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matrix standardization, inferring technique
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$begingroup$
I am reading a research paper in which authors perform a matrix standardization but do not explain the actual procedure. The original 3x3 matrix is:
411; 124; 45
112; 17; 50
20; 378; 285
the standardized is:
0; 0.7036; 1
0.7647; 1; 0.9792
1; 0; 0
the minimum of each column (and row) becomes 1, the maximum of each column becomes 0, but I don't get how the other values are calculated. Any idea? Thank you
linear-algebra statistics matrix-calculus
asked Mar 21 at 15:01
G. M. CampedelliG. M. Campedelli
82
$endgroup$
add a comment |
$begingroup$
I am reading a research paper in which authors perform a matrix standardization but do not explain the actual procedure. The original 3x3 matrix is:
411; 124; 45
112; 17; 50
20; 378; 285
the standardized is:
0; 0.7036; 1
0.7647; 1; 0.9792
1; 0; 0
the minimum of each column (and row) becomes 1, the maximum of each column becomes 0, but I don't get how the other values are calculated. Any idea? Thank you
linear-algebra statistics matrix-calculus
asked Mar 21 at 15:01
G. M. CampedelliG. M. Campedelli
82
$endgroup$
add a comment |
$begingroup$
I am reading a research paper in which authors perform a matrix standardization but do not explain the actual procedure. The original 3x3 matrix is:
411; 124; 45
112; 17; 50
20; 378; 285
the standardized is:
0; 0.7036; 1
0.7647; 1; 0.9792
1; 0; 0
the minimum of each column (and row) becomes 1, the maximum of each column becomes 0, but I don't get how the other values are calculated. Any idea? Thank you
linear-algebra statistics matrix-calculus
asked Mar 21 at 15:01
G. M. CampedelliG. M. Campedelli
82
$endgroup$
I am reading a research paper in which authors perform a matrix standardization but do not explain the actual procedure. The original 3x3 matrix is:
411; 124; 45
112; 17; 50
20; 378; 285
the standardized is:
0; 0.7036; 1
0.7647; 1; 0.9792
1; 0; 0
the minimum of each column (and row) becomes 1, the maximum of each column becomes 0, but I don't get how the other values are calculated. Any idea? Thank you
linear-algebra statistics matrix-calculus
linear-algebra statistics matrix-calculus
asked Mar 21 at 15:01
G. M. CampedelliG. M. Campedelli
82
asked Mar 21 at 15:01
G. M. CampedelliG. M. Campedelli
82
asked Mar 21 at 15:01
G. M. CampedelliG. M. Campedelli
82
asked Mar 21 at 15:01
G. M. CampedelliG. M. Campedelli
82
asked Mar 21 at 15:01
G. M. CampedelliG. M. Campedelli
82
82
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $textbf{a}_j={a_{1j},ldots, a_{ij}, ldots, a_{1m}}$. Then the formula is
$$tilde a_{ij}=frac{maxlimits_i(textbf{a}_j)-a_{ij}}{maxlimits_i(textbf{a}_j)-minlimits_i(textbf{a}_j)}$$
$i$ is the index for the row and $j$ is the index for the column. For example,
$$tilde a_{13}=frac{max{45,50,285}-45}{max{45,50,285}-min{45,50,285}}=frac{285-45}{285-45}=1$$
answered Mar 21 at 17:48
callculuscallculus
18.7k31428
$endgroup$
$begingroup$
Thank you. In the meanwhile I have actually found the solution by myself but thanks again for the clear explanation!
$endgroup$
– G. M. Campedelli
Mar 22 at 9:35
$begingroup$
You´re welcome. Nice that we´ve got the same result.
$endgroup$
– callculus
Mar 22 at 12:08
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $textbf{a}_j={a_{1j},ldots, a_{ij}, ldots, a_{1m}}$. Then the formula is
$$tilde a_{ij}=frac{maxlimits_i(textbf{a}_j)-a_{ij}}{maxlimits_i(textbf{a}_j)-minlimits_i(textbf{a}_j)}$$
$i$ is the index for the row and $j$ is the index for the column. For example,
$$tilde a_{13}=frac{max{45,50,285}-45}{max{45,50,285}-min{45,50,285}}=frac{285-45}{285-45}=1$$
answered Mar 21 at 17:48
callculuscallculus
18.7k31428
$endgroup$
$begingroup$
Thank you. In the meanwhile I have actually found the solution by myself but thanks again for the clear explanation!
