hello for examplei have this matrix w2<-c(0.1,0.2,0.4,0.2,0.4,0.1)aa<-matrix(w1,nrow=3,ncol=3)aa ???? [,1] [,2] [,3] [1,]? 0.4? 0.4? 0.4 [2,]? 0.1? 0.1? 0.1 [3,]? 0.2? 0.2? 0.2 if i use this code matrix(as.numeric(aa)[!as.numeric(aa) %in% diag(aa)],2,3) i will obtaine this matrix[,1] [,2] [,3] [1,]?? NA?? NA?? NA [2,]?? NA?? NA?? NA but me i want this matrix[,1] [,2] [,3] [1,]? 0.1? 0.4? 0.4 [2,]? 0.2? 0.2? 0.1 thank you [[alternative HTML version deleted]]
i) Your code creates w2 but references w1 to create aa. So you needed aa <- matrix(rep(c(0.4, 0.1, 0.2), 3), 3,3) for a working example. ii) This> matrix(as.numeric(aa)[!as.numeric(aa) %in% diag(aa)],2,3)removes any value that is present in the diagonal of aa. Look up ?"%in%" to see what that does; it returns TRUE whenever anything in as.numeric(aa) matches anything in your diagonal. All the values in aa match one of c(0.4, 0.1, 0.2). So since your whole matrix consists of these three numbers, you told R to leave out everything in aa and then create a 2x3 matrix with the result. Hence the NAs iii) If you want to extract odd parts of a matrix explicitly, see ?"[" and particularly the section on indexing using arrays iv) You can use logical indexing. In the special case of the diagonal, you can use diag() to create a matrix of logicals, logically negate that and apply that to your matrix: aa[ !diag(rep(TRUE, 3)) ] and, in twoi rows: matrix( aa[ !diag(rep(TRUE, 3)) ], 2,3)> for examplei have this matrix > w2<-c(0.1,0.2,0.4,0.2,0.4,0.1) > aa<-matrix(w1,nrow=3,ncol=3) > aa > ???? [,1] [,2] [,3] > [1,]? 0.4? 0.4? 0.4 > [2,]? 0.1? 0.1? 0.1 > [3,]? 0.2? 0.2? 0.2 > > if i use this code > matrix(as.numeric(aa)[!as.numeric(aa) %in% diag(aa)],2,3) > > i will obtaine this matrix[,1] [,2] [,3] > [1,]?? NA?? NA?? NA > [2,]?? NA?? NA?? NA > > but me i want this matrix[,1] [,2] [,3] > [1,]? 0.1? 0.4? 0.4 > [2,]? 0.2? 0.2? 0.1 > > thank you > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at 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.******************************************************************* This email and any attachments are confidential. Any use, copying or disclosure other than by the intended recipient is unauthorised. If you have received this message in error, please notify the sender immediately via +44(0)20 8943 7000 or notify postmaster at lgcgroup.com and delete this message and any copies from your computer and network. LGC Limited. Registered in England 2991879. Registered office: Queens Road, Teddington, Middlesex, TW11 0LY, UK
Steve's method is very slick. I think this is a bit easier to understand. A <- matrix(1:9, 3, 3) A B <- matrix(nrow=2, ncol=3) B[lower.tri(B, diag=TRUE)] <- A[lower.tri(A)] B[upper.tri(B, diag=FALSE)] <- A[upper.tri(A)] B> A <- matrix(1:9, 3, 3) > A[,1] [,2] [,3] [1,] 1 4 7 [2,] 2 5 8 [3,] 3 6 9> B <- matrix(nrow=2, ncol=3) > B[lower.tri(B, diag=TRUE)] <- A[lower.tri(A)] > B[upper.tri(B, diag=FALSE)] <- A[upper.tri(A)] > B[,1] [,2] [,3] [1,] 2 4 7 [2,] 3 6 8>On Wed, Nov 14, 2018 at 11:04 AM S Ellison <S.Ellison at lgcgroup.com> wrote:> > i) Your code creates w2 but references w1 to create aa. > > So you needed > aa <- matrix(rep(c(0.4, 0.1, 0.2), 3), 3,3) > for a working example. > > ii) This > > matrix(as.numeric(aa)[!as.numeric(aa) %in% diag(aa)],2,3) > removes any value that is present in the diagonal of aa. Look up ?"%in%" to see what that does; it returns TRUE whenever anything in as.numeric(aa) matches anything in your diagonal. All the values in aa match one of c(0.4, 0.1, 0.2). So since your whole matrix consists of these three numbers, you told R to leave out everything in aa and then create a 2x3 matrix with the result. Hence the NAs > > iii) If you want to extract odd parts of a matrix explicitly, see ?"[" and particularly the section on indexing using arrays > > iv) You can use logical indexing. In the special case of the diagonal, you can use diag() to create a matrix of logicals, logically negate that and apply that to your matrix: > aa[ !diag(rep(TRUE, 3)) ] > > and, in twoi rows: > matrix( aa[ !diag(rep(TRUE, 3)) ], 2,3) > > > for examplei have this matrix > > w2<-c(0.1,0.2,0.4,0.2,0.4,0.1) > > aa<-matrix(w1,nrow=3,ncol=3) > > aa > > [,1] [,2] [,3] > > [1,] 0.4 0.4 0.4 > > [2,] 0.1 0.1 0.1 > > [3,] 0.2 0.2 0.2 > > > > if i use this code > > matrix(as.numeric(aa)[!as.numeric(aa) %in% diag(aa)],2,3) > > > > i will obtaine this matrix[,1] [,2] [,3] > > [1,] NA NA NA > > [2,] NA NA NA > > > > but me i want this matrix[,1] [,2] [,3] > > [1,] 0.1 0.4 0.4 > > [2,] 0.2 0.2 0.1 > > > > thank you > > > > [[alternative HTML version deleted]] > > > > ______________________________________________ > > R-help at 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. > > > ******************************************************************* > This email and any attachments are confidential. Any u...{{dropped:14}}