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Pulling the principal components out of a DimensionReducerFunction?
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?How can I determine the importance of variables from Classify?Why is the classify function not giving the desired output?How to use Mathematica to train a network Using out of core classification?How to train a net for recognize the numberHow to train a network using out of core training when data have different length?FeatureSpacePlot freaks out with LatentSemanticAnalysis on wordsHow can I reproduce the result of DimensionReduction?How can I see the predictor function that Mathematica produces?Machine learning: How to fix part of the weight matrix?Ordinal subset in the set of classes
$begingroup$
Suppose I perform dimension reduction using PCA:
dr = DimensionReduction[1, 2, 3, 2, 3, 5, 3, 5, 8, 4, 5,
8.5, Method -> "PrincipalComponentsAnalysis"]
If I want to see the principal components themselves, in the original space, one thought is to use the "OriginalData" feature of the DimensionReducerFunction, on basis vectors in the new space:
In[8]:= dr[1.0, 0.0, "OriginalData"]
dr[0.0, 1.0, "OriginalData"]
Out[8]= 1.86006, 2.9998, 4.81724
Out[9]= 3.38701, 3.01026, 5.64163
Is this a reasonable thing to do, or am I misinterpreting how the "OriginalData" feature works? And is there a better way to pull out the principal components themselves? People often want to visualize these for various reasons.
(There are several other questions about how to solve a similar problem with the PrincipalComponents function; this is a question about a different function.)
machine-learning dimension-reduction
edited Mar 31 at 22:00
J. M. is away♦
98.9k10311467
asked Mar 31 at 17:23
Michael CurryMichael Curry
786312
$endgroup$
add a comment |
$begingroup$
Suppose I perform dimension reduction using PCA:
dr = DimensionReduction[1, 2, 3, 2, 3, 5, 3, 5, 8, 4, 5,
8.5, Method -> "PrincipalComponentsAnalysis"]
If I want to see the principal components themselves, in the original space, one thought is to use the "OriginalData" feature of the DimensionReducerFunction, on basis vectors in the new space:
In[8]:= dr[1.0, 0.0, "OriginalData"]
dr[0.0, 1.0, "OriginalData"]
Out[8]= 1.86006, 2.9998, 4.81724
Out[9]= 3.38701, 3.01026, 5.64163
Is this a reasonable thing to do, or am I misinterpreting how the "OriginalData" feature works? And is there a better way to pull out the principal components themselves? People often want to visualize these for various reasons.
(There are several other questions about how to solve a similar problem with the PrincipalComponents function; this is a question about a different function.)
machine-learning dimension-reduction
edited Mar 31 at 22:00
J. M. is away♦
98.9k10311467
asked Mar 31 at 17:23
Michael CurryMichael Curry
786312
$endgroup$
add a comment |
$begingroup$
Suppose I perform dimension reduction using PCA:
dr = DimensionReduction[1, 2, 3, 2, 3, 5, 3, 5, 8, 4, 5,
8.5, Method -> "PrincipalComponentsAnalysis"]
If I want to see the principal components themselves, in the original space, one thought is to use the "OriginalData" feature of the DimensionReducerFunction, on basis vectors in the new space:
In[8]:= dr[1.0, 0.0, "OriginalData"]
dr[0.0, 1.0, "OriginalData"]
Out[8]= 1.86006, 2.9998, 4.81724
Out[9]= 3.38701, 3.01026, 5.64163
Is this a reasonable thing to do, or am I misinterpreting how the "OriginalData" feature works? And is there a better way to pull out the principal components themselves? People often want to visualize these for various reasons.
(There are several other questions about how to solve a similar problem with the PrincipalComponents function; this is a question about a different function.)
machine-learning dimension-reduction
edited Mar 31 at 22:00
J. M. is away♦
98.9k10311467
asked Mar 31 at 17:23
Michael CurryMichael Curry
786312
$endgroup$
Suppose I perform dimension reduction using PCA:
dr = DimensionReduction[1, 2, 3, 2, 3, 5, 3, 5, 8, 4, 5,
8.5, Method -> "PrincipalComponentsAnalysis"]
If I want to see the principal components themselves, in the original space, one thought is to use the "OriginalData" feature of the DimensionReducerFunction, on basis vectors in the new space:
In[8]:= dr[1.0, 0.0, "OriginalData"]
dr[0.0, 1.0, "OriginalData"]
Out[8]= 1.86006, 2.9998, 4.81724
Out[9]= 3.38701, 3.01026, 5.64163
Is this a reasonable thing to do, or am I misinterpreting how the "OriginalData" feature works? And is there a better way to pull out the principal components themselves? People often want to visualize these for various reasons.
(There are several other questions about how to solve a similar problem with the PrincipalComponents function; this is a question about a different function.)
machine-learning dimension-reduction
machine-learning dimension-reduction
edited Mar 31 at 22:00
J. M. is away♦
98.9k10311467
asked Mar 31 at 17:23
Michael CurryMichael Curry
786312
edited Mar 31 at 22:00
J. M. is away♦
98.9k10311467
asked Mar 31 at 17:23
Michael CurryMichael Curry
786312
edited Mar 31 at 22:00
J. M. is away♦
98.9k10311467
edited Mar 31 at 22:00
J. M. is away♦
98.9k10311467
edited Mar 31 at 22:00
J. M. is away♦
98.9k10311467
98.9k10311467
asked Mar 31 at 17:23
Michael CurryMichael Curry
786312
asked Mar 31 at 17:23
Michael CurryMichael Curry
786312
asked Mar 31 at 17:23
Michael CurryMichael Curry
786312
786312
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's your data:
data = 1, 2, 3, 2, 3, 5, 3, 5, 8, 4, 5, 8.5;
dr = DimensionReduction[data, Method -> "PrincipalComponentsAnalysis"];
This is not exactly a top level solution, but we can pick apart the DimensionReducerFunction and see inside (try dr[[1]] to see many internal properties).
Looking further, in there we have a matrix:
Transpose[dr[[1, "Model", "Matrix"]]]
-0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163
I think these are the components. We can try to verify:
Transpose[Last[SingularValueDecomposition[Standardize[data], 2]]]
-0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163
answered Mar 31 at 18:58
Chip HurstChip Hurst
23.4k15994
$endgroup$
2
$begingroup$
It doesn't look like you need the pre-multiplication by2/Sqrt[3]; after all, any such rescaling would show itself in the singular values and not the orthogonal factors. The important thing is the shift, whichStandardize[]of course does.
$endgroup$
– J. M. is away♦
Mar 31 at 22:12
$begingroup$
Good catch, I've edited the post. I can't remember what I ran into that led me to such a conclusion.
$endgroup$
– Chip Hurst
Mar 31 at 22:58
add a comment |
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1 Answer
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active
oldest
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1 Answer
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oldest
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oldest
votes
$begingroup$
Here's your data:
data = 1, 2, 3, 2, 3, 5, 3, 5, 8, 4, 5, 8.5;
dr = DimensionReduction[data, Method -> "PrincipalComponentsAnalysis"];
This is not exactly a top level solution, but we can pick apart the DimensionReducerFunction and see inside (try dr[[1]] to see many internal properties).
Looking further, in there we have a matrix:
Transpose[dr[[1, "Model", "Matrix"]]]
-0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163
I think these are the components. We can try to verify:
Transpose[Last[SingularValueDecomposition[Standardize[data], 2]]]
-0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163
answered Mar 31 at 18:58
Chip HurstChip Hurst
23.4k15994
$endgroup$
2
$begingroup$
It doesn't look like you need the pre-multiplication by2/Sqrt[3]; after all, any such rescaling would show itself in the singular values and not the orthogonal factors. The important thing is the shift, whichStandardize[]of course does.
$endgroup$
– J. M. is away♦
Mar 31 at 22:12
$begingroup$
Good catch, I've edited the post. I can't remember what I ran into that led me to such a conclusion.
$endgroup$
– Chip Hurst
Mar 31 at 22:58
add a comment |
$begingroup$
Here's your data:
data = 1, 2, 3, 2, 3, 5, 3, 5, 8, 4, 5, 8.5;
dr = DimensionReduction[data, Method -> "PrincipalComponentsAnalysis"];
This is not exactly a top level solution, but we can pick apart the DimensionReducerFunction and see inside (try dr[[1]] to see many internal properties).
