Hi again, I just wanted to thank folks for their suggestions; they led me to understand the *apply family a little better, and to realize the issue was really a question of how to convert a list of equal length vectors into a matrix. In this case sapply only needs to be asked to identify these vectors individually; I don't know if R has the equivalent of an identity function, but the following solution accomplishes this:
> splitvectors <- sapply(splitlist, function(x) x) > splitvectors [,1] [,2] [1,] "a1" "a2" [2,] "b1" "b2" or, by replacing the anonymous function by c, we obtain a more elegant but more wasteful solution. Thanks again for everyone's help, David Romano On Fri, Sep 7, 2012 at 11:12 AM, David Romano <roma...@grinnell.edu> wrote: > Hi folks, > > Suppose I create the character vector charvec by > > > charvec<-c("a1.b1","a2.b2") > > charvec > [1] "a1.b1" "a2.b2" > > and then I use strsplit on charvec as follows: > > > splitlist<-strsplit(charvec,split=".",fixed=TRUE) > > splitlist > [[1]] > [1] "a1" "b1" > > [[2]] > [1] "a2" "b2" > > > I was wondering whether there is already a function which can extract > the "a" and "b" parts of the list splitlist; that is, that can return > the same vectors as those created by c("a1","a2") and c("b1","b2"). > > Thanks, > David Romano > [[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.