How to extract best parameters from a CrossValidatorModel

问题

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,

回答1:

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
}

回答2:

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

回答3:

This is how you get the chosen parameters

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

回答4:

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.

回答5:

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"))

回答6:

This is the ParamGridBuilder()

paraGrid = ParamGridBuilder().addGrid(
hashingTF.numFeatures, [10, 100, 1000]
).addGrid(
    lr.regParam, [0.1, 0.01, 0.001]
).build()

There are 3 stages in pipeline. It seems we can assess parameters as the following:

for stage in cv_model.bestModel.stages:
    print 'stages: {}'.format(stage)
    print stage.params
    print '\n'

stage: Tokenizer_46ffb9fac5968c6c152b
[Param(parent='Tokenizer_46ffb9fac5968c6c152b', name='inputCol', doc='input column name'), Param(parent='Tokenizer_46ffb9fac5968c6c152b', name='outputCol', doc='output column name')]

stage: HashingTF_40e1af3ba73764848d43
[Param(parent='HashingTF_40e1af3ba73764848d43', name='inputCol', doc='input column name'), Param(parent='HashingTF_40e1af3ba73764848d43', name='numFeatures', doc='number of features'), Param(parent='HashingTF_40e1af3ba73764848d43', name='outputCol', doc='output column name')]

stage: LogisticRegression_451b8c8dbef84ecab7a9
[]

However, there is no parameter in the last stage, logiscRegression.

We can also get weight and intercept parameter from logistregression like the following:

cv_model.bestModel.stages[1].getNumFeatures()
10
cv_model.bestModel.stages[2].intercept
1.5791827733883774
cv_model.bestModel.stages[2].weights
DenseVector([-2.5361, -0.9541, 0.4124, 4.2108, 4.4707, 4.9451, -0.3045, 5.4348, -0.1977, -1.8361])

Full exploration: http://kuanliang.github.io/2016-06-07-SparkML-pipeline/

回答7:

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.

回答8:

If java，see this debug show;

bestModel.parent().extractParamMap()

回答9:

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

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

来源：https://stackoverflow.com/questions/31749593/how-to-extract-best-parameters-from-a-crossvalidatormodel