scala - 如何迭代scalawrappedArray？ ( Spark )

Question

我执行以下操作:

val tempDict = sqlContext.sql("select words.pName_token,collect_set(words.pID) as docids 
                               from words
                               group by words.pName_token").toDF()

val wordDocs = tempDict.filter(newDict("pName_token")===word)

val listDocs = wordDocs.map(t => t(1)).collect()

listDocs: Array

[Any] = Array(WrappedArray(123, 234, 205876618, 456))

我的问题是如何迭代这个包装数组或将其转换为列表？

我为 listDocs 获得的选项是 apply、asInstanceOf、clone、isInstanceOf、length、toString 和 update。

我该如何继续？

Answer 1

这是解决此问题的一种方法。

import org.apache.spark.sql.Row
import org.apache.spark.sql.functions._
import scala.collection.mutable.WrappedArray

val data = Seq((Seq(1,2,3),Seq(4,5,6),Seq(7,8,9)))
val df = sqlContext.createDataFrame(data)
val first = df.first

// use a pattern match to deferral the type
val mapped = first.getAs[WrappedArray[Int]](0)

// now we can use it like normal collection
mapped.mkString("\n")

// get rows where has array
val rows = df.collect.map {
    case Row(a: Seq[Any], b: Seq[Any], c: Seq[Any]) => 
        (a, b, c)
}
rows.mkString("\n")