Question

我的数据框类似于：

+---+-----+-----+
|key|thing|value|
+---+-----+-----+
| u1|  foo|    1|
| u1|  foo|    2|
| u1|  bar|   10|
| u2|  foo|   10|
| u2|  foo|    2|
| u2|  bar|   10|
+---+-----+-----+

并希望得到以下结果：

+---+-----+---------+----+
|key|thing|sum_value|rank|
+---+-----+---------+----+
| u1|  bar|       10|   1|
| u1|  foo|        3|   2|
| u2|  foo|       12|   1|
| u2|  bar|       10|   2|
+---+-----+---------+----+

当前，有类似的代码：

val df = Seq(("u1", "foo", 1), ("u1", "foo", 2), ("u1", "bar", 10), ("u2", "foo", 10), ("u2", "foo", 2), ("u2", "bar", 10)).toDF("key", "thing", "value")

 // calculate sums per key and thing
 val aggregated = df.groupBy("key", "thing").agg(sum("value").alias("sum_value"))

 // get topk items per key
 val k = lit(10)
 val topk = aggregated.withColumn("rank", rank over  Window.partitionBy("key").orderBy(desc("sum_value"))).filter('rank < k)

但是，此代码非常效率低。窗口功能会生成商品的总订单，并导致随机排序。

如何更有效地计算前k个项目？也许使用近似函数，即类似于https://datasketches.github.io/或https://spark.apache.org/docs/latest/ml-frequent-pattern-mining.html

的草图

Answer 1

这是推荐系统的经典算法。

case class Rating(thing: String, value: Int) extends Ordered[Rating] {
  def compare(that: Rating): Int = -this.value.compare(that.value)
}

case class Recommendation(key: Int, ratings: Seq[Rating]) {
  def keep(n: Int) = this.copy(ratings = ratings.sorted.take(n))
}

val TOPK = 10

df.groupBy('key)
  .agg(collect_list(struct('thing, 'value)) as "ratings")
  .as[Recommendation]
  .map(_.keep(TOPK))

您还可以在以下位置查看源代码：

Spotify大数据Rosetta代码/ TopItemsPerUser.scala，此处是针对Spark或Scio的几种解决方案
Spark MLLib / TopByKeyAggregator.scala，被认为是使用推荐算法时的最佳实践，尽管看起来他们的示例仍然使用RDD。

import org.apache.spark.mllib.rdd.MLPairRDDFunctions._

sc.parallelize(Array(("u1", ("foo", 1)), ("u1", ("foo", 2)), ("u1", ("bar", 10)), ("u2", ("foo", 10)),
  ("u2", ("foo", 2)), ("u2", ("bar", 10))))
  .topByKey(10)(Ordering.by(_._2))

Answer 2

RDD进行救援

aggregated.as[(String, String, Long)].rdd.groupBy(_._1).map{ case (thing, it) => (thing, it.map(e=> (e._2, e._3)).toList.sortBy(sorter => sorter._2).take(1))}.toDF.show
+---+----------+
| _1|        _2|
+---+----------+
| u1| [[foo,3]]|
| u2|[[bar,10]]|
+---+----------+

很可能可以使用评论中的建议对此进行改进。即不是从aggregated开始，而是从df开始。这可能类似于：

df.as[(String, String, Long)].rdd.groupBy(_._1).map{case (thing, it) => {
      val aggregatedInner = it.groupBy(e=> (e._2)).mapValues(events=> events.map(value => value._3).sum)
      val topk = aggregatedInner.toArray.sortBy(sorter=> sorter._2).take(1)
      (thing, topk)
    }}.toDF.show

有效计算Spark中的前k个元素

2 个答案: