将RDD [org.apache.spark.sql.Row]转换为RDD [org.apache.spark.mllib.linalg.Vector]

Question

我对Spark和Scala相对较新.

我从以下数据帧开始(单个列由密集的双打矢量组成):

scala> val scaledDataOnly_pruned = scaledDataOnly.select("features")
scaledDataOnly_pruned: org.apache.spark.sql.DataFrame = [features: vector]

scala> scaledDataOnly_pruned.show(5)
+--------------------+
|            features|
+--------------------+
|[-0.0948337274182...|
|[-0.0948337274182...|
|[-0.0948337274182...|
|[-0.0948337274182...|
|[-0.0948337274182...|
+--------------------+

直接转换为RDD会生成org.apache.spark.rdd.RDD [org.apache.spark.sql.Row]的实例:

scala> val scaledDataOnly_rdd = scaledDataOnly_pruned.rdd
scaledDataOnly_rdd: org.apache.spark.rdd.RDD[org.apache.spark.sql.Row] = MapPartitionsRDD[32] at rdd at <console>:66

有谁知道如何将此DF转换为org.apache.spark.rdd.RDD [org.apache.spark.mllib.linalg.Vector]的实例？到目前为止,我的各种尝试都没有成功.

提前感谢您的任何指示!

Answer 1

刚发现:

val scaledDataOnly_rdd = scaledDataOnly_pruned.map{x:Row => x.getAs[Vector](0)}

Answer 2

编辑:使用更复杂的方式来解释Row中的字段.

这对我有用

val featureVectors = features.map(row => {
  Vectors.dense(row.toSeq.toArray.map({
    case s: String => s.toDouble
    case l: Long => l.toDouble
    case _ => 0.0
  }))
})

features是spark SQL的DataFrame.