[GitHub] spark pull request: [SPARK-1406] Mllib pmml model export

[mengxr <gi...@git.apache.org>]

Github user mengxr commented on the pull request:

    https://github.com/apache/spark/pull/3062#issuecomment-72540267
  
    @srowen @selvinsource From user perspective, this is the code they need to export a model to PMML:
    
    ~~~
    import org.apache.spark.mllib.export.ModelExporter
    
    val kmeansModel = ...
    ModelExporter.toPMML(kmeansModel, "/path")
    ~~~
    
    They need to remember two things: 1) `ModelExporter`'s name and its package, 2) whether a model is exportable to PMML or not because `toPMML` takes `Any`. As a user, I would prefer the following
    
    ~~~
    val kmeansModel = ...
    val pmml = kmeansModel.toPMML
    ~~~
    
    We can add a trait called `PMMLExportable`, which contains `toPMML`. (We need to discuss the return type of `toPMML`.) So users know whether a model can be exported to PMML or not. For the implementation, I agree with @srowen that `.pmml` package should be sufficient because there is no other exchangeable formats for predictive models. The real export code for each model should live under this package as private and `toPMML` in each model becomes a simple wrapper.
    
    Due to lacking feature attributes, I'd like to mark everything as private for now until we add ML attributes.


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