How to extract best parameters from a CrossValidatorModel

Question

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,

Answer 1

This is how you get the chosen parameters

println(cvModel.bestModel.getMaxIter)   
println(cvModel.bestModel.getRegParam)  

Answer 2

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
}

Answer 3

this java code should work: cvModel.bestModel().parent().extractParamMap().you can translate it to scala code parent()method will return an estimator, you can get the best params then.

Answer 4

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"))

Answer 5

I am working with Spark Scala 1.6.x and here is a full example of how i can set and fit a CrossValidator and then return the value of the parameter used to get the best model (assuming that training.toDF gives a dataframe ready to be used) :

import org.apache.spark.ml.classification.LogisticRegression
import org.apache.spark.ml.tuning.{CrossValidator, ParamGridBuilder}
import org.apache.spark.ml.evaluation.MulticlassClassificationEvaluator

// Instantiate a LogisticRegression object
val lr = new LogisticRegression()

// Instantiate a ParamGrid with different values for the 'RegParam' parameter of the logistic regression
val paramGrid = new ParamGridBuilder().addGrid(lr.regParam, Array(0.0001, 0.001, 0.01, 0.1, 0.25, 0.5, 0.75, 1)).build()

// Setting and fitting the CrossValidator on the training set, using 'MultiClassClassificationEvaluator' as evaluator
val crossVal = new CrossValidator().setEstimator(lr).setEvaluator(new MulticlassClassificationEvaluator).setEstimatorParamMaps(paramGrid)
val cvModel = crossVal.fit(training.toDF)

// Getting the value of the 'RegParam' used to get the best model
val bestModel = cvModel.bestModel                    // Getting the best model
val paramReference = bestModel.getParam("regParam")  // Getting the reference of the parameter you want (only the reference, not the value)
val paramValue = bestModel.get(paramReference)       // Getting the value of this parameter
print(paramValue)                                    // In my case : 0.001

You can do the same for any parameter or any other type of model.

Answer 6

Building in the solution of @macfeliga, a single liner that works for pipelines:

cvModel.bestModel.asInstanceOf[PipelineModel]
    .stages.foreach(stage => println(stage.extractParamMap))