如何在Apache Spark中基于时间的两个数据集的联合？

Question

Given two datasets of S and R both with a time column (t) as described below:

//snapshot with id at t
case class S(id: String, t: Int)

//reference data at t
case class R(t: Int, fk: String)

//Example test case
val ss: Dataset[S] = Seq(S("a", 1), S("a", 3), S("b", 5), S("b", 7))
      .toDS

    val rs: Dataset[R] = Seq(R(0, "a"), R(2, "a"), R(6, "b"))
      .toDS

    val srs: Dataset[(S, Option[R])] = ss
      .asOfJoin(rs)

    srs.collect() must contain theSameElementsAs
      Seq((S("a", 1), Some(R(0, "a"))), (S("a", 3), Some(R(2, "a"))), (S("b", 5), None), (S("b", 7), Some(R(6, "b"))))

Goal is to find the most recent row in R that matches E's id if possible ie R can be optional in the output.

asOfJoin is defined as below:

  implicit class SOps(ss: Dataset[S]) {
    def asOfJoin(rs: Dataset[R])(implicit spark: SparkSession): Dataset[(S, Option[R])] = ???
  }

One solution using Dataset API is as follows:

def asOfJoin(rs: Dataset[R])(implicit spark: SparkSession): Dataset[(S, Option[R])] = {
      import spark.implicits._

      ss
        .joinWith(
          rs,
          ss("id") === rs("fk") && ss("t") >= rs("t"),
          "left_outer")
       .map { case (l, r) => (l, Option(r)) }
       .groupByKey { case (s, _) => s }
       .reduceGroups { (x, y) =>
         (x, y) match {
           case ((_, Some(R(tx, _))), (_, Some(R(ty, _)))) => if (tx > ty) x else y
           case _ => x
         }
       }
       .map { case (_, r) => r }
}

Answer 1

I'm not sure about the size of the dataset S and dataset R. But from your codes, I can see that the efficiency of the join(with unequal expressions) is bad, and I can give some suggestions based on different specific scenarios:

Either Dataset R or Dataset S doesn't have too much data.

I suggest that you can broadcast the smaller dataset and finish the business logic in a spark udf with the help of broadcast variable. In this way, you don't need the shuffle(join) process, which helps you save a lot of time and resources.

For every unique id, count(distinct t) is not big.

I suggest that you can do a pre-aggregation by grouping id and collect_set(t) like this:

select id,collect_set(t) as t_set from S 

In this way, you can remove the unequal expression(ss("t") >= rs("t")) in the join. And write your business logic with two t_sets from dataset S and dataset R.

For other scenarios:

I suggest that you optimize your codes with a equal join and a window function. Since I'm more familiar with SQL, I write SQL here, which can be transformed to dataset API:

select
  sid,
  st,
  rt
from
(
    select 
      S.id as sid,
      S.t as st,
      R.t as rt,
      row_number() over (partition by S.id order by (S.t - NVL(R.t, 0)) rn
    from
      S
    left join R on S.id = R.fk) tbl
where tbl.rn = 1

Answer 2

I took @bupt_ljy 's comment about avoiding a theta join and following seems to scale really well:

def asOfJoin(rs: Dataset[R])(implicit spark: SparkSession): Dataset[(S, Option[R])] = {
  import spark.implicits._

  ss
    .joinWith(
      rs.sort(rs("fk"), rs("t")),
      ss("id") === rs("fk"),
      "left_outer")
    .map { case (l, r) => (l, Option(r)) }
    .groupByKey { case (s, _) => s }
    .flatMapGroups { (k, vs) =>
      new Iterator[(S, Option[R])] {
        private var didNotStart: Boolean = true

        override def hasNext: Boolean = didNotStart

        override def next(): (S, Option[R]) = {
          didNotStart = false
          vs
            .find { case (l, rOpt) =>
              rOpt match {
                case Some(r) => l.t >= r.t
                case _ => false
              }
            }.getOrElse((k, None))
        }
      }
    }
}

However, still super imperative code and there must be a better way...