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Pulling the principal components out of a DimensionReducerFunction?

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Pulling the principal components out of a DimensionReducerFunction?



The Next CEO of Stack OverflowHow 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










3












$begingroup$


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











$endgroup$
















    3












    $begingroup$


    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











    $endgroup$














      3












      3








      3


      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











      $endgroup$




      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 14 mins ago









      J. M. is slightly pensive♦

      98.8k10311467




      98.8k10311467










      asked 4 hours ago









      Michael CurryMichael Curry

      781312




      781312




















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          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[2.0/Sqrt[3] Standardize[data], 2]]]



          -0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163



          (I think we need the 2.0/Sqrt[3] because DimensionReduction appears to use biased standard deviation whereas Standardize uses the unbiased form. I'm not sure though.)






          share|improve this answer











          $endgroup$












          • $begingroup$
            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 slightly pensive♦
            3 mins ago












          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          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[2.0/Sqrt[3] Standardize[data], 2]]]



          -0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163



          (I think we need the 2.0/Sqrt[3] because DimensionReduction appears to use biased standard deviation whereas Standardize uses the unbiased form. I'm not sure though.)






          share|improve this answer











          $endgroup$












          • $begingroup$
            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 slightly pensive♦
            3 mins ago
















          4












          $begingroup$

          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[2.0/Sqrt[3] Standardize[data], 2]]]



          -0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163



          (I think we need the 2.0/Sqrt[3] because DimensionReduction appears to use biased standard deviation whereas Standardize uses the unbiased form. I'm not sure though.)






          share|improve this answer











          $endgroup$












          • $begingroup$
            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 slightly pensive♦
            3 mins ago














          4












          4








          4





          $begingroup$

          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[2.0/Sqrt[3] Standardize[data], 2]]]



          -0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163



          (I think we need the 2.0/Sqrt[3] because DimensionReduction appears to use biased standard deviation whereas Standardize uses the unbiased form. I'm not sure though.)






          share|improve this answer











          $endgroup$



          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[2.0/Sqrt[3] Standardize[data], 2]]]



          -0.572383, -0.577502, -0.582125, 0.793367, -0.56945, -0.215163



          (I think we need the 2.0/Sqrt[3] because DimensionReduction appears to use biased standard deviation whereas Standardize uses the unbiased form. I'm not sure though.)







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 54 mins ago









          Michael Curry

          781312




          781312










          answered 3 hours ago









          Chip HurstChip Hurst

          22.8k15892




          22.8k15892











          • $begingroup$
            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 slightly pensive♦
            3 mins ago

















          • $begingroup$
            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 slightly pensive♦
            3 mins ago
















          $begingroup$
          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 slightly pensive♦
          3 mins ago





          $begingroup$
          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 slightly pensive♦
          3 mins ago


















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