Completing MDS manually in R
$begingroup$
Given a matrix A, I want to complete Multidimensional Scaling by hand, instead of using any given R functions.
As such, I have calculated the centered matrix B with the following code:
n<-nrow(A)
id<-diag(n)
e<-diag(id)
H <- id - (1/n)*e %*% etranspose
B <- (-1/2)* H %*% A %*% H
My question is: how can I use my B matrix to complete multidimensional scaling on my A matrix, without the cmdscale function or anything along those lines?
machine-learning statistics dimensionality-reduction
edited Dec 28 '18 at 0:50
Alex L
1478
asked Nov 16 '15 at 1:53
potatosouppotatosoup
262
$endgroup$
bumped to the homepage by Community♦ 4 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
Given a matrix A, I want to complete Multidimensional Scaling by hand, instead of using any given R functions.
As such, I have calculated the centered matrix B with the following code:
n<-nrow(A)
id<-diag(n)
e<-diag(id)
H <- id - (1/n)*e %*% etranspose
B <- (-1/2)* H %*% A %*% H
My question is: how can I use my B matrix to complete multidimensional scaling on my A matrix, without the cmdscale function or anything along those lines?
machine-learning statistics dimensionality-reduction
edited Dec 28 '18 at 0:50
Alex L
1478
asked Nov 16 '15 at 1:53
potatosouppotatosoup
262
$endgroup$
bumped to the homepage by Community♦ 4 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
Given a matrix A, I want to complete Multidimensional Scaling by hand, instead of using any given R functions.
As such, I have calculated the centered matrix B with the following code:
n<-nrow(A)
id<-diag(n)
e<-diag(id)
H <- id - (1/n)*e %*% etranspose
B <- (-1/2)* H %*% A %*% H
My question is: how can I use my B matrix to complete multidimensional scaling on my A matrix, without the cmdscale function or anything along those lines?
machine-learning statistics dimensionality-reduction
edited Dec 28 '18 at 0:50
Alex L
1478
asked Nov 16 '15 at 1:53
potatosouppotatosoup
262
$endgroup$
Given a matrix A, I want to complete Multidimensional Scaling by hand, instead of using any given R functions.
As such, I have calculated the centered matrix B with the following code:
n<-nrow(A)
id<-diag(n)
e<-diag(id)
H <- id - (1/n)*e %*% etranspose
B <- (-1/2)* H %*% A %*% H
My question is: how can I use my B matrix to complete multidimensional scaling on my A matrix, without the cmdscale function or anything along those lines?
machine-learning statistics dimensionality-reduction
machine-learning statistics dimensionality-reduction
edited Dec 28 '18 at 0:50
Alex L
1478
asked Nov 16 '15 at 1:53
potatosouppotatosoup
262
edited Dec 28 '18 at 0:50
Alex L
1478
asked Nov 16 '15 at 1:53
potatosouppotatosoup
262
edited Dec 28 '18 at 0:50
Alex L
1478
edited Dec 28 '18 at 0:50
Alex L
1478
edited Dec 28 '18 at 0:50
Alex L
1478
1478
asked Nov 16 '15 at 1:53
potatosouppotatosoup
262
asked Nov 16 '15 at 1:53
potatosouppotatosoup
262
asked Nov 16 '15 at 1:53
potatosouppotatosoup
262
262
bumped to the homepage by Community♦ 4 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 4 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
1 Answer
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$begingroup$
If you're interested in the classic MDS algorithm, it is spelled out very nicely on the Wikipedia page:
Classical MDS uses the fact that the coordinate matrix can be derived
by eigenvalue decomposition from $B=XX'$. And the matrix $B$ can be
computed from proximity matrix $D$ by using double centering.
- Set up the squared proximity matrix $D^{(2)}=[d_{ij}^{2}]$
Apply double centering: $B=-{frac {1}{2}}JD^{(2)}J$ using the centering matrix $J=I-{frac {1}{n}}11'$, where $n$ is the number of
objects.
Determine the $m$ largest eigenvalues $lambda _{1},lambda _{2},...,lambda _{m}$ and corresponding eigenvectors $e_{1},e_{2},...,e_{m}$ of $B$ (where $m$ is the number of dimensions
desired for the output).
Now, $X=E_{m}Lambda _{m}^{1/2}$, where $E_{m}$ is the matrix of $m$ eigenvectors and $Lambda _{m}$ is the diagonal matrix of $m$
eigenvalues of $B$.
I can't really tell what your $A$ matrix is, but if each entry is a measure of the distance from the $i^{th}$ to $j^{th}$ entry, then that would be your proximity matrix.
Also, $11'$ is an n-by-n matrix of all 1's. This should be fairly self-explanatory, however most of these terms can be Googled if you're unsure. It really shouldn't be any more than about 10 lines of code.
answered Dec 27 '18 at 22:48
Alex LAlex L
1478
$endgroup$
add a comment |
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$begingroup$
If you're interested in the classic MDS algorithm, it is spelled out very nicely on the Wikipedia page:
Classical MDS uses the fact that the coordinate matrix can be derived
by eigenvalue decomposition from $B=XX'$. And the matrix $B$ can be
computed from proximity matrix $D$ by using double centering.
- Set up the squared proximity matrix $D^{(2)}=[d_{ij}^{2}]$
Apply double centering: $B=-{frac {1}{2}}JD^{(2)}J$ using the centering matrix $J=I-{frac {1}{n}}11'$, where $n$ is the number of
objects.
