scala 如何迭代ScalawrappedArray？（火花）
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How to iterate scala wrappedArray? (Spark)
提问by boY
I perform the following operations:
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))
My question is how do I iterate over this wrapped array or convert this into a list?
我的问题是如何遍历这个包装的数组或将其转换为列表？
The options I get for the listDocsare apply, asInstanceOf, clone, isInstanceOf, length, toString, and update.
我得到的选项listDocs是apply， asInstanceOf， clone， isInstanceOf， length， toString，和 update。
How do I proceed?
我该如何进行？
采纳答案by Rockie Yang
Here is one way to solve this.
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")
