Always amazes me how many ways there are to do these things, none of which I was able to find myself. Thanks! I think the key here was ‘pmin,’ which I didn’t know before.
Michael Ashton, CFA Managing Principal Enduring Investments LLC W: 973.457.4602 C: 551.655.8006 Schedule a Call: https://calendly.com/m-ashton From: Bert Gunter [mailto:bgunter.4...@gmail.com] Sent: Wednesday, May 27, 2020 2:22 PM To: Rui Barradas Cc: Michael Ashton; r-help@r-project.org Subject: Re: [R] struggling with apply Better, I think (no indexing): t(apply(somematrix,1,function(x)pmin(x,UB))) Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Wed, May 27, 2020 at 10:56 AM Rui Barradas <ruipbarra...@sapo.pt<mailto:ruipbarra...@sapo.pt>> wrote: Hello, Try pmin. And loop by column/UB index with sapply/seq_along. sapply(seq_along(UB), function(i) pmin(UB[i], somematrix[,i])) # [,1] [,2] [,3] [,4] #[1,] 1.0 5.5 8.5 7.0 #[2,] 2.5 3.0 8.0 10.5 #[3,] 2.5 5.5 5.0 10.5 Hope this helps, Rui Barradas Às 18:46 de 27/05/20, Michael Ashton escreveu: > Hi - > > I have a matrix of n rows and 4 columns. > > I want to cap the value in each column by a different upper bound. So, > suppose my matrix is > > somematrix <- matrix(c(1,4,3,6,3,9,12,8,5,7,11,11),nrow=3,ncol=4) >> somematrix > [,1] [,2] [,3] [,4] > [1,] 1 6 12 7 > [2,] 4 3 8 11 > [3,] 3 9 5 11 > > Now I want to have the maximum value in each column described by > UB=c(2.5, 5.5, 8.5, 10.5) > > So that the right answer will look like: > [,1] [,2] [,3] [,4] > [1,] 1 5.5 8.5 7 > [2,] 2.5 3 8 10.5 > [3,] 2.5 5.5 5 10.5 > > I've tried a few things, like: > newmatrix <- apply(somematrix,c(1,2),function(x) min(UB,x)) > > but I can't figure out to apply the relevant element of the UB list to the > right element of the matrix. When I run the above, for example, it takes > min(UB,x) over all UB, so I get: > > newmatrix > [,1] [,2] [,3] [,4] > [1,] 1.0 2.5 2.5 2.5 > [2,] 2.5 2.5 2.5 2.5 > [3,] 2.5 2.5 2.5 2.5 > > I'm sure there's a simple and elegant solution but I don't know what it is! > > Thanks in advance, > > Mike > > Michael Ashton, CFA > Managing Principal > > Enduring Investments LLC > W: 973.457.4602 > C: 551.655.8006 > Schedule a Call: https://calendly.com/m-ashton > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help@r-project.org<mailto: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. > ______________________________________________ R-help@r-project.org<mailto: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. [[alternative HTML version deleted]] ______________________________________________ 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.
