I want to find the parameters of ParamGridBuilder that make the best model in CrossValidator in Spark 1.4.x,
In Pipeline Example in Spark documentation, they add different parameters (numFeatures, regParam) by using ParamGridBuilder in the Pipeline. Then by the following line of code they make the best model:
val cvModel = crossval.fit(training.toDF)
Now, I want to know what are the parameters (numFeatures, regParam) from ParamGridBuilder that produces the best model.
I already used the following commands without success:
cvModel.bestModel.extractParamMap().toString()
cvModel.params.toList.mkString("(", ",", ")")
cvModel.estimatorParamMaps.toString()
cvModel.explainParams()
cvModel.getEstimatorParamMaps.mkString("(", ",", ")")
cvModel.toString()
Any help?
Thanks in advance,
One method to get a proper ParamMap object is to use CrossValidatorModel.avgMetrics: Array[Double] to find the argmax ParamMap:
implicit class BestParamMapCrossValidatorModel(cvModel: CrossValidatorModel) { def bestEstimatorParamMap: ParamMap = { cvModel.getEstimatorParamMaps .zip(cvModel.avgMetrics) .maxBy(_._2) ._1 } } When run on the CrossValidatorModel trained in the Pipeline Example you cited gives:
scala> println(cvModel.bestEstimatorParamMap) { hashingTF_2b0b8ccaeeec-numFeatures: 100, logreg_950a13184247-regParam: 0.1 }
val bestPipelineModel = cvModel.bestModel.asInstanceOf[PipelineModel]
val stages = bestPipelineModel.stages
val hashingStage = stages(1).asInstanceOf[HashingTF]
println("numFeatures = " + hashingStage.getNumFeatures)
val lrStage = stages(2).asInstanceOf[LogisticRegressionModel]
println("regParam = " + lrStage.getRegParam)
source
To print everything in paramMap, you actually don't have to call parent:
cvModel.bestModel.extractParamMap()
To answer OP's question, to get a single best parameter, for example regParam:
cvModel.bestModel.extractParamMap().apply(cvModel.bestModel.getParam("regParam"))
This is how you get the chosen parameters
println(cvModel.bestModel.getMaxIter)
println(cvModel.bestModel.getRegParam)
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