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Getting scores from PCA

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4

1

$begingroup$

I'm not super familiar with Principal Component Analysis, but from what I understand, it sorts a vector in order of decreasing variance, and uses that to transform the vector to correlate variables linearly.

I'm not trying to transform the data I have - I'm simply trying, for a list of 784-length feature vectors, to get the PCA "scores" for each feature. Mathematica returns them sorted - I need to be able to correlate them with the actual features, so I need them in their original order. PrincipalComponents doesn't seem to be able to return anything in this format. Is there a way to?

matrix linear-algebra

share|improve this question

edited 10 hours ago

TreFox

asked 11 hours ago

TreFoxTreFox

1,55077 silver badges2323 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  does PrincipalComponents[data][[Ordering @ data]] or PrincipalComponents[data][[Ordering[Variance/@ data]]] give what you need?
  $endgroup$
  – kglr
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kglr Since I'm passing PrincipalComponents a 10x784 matrix, that seems to just reorder the 784-length sublists, rather than the values within those lists. I want to correlate the PCA scores with the feature (value in the 784-length list) they actually correspond to.
  $endgroup$
  – TreFox
  10 hours ago
  
  
  

  
  
  
  

add a comment | 

4

1

$begingroup$

I'm not super familiar with Principal Component Analysis, but from what I understand, it sorts a vector in order of decreasing variance, and uses that to transform the vector to correlate variables linearly.

I'm not trying to transform the data I have - I'm simply trying, for a list of 784-length feature vectors, to get the PCA "scores" for each feature. Mathematica returns them sorted - I need to be able to correlate them with the actual features, so I need them in their original order. PrincipalComponents doesn't seem to be able to return anything in this format. Is there a way to?

matrix linear-algebra

share|improve this question

edited 10 hours ago

TreFox

asked 11 hours ago

TreFoxTreFox

1,55077 silver badges2323 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  does PrincipalComponents[data][[Ordering @ data]] or PrincipalComponents[data][[Ordering[Variance/@ data]]] give what you need?
  $endgroup$
  – kglr
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kglr Since I'm passing PrincipalComponents a 10x784 matrix, that seems to just reorder the 784-length sublists, rather than the values within those lists. I want to correlate the PCA scores with the feature (value in the 784-length list) they actually correspond to.
  $endgroup$
  – TreFox
  10 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I'm not super familiar with Principal Component Analysis, but from what I understand, it sorts a vector in order of decreasing variance, and uses that to transform the vector to correlate variables linearly.

I'm not trying to transform the data I have - I'm simply trying, for a list of 784-length feature vectors, to get the PCA "scores" for each feature. Mathematica returns them sorted - I need to be able to correlate them with the actual features, so I need them in their original order. PrincipalComponents doesn't seem to be able to return anything in this format. Is there a way to?

matrix linear-algebra

share|improve this question

edited 10 hours ago

TreFox

asked 11 hours ago

TreFoxTreFox

1,55077 silver badges2323 bronze badges

$endgroup$

share|improve this question

edited 10 hours ago

TreFox

asked 11 hours ago

TreFoxTreFox

1,55077 silver badges2323 bronze badges

share|improve this question

edited 10 hours ago

TreFox

asked 11 hours ago

TreFoxTreFox

1,55077 silver badges2323 bronze badges

asked 11 hours ago

TreFoxTreFox

1,55077 silver badges2323 bronze badges

asked 11 hours ago

TreFoxTreFox

1,55077 silver badges2323 bronze badges

1 Answer
1

active

oldest

votes

6

$begingroup$

You're in luck, because I recently waded through this problem myself. If I understand your question correctly, you want to know the matrix that transforms the data into the output of PrincipalComponents. The answer to this: that matrix is just the eigenvectors of the correlation matrix.

Simple 2D example:

data = RandomVariate[BinormalDistribution[{-1, 2}, {1, 2}, 0.9], 100];

eig = Eigensystem[Covariance[data]];

ListPlot[

 {

  PrincipalComponents[data],

  Standardize[data, Mean, 1 &].Transpose[eig[[2]]]

  }

 ]

As you can see, the result is the same except for the two clouds being mirrored in the x-axis. This makes sense, since principle component analysis is about transforming the data such that the covariance matrix become diagonal (with the diagonal decreasing towards the bottom right) and flipping the data along an axis leaves the covariance invariant.

As a bonus, the eigenvalues of the covariance matrix tells you how much variance each principle component accounts for, so you don't have to calculate that separately:

eig[[1]]

Variance[PrincipalComponents[data]]



Out[142]= {4.62687, 0.137012}



Out[143]= {4.62687, 0.137012}

share|improve this answer

edited 9 hours ago

answered 10 hours ago

Sjoerd SmitSjoerd Smit

6,3801212 silver badges2424 bronze badges

$endgroup$

add a comment | 

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6

$begingroup$

You're in luck, because I recently waded through this problem myself. If I understand your question correctly, you want to know the matrix that transforms the data into the output of PrincipalComponents. The answer to this: that matrix is just the eigenvectors of the correlation matrix.

