Scala随机森林特征重要性提取与名称（标签）答案

【问题标题】：Scala Random forest feature importance extraction with names (labels)Scala随机森林特征重要性提取与名称（标签）
【发布时间】：2018-04-13 02:52:51
【问题描述】：

有没有办法从模型中提取特征重要性并附加featureCols 名称以便于分析？

我有类似的东西：

val featureCols = Array("a","b","c".......... like 67 more)

val assembler = new VectorAssembler().setInputCols(featureCols).setOutputCol("features")
val df2 = assembler.transform(modeling_db)
val labelIndexer = new StringIndexer().setInputCol("def").setOutputCol("label")
val df3 = labelIndexer.fit(df2).transform(df2)
val splitSeed = 5043
val Array(trainingData, testDataCE) = df3.randomSplit(Array(0.7, 0.3), splitSeed)
val classifier = new RandomForestClassifier().setImpurity("gini").setMaxDepth(19).setNumTrees(57).setFeatureSubsetStrategy("auto").setSeed(5043)
val model = classifier.fit(trainingData)

之后，我们尝试提取重要性：

model.featureImportances

而答案真的很难分析：

res14: org.apache.spark.mllib.linalg.Vector = (71,[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,20,23,25,27,33,34,35,38,39,41,42,45,47,48,49,50,51,52,53,54,55,56,57,58,60,61,62,63,64,65,66,67,68,69,70],[0.22362951804309808,0.1830148359365108,0.10246542303449771,0.1699399958851977,0.06486419413350401,0.05187244974385025,0.02627047699833213,0.014498050071723645,0.026182513062665076,0.007126662761055224,0.012375060477018274,0.004354513006816487,0.004361008357237427,0.008435852744278544,0.003195472326415685,0.0023071401643885753,0.004602370417578224,0.0030394399903992345,6.92408316823549E-4,0.011207695216651398,7.609910745572573E-4,8.316382113306638E-4,0.0021506289318167916,0.0013468620354363688,0.006968754359778437,0.018796331618729723,0.0024516591941419444,0.005980997035580654,0.0027983...

有没有办法“uppack”这个答案并将其附加到原始标签名称中？

【问题讨论】：

标签： scala apache-spark machine-learning random-forest feature-selection

【解决方案1】：

您在featureCols 中有原始列名，并且似乎没有涉及任何向量，因此您可以简单地将zip 两个数组放在一起。对于这样的输入数据：

val featureCols = Array("a", "b", "c", "d", "e")
val featureImportance = Vectors.dense(Array(0.15, 0.25, 0.1, 0.35, 0.15)).toSparse

简单地做

val res = featureCols.zip(featureImportance.toArray).sortBy(-_._2)

通过打印将导致

(d,0.35)
(b,0.25)
(a,0.15)
(e,0.15)
(c,0.1)

【讨论】：