> On 10 Oct 2015, at 13:39, peter dalgaard <pda...@gmail.com> wrote: > >> >> On 10 Oct 2015, at 10:49 , Berend Hasselman <b...@xs4all.nl> wrote: >> >>> >>> On 10 Oct 2015, at 10:00, David Winsemius <dwinsem...@comcast.net> wrote: >>> >>> >>> On Oct 9, 2015, at 10:57 PM, Steven Yen wrote: >>> >>>> Dear >>>> How do you construct a lower triangular matrix from a vector. >>>> >>>> I want to make vector >>>> >>>> a <- 1:10 >>>> >>>> into a triangular matrix >>>> >>>> 1 0 0 0 >>>> 2 3 0 0 >>>> 4 5 6 0 >>>> 7 8 9 10 >>>> >>> >>> I'm not sure this method with logical indexing will be the most elegant: >>> >>> ?lower.tri >>> ?col >>> >>>> b=matrix(0, sqrt(10)+1,sqrt(10)+1) >>> >>>> b[lower.tri(b)| row(b)==col(b)] <- 1:10 >>>> b >>> [,1] [,2] [,3] [,4] >>> [1,] 1 0 0 0 >>> [2,] 2 5 0 0 >>> [3,] 3 6 8 0 >>> [4,] 4 7 9 10 >>> >> >> That doesn’t seem to be what the OP wanted. >> >> This should do it. >> >> a <- 1:10 >> C <- matrix(0, sqrt(length(a))+1,sqrt(length(a))+1) >> i.upr <- which(upper.tri(C, diag = TRUE), arr.ind=TRUE) >> C[i.upr] <- a >> t(C) >> >> resulting in >> >> [,1] [,2] [,3] [,4] >> [1,] 1 0 0 0 >> [2,] 2 3 0 0 >> [3,] 4 5 6 0 >> [4,] 7 8 9 10 >> >> I found this here: >> http://stackoverflow.com/questions/24472060/indexing-upper-or-lower-triangle-in-matrix-with-diagonal > > It's crossing the creek a couple of times too many, though. This'll do: > >> M <- matrix(0,4,4) >> M[upper.tri(M,TRUE)] <- 1:10 >> t(M) > [,1] [,2] [,3] [,4] > [1,] 1 0 0 0 > [2,] 2 3 0 0 > [3,] 4 5 6 0 > [4,] 7 8 9 10 > > I'm also not buying the sqrt(length(a))+1 bit --- floor(2*length(a)) is more > like it. >
Shouldn’t that be m <- sqrt(floor(2*length(a))) M <- matrix(0,m,m) for the general case? Berend ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
