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
extrat non diagonal
5 messages · malika yassa, S Ellison, Richard M. Heiberger +1 more
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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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]]
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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 <rmh at 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.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]]
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******************************************************************* 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
______________________________________________ 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.
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 <rmh at 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 <rmh at 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.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 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
______________________________________________ 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.
______________________________________________ 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.