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]
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.
