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FeatureExtract function with "StandardizedVector" do not show the standard devaition of one

Sangdon Lee
Posted 1 year ago

I am wondering what kind of scaling is applied to the FeatureExtract function with "StandardizedVector" option. The output does not show the standard devaition of one for each columns. This function is used in NetEncoder["FeatureExtractor"] also. 

x={{1., 2., 3.}, {4., 5., 6.}, {7., 8., 9.}, {10., 11., 12.}, {13., 14.,15.}} 

xS01 = FeatureExtract[x, "StandardizedVector"];
xS02 = Standardize[x];
StandardDeviation /@ {xS01, xS02}
Mean /@ {xS01, xS02} // Chop
{{1.11803, 1.11803, 1.11803}, {1., 1., 1.}}

Thanks.

Attachments:
FeatureExtractor_V1.nb
POSTED BY: Sangdon Lee
2 Replies
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Sangdon Lee
Sangdon Lee
Posted 1 year ago

Thank you for clarification. Could you please provide few references? I am not aware of this biased estimator for the variance.

POSTED BY: Sangdon Lee

It uses the biased estimator for the variance

x = {{1., 2., 3.}, {4., 5., 6.}, {7., 8., 9.}, {10., 11., 
   12.}, {13., 14., 15.}}

xS01 = FeatureExtract[x, "StandardizedVector"];
xS02 = Standardize[x];
xS03 = Sqrt[Length[x]/(Length[x] - 1)] xS02;
StandardDeviation /@ {xS01, xS02, xS03}
Mean /@ {xS01, xS02, xS03} // Chop

(*
 {{1., 2., 3.}, {4., 5., 6.}, {7., 8., 9.}, {10., 11., 12.}, {13., 14., 15.}}

 {{1.11803, 1.11803, 1.11803}, {1., 1., 1.}, {1.11803, 1.11803, 1.11803}}

  {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}
*)
POSTED BY: Giulio Alessandrini
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