Jaccard index, matrix notation
$begingroup$
I have a matrix with rows representing events and columns representing users. The elements of the matrix are binary values indicating if a user has attended the event or not.
begin{bmatrix}1&1&0&1&1\1&1&0&0&1\ 1&0&0&1&1end{bmatrix}
I need matrix notation to compute the Jaccard distances between users.
begin{align}
J(U_1,U_2)=frac{|U_1cap U_2|}{|U_1cup U_2|}
end{align}
To compute the numerator I can use the matrix operation
begin{align}
A^Ttimes A
end{align}
Now my question is how to get the denominator of Jaccard index using the matrix notation.
matrices
asked Jan 29 '17 at 12:43
MortyMorty
102
$endgroup$
add a comment |
$begingroup$
I have a matrix with rows representing events and columns representing users. The elements of the matrix are binary values indicating if a user has attended the event or not.
begin{bmatrix}1&1&0&1&1\1&1&0&0&1\ 1&0&0&1&1end{bmatrix}
I need matrix notation to compute the Jaccard distances between users.
begin{align}
J(U_1,U_2)=frac{|U_1cap U_2|}{|U_1cup U_2|}
end{align}
To compute the numerator I can use the matrix operation
begin{align}
A^Ttimes A
end{align}
Now my question is how to get the denominator of Jaccard index using the matrix notation.
matrices
asked Jan 29 '17 at 12:43
MortyMorty
102
$endgroup$
add a comment |
$begingroup$
I have a matrix with rows representing events and columns representing users. The elements of the matrix are binary values indicating if a user has attended the event or not.
begin{bmatrix}1&1&0&1&1\1&1&0&0&1\ 1&0&0&1&1end{bmatrix}
I need matrix notation to compute the Jaccard distances between users.
begin{align}
J(U_1,U_2)=frac{|U_1cap U_2|}{|U_1cup U_2|}
end{align}
To compute the numerator I can use the matrix operation
begin{align}
A^Ttimes A
end{align}
Now my question is how to get the denominator of Jaccard index using the matrix notation.
matrices
asked Jan 29 '17 at 12:43
MortyMorty
102
$endgroup$
I have a matrix with rows representing events and columns representing users. The elements of the matrix are binary values indicating if a user has attended the event or not.
begin{bmatrix}1&1&0&1&1\1&1&0&0&1\ 1&0&0&1&1end{bmatrix}
I need matrix notation to compute the Jaccard distances between users.
begin{align}
J(U_1,U_2)=frac{|U_1cap U_2|}{|U_1cup U_2|}
end{align}
To compute the numerator I can use the matrix operation
begin{align}
A^Ttimes A
end{align}
Now my question is how to get the denominator of Jaccard index using the matrix notation.
matrices
matrices
asked Jan 29 '17 at 12:43
MortyMorty
102
asked Jan 29 '17 at 12:43
MortyMorty
102
asked Jan 29 '17 at 12:43
MortyMorty
102
asked Jan 29 '17 at 12:43
MortyMorty
102
asked Jan 29 '17 at 12:43
MortyMorty
102
102
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Define the vectors
$$eqalign{
b &= A^T1 cr
p &= exp(b) cr
}$$
and the matrices
$$eqalign{
L &= log(pp^T) cr
J &= (A^TA)oslash(L-A^TA) cr
}$$
Note that the log and exp functions are applied elementwise, $oslash$ represents elementwise division, and $1$ is a vector of all ones.
The elements of the $J$-matrix are the Jaccard distances, i.e.
$,,J_{ik} = J(U_i,U_k)$
The third column of your $A$-matrix is problematic since it results in $,J_{33}=frac{0}{0}$
There may be better ways of generating the $L$-matrix. However it is done, its elements must satisfy $$L_{ik} = b_i+b_k$$
Update
A much better way to calculate the $L$-matrix is
$$L = b1^T + 1b^T = A^TU + U^TA$$
where $U$ is a matrix of all ones which has the same shape as $A$.
Now the $J$-matrix can be written as
$$J=(A^TA)oslash(A^TU + U^TA - A^TA)$$
answered Dec 9 '18 at 3:27
greggreg
7,8701821
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
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votes
$begingroup$
Define the vectors
$$eqalign{
b &= A^T1 cr
p &= exp(b) cr
}$$
and the matrices
$$eqalign{
L &= log(pp^T) cr
J &= (A^TA)oslash(L-A^TA) cr
}$$
Note that the log and exp functions are applied elementwise, $oslash$ represents elementwise division, and $1$ is a vector of all ones.
The elements of the $J$-matrix are the Jaccard distances, i.e.
$,,J_{ik} = J(U_i,U_k)$
The third column of your $A$-matrix is problematic since it results in $,J_{33}=frac{0}{0}$
There may be better ways of generating the $L$-matrix. However it is done, its elements must satisfy $$L_{ik} = b_i+b_k$$
Update
A much better way to calculate the $L$-matrix is
$$L = b1^T + 1b^T = A^TU + U^TA$$
where $U$ is a matrix of all ones which has the same shape as $A$.
Now the $J$-matrix can be written as
$$J=(A^TA)oslash(A^TU + U^TA - A^TA)$$
answered Dec 9 '18 at 3:27
greggreg
7,8701821
$endgroup$
add a comment |
$begingroup$
Define the vectors
$$eqalign{
b &= A^T1 cr
p &= exp(b) cr
}$$
and the matrices
$$eqalign{
L &= log(pp^T) cr
J &= (A^TA)oslash(L-A^TA) cr
}$$
Note that the log and exp functions are applied elementwise, $oslash$ represents elementwise division, and $1$ is a vector of all ones.
The elements of the $J$-matrix are the Jaccard distances, i.e.
$,,J_{ik} = J(U_i,U_k)$
The third column of your $A$-matrix is problematic since it results in $,J_{33}=frac{0}{0}$
There may be better ways of generating the $L$-matrix. However it is done, its elements must satisfy $$L_{ik} = b_i+b_k$$
Update
A much better way to calculate the $L$-matrix is
$$L = b1^T + 1b^T = A^TU + U^TA$$
where $U$ is a matrix of all ones which has the same shape as $A$.
Now the $J$-matrix can be written as
$$J=(A^TA)oslash(A^TU + U^TA - A^TA)$$
answered Dec 9 '18 at 3:27
greggreg
7,8701821
$endgroup$
add a comment |
$begingroup$
Define the vectors
$$eqalign{
b &= A^T1 cr
p &= exp(b) cr
}$$
and the matrices
$$eqalign{
L &= log(pp^T) cr
J &= (A^TA)oslash(L-A^TA) cr
}$$
Note that the log and exp functions are applied elementwise, $oslash$ represents elementwise division, and $1$ is a vector of all ones.
The elements of the $J$-matrix are the Jaccard distances, i.e.
$,,J_{ik} = J(U_i,U_k)$
The third column of your $A$-matrix is problematic since it results in $,J_{33}=frac{0}{0}$
There may be better ways of generating the $L$-matrix. However it is done, its elements must satisfy $$L_{ik} = b_i+b_k$$
Update
A much better way to calculate the $L$-matrix is
$$L = b1^T + 1b^T = A^TU + U^TA$$
where $U$ is a matrix of all ones which has the same shape as $A$.
Now the $J$-matrix can be written as
$$J=(A^TA)oslash(A^TU + U^TA - A^TA)$$
answered Dec 9 '18 at 3:27
greggreg
7,8701821
$endgroup$
Define the vectors
$$eqalign{
b &= A^T1 cr
p &= exp(b) cr
}$$
and the matrices
$$eqalign{
L &= log(pp^T) cr
J &= (A^TA)oslash(L-A^TA) cr
}$$
Note that the log and exp functions are applied elementwise, $oslash$ represents elementwise division, and $1$ is a vector of all ones.
The elements of the $J$-matrix are the Jaccard distances, i.e.
$,,J_{ik} = J(U_i,U_k)$
The third column of your $A$-matrix is problematic since it results in $,J_{33}=frac{0}{0}$
There may be better ways of generating the $L$-matrix. However it is done, its elements must satisfy $$L_{ik} = b_i+b_k$$
Update
A much better way to calculate the $L$-matrix is
$$L = b1^T + 1b^T = A^TU + U^TA$$
where $U$ is a matrix of all ones which has the same shape as $A$.
Now the $J$-matrix can be written as
$$J=(A^TA)oslash(A^TU + U^TA - A^TA)$$
answered Dec 9 '18 at 3:27
greggreg
7,8701821
edited Dec 9 '18 at 16:39
answered Dec 9 '18 at 3:27
greggreg
7,8701821
answered Dec 9 '18 at 3:27
greggreg
7,8701821
answered Dec 9 '18 at 3:27
greggreg
7,8701821
7,8701821
add a comment |
add a comment |
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