Determine the $m$ largest eigenvalues $lambda _{1},lambda _{2},...,lambda _{m}$ and corresponding eigenvectors $e_{1},e_{2},...,e_{m}$ of $B$ (where $m$ is the number of dimensions
desired for the output).
Now, $X=E_{m}Lambda _{m}^{1/2}$, where $E_{m}$ is the matrix of $m$ eigenvectors and $Lambda _{m}$ is the diagonal matrix of $m$
eigenvalues of $B$.
I can't really tell what your $A$ matrix is, but if each entry is a measure of the distance from the $i^{th}$ to $j^{th}$ entry, then that would be your proximity matrix.
Also, $11'$ is an n-by-n matrix of all 1's. This should be fairly self-explanatory, however most of these terms can be Googled if you're unsure. It really shouldn't be any more than about 10 lines of code.
answered Dec 27 '18 at 22:48
Alex LAlex L
1478
$endgroup$
add a comment |
$begingroup$
If you're interested in the classic MDS algorithm, it is spelled out very nicely on the Wikipedia page:
Classical MDS uses the fact that the coordinate matrix can be derived
by eigenvalue decomposition from $B=XX'$. And the matrix $B$ can be
computed from proximity matrix $D$ by using double centering.
- Set up the squared proximity matrix $D^{(2)}=[d_{ij}^{2}]$
Apply double centering: $B=-{frac {1}{2}}JD^{(2)}J$ using the centering matrix $J=I-{frac {1}{n}}11'$, where $n$ is the number of
objects.
Determine the $m$ largest eigenvalues $lambda _{1},lambda _{2},...,lambda _{m}$ and corresponding eigenvectors $e_{1},e_{2},...,e_{m}$ of $B$ (where $m$ is the number of dimensions
desired for the output).
Now, $X=E_{m}Lambda _{m}^{1/2}$, where $E_{m}$ is the matrix of $m$ eigenvectors and $Lambda _{m}$ is the diagonal matrix of $m$
eigenvalues of $B$.
I can't really tell what your $A$ matrix is, but if each entry is a measure of the distance from the $i^{th}$ to $j^{th}$ entry, then that would be your proximity matrix.
Also, $11'$ is an n-by-n matrix of all 1's. This should be fairly self-explanatory, however most of these terms can be Googled if you're unsure. It really shouldn't be any more than about 10 lines of code.
answered Dec 27 '18 at 22:48
Alex LAlex L
1478
$endgroup$
add a comment |
$begingroup$
If you're interested in the classic MDS algorithm, it is spelled out very nicely on the Wikipedia page:
Classical MDS uses the fact that the coordinate matrix can be derived
by eigenvalue decomposition from $B=XX'$. And the matrix $B$ can be
computed from proximity matrix $D$ by using double centering.
- Set up the squared proximity matrix $D^{(2)}=[d_{ij}^{2}]$
Apply double centering: $B=-{frac {1}{2}}JD^{(2)}J$ using the centering matrix $J=I-{frac {1}{n}}11'$, where $n$ is the number of
objects.
Determine the $m$ largest eigenvalues $lambda _{1},lambda _{2},...,lambda _{m}$ and corresponding eigenvectors $e_{1},e_{2},...,e_{m}$ of $B$ (where $m$ is the number of dimensions
desired for the output).
Now, $X=E_{m}Lambda _{m}^{1/2}$, where $E_{m}$ is the matrix of $m$ eigenvectors and $Lambda _{m}$ is the diagonal matrix of $m$
eigenvalues of $B$.
I can't really tell what your $A$ matrix is, but if each entry is a measure of the distance from the $i^{th}$ to $j^{th}$ entry, then that would be your proximity matrix.
Also, $11'$ is an n-by-n matrix of all 1's. This should be fairly self-explanatory, however most of these terms can be Googled if you're unsure. It really shouldn't be any more than about 10 lines of code.
answered Dec 27 '18 at 22:48
Alex LAlex L
1478
$endgroup$
If you're interested in the classic MDS algorithm, it is spelled out very nicely on the Wikipedia page:
Classical MDS uses the fact that the coordinate matrix can be derived
by eigenvalue decomposition from $B=XX'$. And the matrix $B$ can be
computed from proximity matrix $D$ by using double centering.
- Set up the squared proximity matrix $D^{(2)}=[d_{ij}^{2}]$
Apply double centering: $B=-{frac {1}{2}}JD^{(2)}J$ using the centering matrix $J=I-{frac {1}{n}}11'$, where $n$ is the number of
objects.
Determine the $m$ largest eigenvalues $lambda _{1},lambda _{2},...,lambda _{m}$ and corresponding eigenvectors $e_{1},e_{2},...,e_{m}$ of $B$ (where $m$ is the number of dimensions
desired for the output).
Now, $X=E_{m}Lambda _{m}^{1/2}$, where $E_{m}$ is the matrix of $m$ eigenvectors and $Lambda _{m}$ is the diagonal matrix of $m$
eigenvalues of $B$.
I can't really tell what your $A$ matrix is, but if each entry is a measure of the distance from the $i^{th}$ to $j^{th}$ entry, then that would be your proximity matrix.
Also, $11'$ is an n-by-n matrix of all 1's. This should be fairly self-explanatory, however most of these terms can be Googled if you're unsure. It really shouldn't be any more than about 10 lines of code.
answered Dec 27 '18 at 22:48
Alex LAlex L
1478
answered Dec 27 '18 at 22:48
Alex LAlex L
1478
answered Dec 27 '18 at 22:48
Alex LAlex L
1478
answered Dec 27 '18 at 22:48
Alex LAlex L
1478
1478
add a comment |
add a comment |
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