3
$\begingroup$

For the below code for Mathematica 11.3, I got {-0.153792, 1., 1.} as output.

fe = FeatureExtraction[{{1.4, "A"}, {1.5, "A"}, {2.3, "B"}, {5.4, "B"}}]
fe[{2.4, "A"}]

How can I reconstruct the result of FeatureExtraction manually and obtained {-0.153792, 1., 1.} as output?

Many thanks!

$\endgroup$
2
$\begingroup$

This can be done by some experiments.

First: There are many automatic options of Mathematica. In Mathematica 11.3 we can get more informations about functions, for example we can do Options[fe] to find out more details.

Let's make things easier.

So we reconstruct a method named "NumericVector" in 1-Dimension[FeatureSpace] data maybe without the effect of DimensionReduction. In high-dimension data maybe done Standardize+PCA[Linear DimensionReduce] by Mathematica.

In your example, A and B are treated as Nominal Feature.

Let's concentrate on the numeric part.

data={{1},{0},{-1}};
fe=FeatureExtraction[data,{"NumericVector"}];
fe@data
Standardize@data
Standardize[data,Mean,.816497&]
{{1.22474},{0.},{-1.22474}}
{{1},{0},{-1}}
{{1.22474},{0.},{-1.22474}}

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.