [[4]] > [1] "two" > [[5]] > [1] "one" > [[6]] > [1] "six" > [[7]] > [1] "five" "four" "three" "two" "one" "six" > [[8]] > [1] "four" "three" "two" "one" "six" > [[9]] > [1] "three" "two" "one" "six" > [[10]] > [1] "two" "one" "six" > > > and the other is something like: > > >values > numbers > five 50 > four 40 > three 30 > two 10 > one 10 > six 10 > > > > > > I'm trying to get a list similar to "v", but in which the entries "three", > "two", and so on are replaced by the values in the second list. Any help will > be greatly appreciated. > > best regards, > > Manoel > > > > [[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.