Simple 2D example:

data = RandomVariate[BinormalDistribution[{-1, 2}, {1, 2}, 0.9], 100];

eig = Eigensystem[Covariance[data]];

ListPlot[

 {

  PrincipalComponents[data],

  Standardize[data, Mean, 1 &].Transpose[eig[[2]]]

  }

 ]

As you can see, the result is the same except for the two clouds being mirrored in the x-axis. This makes sense, since principle component analysis is about transforming the data such that the covariance matrix become diagonal (with the diagonal decreasing towards the bottom right) and flipping the data along an axis leaves the covariance invariant.

As a bonus, the eigenvalues of the covariance matrix tells you how much variance each principle component accounts for, so you don't have to calculate that separately:

eig[[1]]

Variance[PrincipalComponents[data]]



Out[142]= {4.62687, 0.137012}



Out[143]= {4.62687, 0.137012}

share|improve this answer

edited 9 hours ago

answered 10 hours ago

Sjoerd SmitSjoerd Smit

6,3801212 silver badges2424 bronze badges

$endgroup$

add a comment | 

6

$begingroup$

You're in luck, because I recently waded through this problem myself. If I understand your question correctly, you want to know the matrix that transforms the data into the output of PrincipalComponents. The answer to this: that matrix is just the eigenvectors of the correlation matrix.

Simple 2D example:

data = RandomVariate[BinormalDistribution[{-1, 2}, {1, 2}, 0.9], 100];

eig = Eigensystem[Covariance[data]];

ListPlot[

 {

  PrincipalComponents[data],

  Standardize[data, Mean, 1 &].Transpose[eig[[2]]]

  }

 ]

As you can see, the result is the same except for the two clouds being mirrored in the x-axis. This makes sense, since principle component analysis is about transforming the data such that the covariance matrix become diagonal (with the diagonal decreasing towards the bottom right) and flipping the data along an axis leaves the covariance invariant.

As a bonus, the eigenvalues of the covariance matrix tells you how much variance each principle component accounts for, so you don't have to calculate that separately:

eig[[1]]

Variance[PrincipalComponents[data]]



Out[142]= {4.62687, 0.137012}



Out[143]= {4.62687, 0.137012}

share|improve this answer

edited 9 hours ago

answered 10 hours ago

Sjoerd SmitSjoerd Smit

6,3801212 silver badges2424 bronze badges

$endgroup$

add a comment | 

$begingroup$

You're in luck, because I recently waded through this problem myself. If I understand your question correctly, you want to know the matrix that transforms the data into the output of PrincipalComponents. The answer to this: that matrix is just the eigenvectors of the correlation matrix.

Simple 2D example:

data = RandomVariate[BinormalDistribution[{-1, 2}, {1, 2}, 0.9], 100];

eig = Eigensystem[Covariance[data]];

ListPlot[

 {

  PrincipalComponents[data],

  Standardize[data, Mean, 1 &].Transpose[eig[[2]]]

  }

 ]

As you can see, the result is the same except for the two clouds being mirrored in the x-axis. This makes sense, since principle component analysis is about transforming the data such that the covariance matrix become diagonal (with the diagonal decreasing towards the bottom right) and flipping the data along an axis leaves the covariance invariant.

As a bonus, the eigenvalues of the covariance matrix tells you how much variance each principle component accounts for, so you don't have to calculate that separately:

eig[[1]]

Variance[PrincipalComponents[data]]



Out[142]= {4.62687, 0.137012}



Out[143]= {4.62687, 0.137012}

share|improve this answer

edited 9 hours ago

answered 10 hours ago

Sjoerd SmitSjoerd Smit

6,3801212 silver badges2424 bronze badges

$endgroup$

data = RandomVariate[BinormalDistribution[{-1, 2}, {1, 2}, 0.9], 100];

eig = Eigensystem[Covariance[data]];

ListPlot[

 {

  PrincipalComponents[data],

  Standardize[data, Mean, 1 &].Transpose[eig[[2]]]

  }

 ]

eig[[1]]

Variance[PrincipalComponents[data]]



Out[142]= {4.62687, 0.137012}



Out[143]= {4.62687, 0.137012}

share|improve this answer

edited 9 hours ago

answered 10 hours ago

Sjoerd SmitSjoerd Smit

6,3801212 silver badges2424 bronze badges

answered 10 hours ago

Sjoerd SmitSjoerd Smit

6,3801212 silver badges2424 bronze badges

answered 10 hours ago

Sjoerd SmitSjoerd Smit

6,3801212 silver badges2424 bronze badges