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Pulling the principal components out of a DimensionReducerFunction?

4

2

$begingroup$

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

share|improve this question

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 | 

4

2

$begingroup$

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

share|improve this question

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

share|improve this question

edited Mar 31 at 22:00

J. M. is away♦

98.9k10311467

asked Mar 31 at 17:23

Michael CurryMichael Curry

786312

$endgroup$

dr = DimensionReduction[1, 2, 3, 2, 3, 5, 3, 5, 8, 4, 5, 
 8.5, Method -> "PrincipalComponentsAnalysis"]

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

share|improve this question

edited Mar 31 at 22:00

J. M. is away♦

98.9k10311467

asked Mar 31 at 17:23

Michael CurryMichael Curry

786312

share|improve this question

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

asked Mar 31 at 17:23

Michael CurryMichael Curry

786312

asked Mar 31 at 17:23

Michael CurryMichael Curry

786312

1 Answer
1

active

oldest

votes

6

$begingroup$

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

share|improve this answer

edited Mar 31 at 22:56

answered Mar 31 at 18:58

Chip HurstChip Hurst

23.4k15994

$endgroup$

• 
  
  
  

  
  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
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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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6

$begingroup$

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

share|improve this answer

edited Mar 31 at 22:56

answered Mar 31 at 18:58

Chip HurstChip Hurst

23.4k15994

$endgroup$

• 
  
  
  

  
  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
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

6

$begingroup$

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

share|improve this answer

edited Mar 31 at 22:56

answered Mar 31 at 18:58

Chip HurstChip Hurst

23.4k15994

$endgroup$

• 
  
  
  

  
  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
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

share|improve this answer

edited Mar 31 at 22:56

answered Mar 31 at 18:58

Chip HurstChip Hurst

23.4k15994

$endgroup$

data = 1, 2, 3, 2, 3, 5, 3, 5, 8, 4, 5, 8.5;
dr = DimensionReduction[data, Method -> "PrincipalComponentsAnalysis"];

Transpose[dr[[1, "Model", "Matrix"]]]

Transpose[Last[SingularValueDecomposition[Standardize[data], 2]]]

share|improve this answer

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