Converting Spark Dataframe(with WrappedArray) to R

Question

I am new to Scala and I want to convert dataframe to rdd. let the label, features convert to RDD[labelPoint] for the input of MLlib. But I can't find out the way to deal with WrappedArray.

scala> test.printSchema
root
 |-- user_id: long (nullable = true)
 |-- brand_store_sn: string (nullable = true)
 |-- label: integer (nullable = true)
 |-- money_score: double (nullable = true)
 |-- normal_score: double (nullable = true)
 |-- action_score: double (nullable = true)
 |-- features: array (nullable = true)
 |    |-- element: string (containsNull = true)
 |-- flag: string (nullable = true)
 |-- dt: string (nullable = true)


scala> test.head
res21: org.apache.spark.sql.Row = [2533,10005072,1,2.0,1.0,1.0,WrappedArray(["d90_pv_1sec:1.4471580313422192", "d3_pv_1sec:0.9030899869919435", "d7_pv_1sec:0.9030899869919435", "d30_pv_1sec:1.414973347970818", "d90_pv_week_decay:1.4235871662780681", "d1_pv_1sec:0.9030899869919435", "d120_pv_1sec:1.4471580313422192"]),user_positive,20161130]

Answer 1

First - since LabeledPoint expects a Vector of Doubles, I'm assuming you also want to split each element in every features array by colon (:), and treat the right-hand side of it as the double, e.g.:

 "d90_pv_1sec:1.4471580313422192" --> 1.4471580313422192

If so - here's the transformation:

import org.apache.spark.mllib.linalg.{Vector, Vectors}
import org.apache.spark.mllib.regression.LabeledPoint

// sample data - DataFrame with label, features and other columns
val df = Seq(
  (1, Array("d90_pv_1sec:1.4471580313422192", "d3_pv_1sec:0.9030899869919435"), 4.0),
  (2, Array("d7_pv_1sec:0.9030899869919435", "d30_pv_1sec:1.414973347970818"), 5.0)
).toDF("label", "features", "ignored")

// extract relevant fields from Row and convert WrappedArray[String] into Vector:
val result = df.rdd.map(r => {
  val label = r.getAs[Int]("label")
  val featuresArray = r.getAs[mutable.WrappedArray[String]]("features")
  val features: Vector = Vectors.dense(
    featuresArray.map(_.split(":")(1).toDouble).toArray
  )
  LabeledPoint(label, features)
})

result.foreach(println)
// (1.0,[1.4471580313422192,0.9030899869919435])
// (2.0,[0.9030899869919435,1.414973347970818])

EDIT: per clarification, now assuming each item in the input array contains the expected index in a resulting sparse vector:

"d90_pv_1sec:1.4471580313422192" --> index = 90; value = 1.4471580313422192

The modified code would be:

val vectorSize: Int = 100 // just a guess - should be the maximum index + 1

val result = df.rdd.map(r => {
  val label = r.getAs[Int]("label")
  val arr = r.getAs[mutable.WrappedArray[String]]("features").toArray
  // parse each item into (index, value) tuple to use in sparse vector
  val elements = arr.map(_.split(":")).map {
    case Array(s, d) => (s.replaceAll("d|_pv_1sec","").toInt, d.toDouble)
  }
  LabeledPoint(label, Vectors.sparse(vectorSize, elements))
})

result.foreach(println)
// (1.0,(100,[3,90],[0.9030899869919435,1.4471580313422192]))
// (2.0,(100,[7,30],[0.9030899869919435,1.414973347970818]))

NOTE: Using s.replaceAll("d|_pv_1sec","") might be a bit slow, as it compiles a regular expression for each item separately. If that's the case, it can be replaced by the faster (yet uglier) s.replace("d", "").replace("_pv_1sec", "") which doesn't use regular expressions.