Looking further, in there we have a matrix:
Transpose[dr[[1, "Model", "Matrix"]]]
-0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163
I think these are the components. We can try to verify:
Transpose[Last[SingularValueDecomposition[Standardize[data], 2]]]
-0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163
answered Mar 31 at 18:58
Chip HurstChip Hurst
23.4k15994
$endgroup$
2
$begingroup$
It doesn't look like you need the pre-multiplication by2/Sqrt[3]; after all, any such rescaling would show itself in the singular values and not the orthogonal factors. The important thing is the shift, whichStandardize[]of course does.
$endgroup$
– J. M. is away♦
Mar 31 at 22:12
$begingroup$
Good catch, I've edited the post. I can't remember what I ran into that led me to such a conclusion.
$endgroup$
– Chip Hurst
Mar 31 at 22:58
add a comment |
$begingroup$
Here's your data:
data = 1, 2, 3, 2, 3, 5, 3, 5, 8, 4, 5, 8.5;
dr = DimensionReduction[data, Method -> "PrincipalComponentsAnalysis"];
This is not exactly a top level solution, but we can pick apart the DimensionReducerFunction and see inside (try dr[[1]] to see many internal properties).
Looking further, in there we have a matrix:
Transpose[dr[[1, "Model", "Matrix"]]]
-0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163
I think these are the components. We can try to verify:
Transpose[Last[SingularValueDecomposition[Standardize[data], 2]]]
-0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163
answered Mar 31 at 18:58
Chip HurstChip Hurst
23.4k15994
$endgroup$
Here's your data:
data = 1, 2, 3, 2, 3, 5, 3, 5, 8, 4, 5, 8.5;
dr = DimensionReduction[data, Method -> "PrincipalComponentsAnalysis"];
This is not exactly a top level solution, but we can pick apart the DimensionReducerFunction and see inside (try dr[[1]] to see many internal properties).
Looking further, in there we have a matrix:
Transpose[dr[[1, "Model", "Matrix"]]]
-0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163
I think these are the components. We can try to verify:
Transpose[Last[SingularValueDecomposition[Standardize[data], 2]]]
-0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163
answered Mar 31 at 18:58
Chip HurstChip Hurst
23.4k15994
edited Mar 31 at 22:56
answered Mar 31 at 18:58
Chip HurstChip Hurst
23.4k15994
answered Mar 31 at 18:58
Chip HurstChip Hurst
23.4k15994
answered Mar 31 at 18:58
Chip HurstChip Hurst
23.4k15994
23.4k15994
2
$begingroup$
It doesn't look like you need the pre-multiplication by2/Sqrt[3]; after all, any such rescaling would show itself in the singular values and not the orthogonal factors. The important thing is the shift, whichStandardize[]of course does.
$endgroup$
– J. M. is away♦
Mar 31 at 22:12
$begingroup$
Good catch, I've edited the post. I can't remember what I ran into that led me to such a conclusion.
$endgroup$
– Chip Hurst
Mar 31 at 22:58
add a comment |
2
$begingroup$
It doesn't look like you need the pre-multiplication by2/Sqrt[3]; after all, any such rescaling would show itself in the singular values and not the orthogonal factors. The important thing is the shift, whichStandardize[]of course does.
$endgroup$
– J. M. is away♦
Mar 31 at 22:12
$begingroup$
Good catch, I've edited the post. I can't remember what I ran into that led me to such a conclusion.
$endgroup$
– Chip Hurst
Mar 31 at 22:58
2
2
$begingroup$
It doesn't look like you need the pre-multiplication by
2/Sqrt[3]; after all, any such rescaling would show itself in the singular values and not the orthogonal factors. The important thing is the shift, which Standardize[] of course does.$endgroup$
– J. M. is away♦
Mar 31 at 22:12
$begingroup$
It doesn't look like you need the pre-multiplication by
2/Sqrt[3]; after all, any such rescaling would show itself in the singular values and not the orthogonal factors. The important thing is the shift, which Standardize[] of course does.$endgroup$
– J. M. is away♦
Mar 31 at 22:12
$begingroup$
Good catch, I've edited the post. I can't remember what I ran into that led me to such a conclusion.
$endgroup$
– Chip Hurst
Mar 31 at 22:58
$begingroup$
Good catch, I've edited the post. I can't remember what I ran into that led me to such a conclusion.
$endgroup$
– Chip Hurst
Mar 31 at 22:58
add a comment |
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