如何在ML Pipeline中访问基础模型的参数? [英] How to access parameters of the underlying model in ML Pipeline?

问题描述

我有一个用LinearRegression处理过的DataFrame.如果直接进行操作(如下所示)，则可以显示模型的详细信息:

I have a DataFrame that is processed with LinearRegression. If I do it directly, like below, I can display the details of the model:

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(例如下面的简化示例)，

However, if I use it inside a pipeline (like in the simplified example below),

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

然后出现以下错误.

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}")

我理解它的含义(很明显，由于管道的原因，我得到了一个不同的类)，但是不知道如何获得真正的基础模型.

I understand what it means (it's obvious I got a different class, because of the pipeline), but don't know how to get to the real underlying model.

推荐答案

LinearRegressionModel应该位于stages内部，且与相应的LinearRegression完全相同.

LinearRegressionModel should be inside stages at the exact same index as its corresponding LinearRegression.

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

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