提问者:小点点

如何使用分解和自定义修改同时转换到其他单列


我有如下数据集:

Input Dataset

Id, Parent_id, Data
-----------------------
1, NULL, favorite: 3
2, NULL, favorite: 4
Output Dataset

Id, Parent_Id, Data
------------------------
1, NULL, favorite: 3
1_t1, 1, favorite: 3
1_t2, 1, favorite: 3
1_t3, 1, favorite: 3
2, NULL, favorite: 4
2_t1, 2, favorite: 4
2_t2, 2, favorite: 4
2_t3, 2, favorite: 4
2_t4, 2, favorite: 4

正如您在上面看到的,我正在尝试将数据列收藏夹计数属性分解为它们自己的单独行,并使用parent_id列来表示其根记录。

到目前为止,我尝试使用Spark SQL Explode函数来尝试执行此操作,但是我无法使其正常工作。


共1个答案

匿名用户

如果我正确理解了您的问题,那么您正试图使用数据列中的数字从现有行生成/创建新行,并希望通过指向原始记录的新的<code>id</code>和<code>parent_id</code>生成这么多的新行

如果是这种情况,则可以使用mapflatMap操作执行此操作,如下所示:

import org.apache.spark.sql.Row

import scala.collection.mutable.ArrayBuffer

import sparkSession.sqlContext.implicits._

val input = Seq(("1", "NULL", "favorite:3"), ("2", "NULL", "favorite:4")).toDF("id", "parent_id", "data")

input.printSchema()
input.show(false)

val resultRDD = input.rdd.map(row => {
  val list = new ArrayBuffer[Row]()
  list += row

  val pointer = row.getAs[String]("data").split(":")(1).toInt

  for (index <- 1 to pointer) {
    val newId = s"${row.getAs[String]("id")}_t$index"
    list += Row.fromSeq(Seq(newId, row.getAs[String]("id"), row.getAs[String]("data")))
  }

  list
}).flatMap(_.toIterator)


val resultDF = sparkSession.createDataFrame(resultRDD, input.schema)
resultDF.show(false)

结果将是:

root
 |-- id: string (nullable = true)
 |-- parent_id: string (nullable = true)
 |-- data: string (nullable = true)

+---+---------+----------+
|id |parent_id|data      |
+---+---------+----------+
|1  |NULL     |favorite:3|
|2  |NULL     |favorite:4|
+---+---------+----------+

+----+---------+----------+
|id  |parent_id|data      |
+----+---------+----------+
|1   |NULL     |favorite:3|
|1_t1|1        |favorite:3|
|1_t2|1        |favorite:3|
|1_t3|1        |favorite:3|
|2   |NULL     |favorite:4|
|2_t1|2        |favorite:4|
|2_t2|2        |favorite:4|
|2_t3|2        |favorite:4|
|2_t4|2        |favorite:4|
+----+---------+----------+