Another way:
> A <- matrix(1:9,3,3,
dimnames=list(Row=paste0("r",1:3),Col=paste0("c",1:3)))
> A
Col
Row c1 c2 c3
r1 1 4 7
r2 2 5 8
r3 3 6 9
> matrix( A[row(A)!=col(A)], nrow(A)-1, ncol(A), dimnames=list(NULL,
colnames(A)))
c1 c2 c3
[1,] 2 4 7
[2,] 3 6 8
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Nov 14, 2018 at 2:32 PM, Richard M. Heiberger <r...@temple.edu>
wrote:
> An even better solution because it has fewer steps.
>
> A <- matrix(1:9, 3, 3)
> A
> B <- A[-1, ]
> 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 <- A[-1, ]
> > 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 2:09 PM Richard M. Heiberger <r...@temple.edu>
> wrote:
> >
> > 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.elli...@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]]
> > > >
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> ______________________________________________
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