如何在ML Pipeline中访问底层模型的参数？

Question

我有一个使用LinearRegression处理的DataFrame.如果我直接进行,如下所示,我可以显示模型的详细信息:

val lr = new LinearRegression()
val lrModel = lr.fit(df)

lrModel: org.apache.spark.ml.regression.LinearRegressionModel = linReg_b22a7bb88404

println(s"Coefficients: ${lrModel.coefficients} Intercept: ${lrModel.intercept}")
Coefficients: [0.9705748115939526] Intercept: 0.31041486689532866

但是,如果我在管道中使用它(如下面的简化示例),

val pipeline = new Pipeline().setStages(Array(lr))
val lrModel = pipeline.fit(df)

然后我收到以下错误.

scala> lrModel
res9: org.apache.spark.ml.PipelineModel = pipeline_99ca9cba48f8

scala> println(s"Coefficients: ${lrModel.coefficients} Intercept: ${lrModel.intercept}")
<console>:68: error: value coefficients is not a member of org.apache.spark.ml.PipelineModel
       println(s"Coefficients: ${lrModel.coefficients} Intercept: ${lrModel.intercept}")
                                         ^
<console>:68: error: value intercept is not a member of org.apache.spark.ml.PipelineModel
       println(s"Coefficients: ${lrModel.coefficients} Intercept: ${lrModel.intercept}")

我理解它意味着什么(显然我有一个不同的类,因为管道),但不知道如何到达真正的底层模型.

Answer 1

LinearRegressionModel应该stages在与其对应的索引完全相同的内部LinearRegression.

import org.apache.spark.ml.regressio??n.LinearRegressionMo??del
lrModel.stages(0).asInstanceOf[LinearRegressionMo??del]