Spark - group and aggregate only several smallest items - scala

Question

In short
I have cartesian-product (cross-join) of two dataframes and function which gives some score for given element of this product. I want now to get few "best matched" elements of the second DF for every member of the first DF.
In details
What follows is a simplified example as my real code is somewhat bloated with additional fields and filters.
Given two sets of data, each having some id and value:
// simple rdds of tuples
val rdd1 = sc.parallelize(Seq(("a", 31),("b", 41),("c", 59),("d", 26),("e",53),("f",58)))
val rdd2 = sc.parallelize(Seq(("z", 16),("y", 18),("x",3),("w",39),("v",98), ("u", 88)))
// convert them to dataframes:
val df1 = spark.createDataFrame(rdd1).toDF("id1", "val1")
val df2 = spark.createDataFrame(rdd2).toDF("id2", "val2")
and some function which for pair of the elements from the first and second dataset gives their "matching score":
def f(a:Int, b:Int):Int = (a * a + b * b * b) % 17
// convert it to udf
val fu = udf((a:Int, b:Int) => f(a, b))
we can create the product of two sets and calculate score for every pair:
val dfc = df1.crossJoin(df2)
val r = dfc.withColumn("rez", fu(col("val1"), col("val2")))
r.show
+---+----+---+----+---+
|id1|val1|id2|val2|rez|
+---+----+---+----+---+
| a| 31| z| 16| 8|
| a| 31| y| 18| 10|
| a| 31| x| 3| 2|
| a| 31| w| 39| 15|
| a| 31| v| 98| 13|
| a| 31| u| 88| 2|
| b| 41| z| 16| 14|
| c| 59| z| 16| 12|
...
And now we want to have this result grouped by id1:
r.groupBy("id1").agg(collect_set(struct("id2", "rez")).as("matches")).show
+---+--------------------+
|id1| matches|
+---+--------------------+
| f|[[v,2], [u,8], [y...|
| e|[[y,5], [z,3], [x...|
| d|[[w,2], [x,6], [v...|
| c|[[w,2], [x,6], [v...|
| b|[[v,2], [u,8], [y...|
| a|[[x,2], [y,10], [...|
+---+--------------------+
But really we want only to retain only few (say 3) of "matches", those having the best score (say, least score).
The question is
How to get the "matches" sorted and reduced to top-N elements? Probably it is something about collect_list and sort_array, though I don't know how to sort by inner field.
Is there a way to ensure optimization in case of large input DFs - e.g. choosing minimums directly while aggregating. I know it could be done easily if I wrote the code without spark - keeping small array or priority queue for every id1 and adding element where it should be, possibly dropping out some previously added.
E.g. it's ok that cross-join is costly operation, but I want to avoid wasting memory on the results most of which I'm going to drop in the next step. My real use case deals with DFs with less than 1 mln entries so cross-join is yet viable but as we want to select only 10-20 top matches for each id1 it seems to be quite desirable not to keep unnecessary data between steps.

Answer 1

For start we need to take only the first n rows. To do this we are partitioning the DF by 'id1' and sorting the groups by the res. We use it to add row number column to the DF, like that we can use where function to take the first n rows. Than you can continue doing the same code your wrote. Grouping by 'id1' and collecting the list. Only now you already have the highest rows.
import org.apache.spark.sql.expressions.Window
import org.apache.spark.sql.functions._
val n = 3
val w = Window.partitionBy($"id1").orderBy($"res".desc)
val res = r.withColumn("rn", row_number.over(w)).where($"rn" <= n).groupBy("id1").agg(collect_set(struct("id2", "res")).as("matches"))
A second option that might be better because you won't need to group the DF twice:
val sortTakeUDF = udf{(xs: Seq[Row], n: Int)} => xs.sortBy(_.getAs[Int]("res")).reverse.take(n).map{case Row(x: String, y:Int)}}
r.groupBy("id1").agg(sortTakeUDF(collect_set(struct("id2", "res")), lit(n)).as("matches"))
In here we create a udf that take the array column and an integer value n. The udf sorts the array by your 'res' and returns only the first n elements.