$endgroup$
– G. M. Campedelli
Mar 22 at 9:35
$begingroup$
You´re welcome. Nice that we´ve got the same result.
$endgroup$
– callculus
Mar 22 at 12:08
add a comment |
$begingroup$
Let $textbf{a}_j={a_{1j},ldots, a_{ij}, ldots, a_{1m}}$. Then the formula is
$$tilde a_{ij}=frac{maxlimits_i(textbf{a}_j)-a_{ij}}{maxlimits_i(textbf{a}_j)-minlimits_i(textbf{a}_j)}$$
$i$ is the index for the row and $j$ is the index for the column. For example,
$$tilde a_{13}=frac{max{45,50,285}-45}{max{45,50,285}-min{45,50,285}}=frac{285-45}{285-45}=1$$
answered Mar 21 at 17:48
callculuscallculus
18.7k31428
$endgroup$
$begingroup$
Thank you. In the meanwhile I have actually found the solution by myself but thanks again for the clear explanation!
$endgroup$
– G. M. Campedelli
Mar 22 at 9:35
$begingroup$
You´re welcome. Nice that we´ve got the same result.
$endgroup$
– callculus
Mar 22 at 12:08
add a comment |
$begingroup$
Let $textbf{a}_j={a_{1j},ldots, a_{ij}, ldots, a_{1m}}$. Then the formula is
$$tilde a_{ij}=frac{maxlimits_i(textbf{a}_j)-a_{ij}}{maxlimits_i(textbf{a}_j)-minlimits_i(textbf{a}_j)}$$
$i$ is the index for the row and $j$ is the index for the column. For example,
$$tilde a_{13}=frac{max{45,50,285}-45}{max{45,50,285}-min{45,50,285}}=frac{285-45}{285-45}=1$$
answered Mar 21 at 17:48
callculuscallculus
18.7k31428
$endgroup$
Let $textbf{a}_j={a_{1j},ldots, a_{ij}, ldots, a_{1m}}$. Then the formula is
$$tilde a_{ij}=frac{maxlimits_i(textbf{a}_j)-a_{ij}}{maxlimits_i(textbf{a}_j)-minlimits_i(textbf{a}_j)}$$
$i$ is the index for the row and $j$ is the index for the column. For example,
$$tilde a_{13}=frac{max{45,50,285}-45}{max{45,50,285}-min{45,50,285}}=frac{285-45}{285-45}=1$$
answered Mar 21 at 17:48
callculuscallculus
18.7k31428
answered Mar 21 at 17:48
callculuscallculus
18.7k31428
answered Mar 21 at 17:48
callculuscallculus
18.7k31428
answered Mar 21 at 17:48
callculuscallculus
18.7k31428
18.7k31428
$begingroup$
Thank you. In the meanwhile I have actually found the solution by myself but thanks again for the clear explanation!
$endgroup$
– G. M. Campedelli
Mar 22 at 9:35
$begingroup$
You´re welcome. Nice that we´ve got the same result.
$endgroup$
– callculus
Mar 22 at 12:08
add a comment |
$begingroup$
Thank you. In the meanwhile I have actually found the solution by myself but thanks again for the clear explanation!
$endgroup$
– G. M. Campedelli
Mar 22 at 9:35
$begingroup$
You´re welcome. Nice that we´ve got the same result.
$endgroup$
– callculus
Mar 22 at 12:08
$begingroup$
Thank you. In the meanwhile I have actually found the solution by myself but thanks again for the clear explanation!
$endgroup$
– G. M. Campedelli
Mar 22 at 9:35
$begingroup$
Thank you. In the meanwhile I have actually found the solution by myself but thanks again for the clear explanation!
$endgroup$
– G. M. Campedelli
Mar 22 at 9:35
$begingroup$
You´re welcome. Nice that we´ve got the same result.
$endgroup$
– callculus
Mar 22 at 12:08
$begingroup$
You´re welcome. Nice that we´ve got the same result.
$endgroup$
– callculus
Mar 22 at 12:08
add a comment |
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