CODEWITHSUNDEEP
scalasortingcollections
DimParf
DimParf

Reputation: 295

How to convert, group and sort java.util.List[java.util.Map[String, Object]]?

I convert this:

import scala.collection.JavaConverters._
import scala.collection.JavaConverters._
val list:java.util.List[java.util.Map[String, Object]] = new java.util.ArrayList[java.util.Map[String, Object]]()
val map1:java.util.Map[String, AnyRef] = new java.util.HashMap[String,AnyRef]()
map1.put("payout", 3.asInstanceOf[AnyRef])
list.add(map1)
val map2:java.util.Map[String, AnyRef] = new java.util.HashMap[String, AnyRef]()
map2.put("payout", 2.asInstanceOf[AnyRef])
list.add(map2)
val map3:java.util.Map[String, AnyRef] = new java.util.HashMap[String, AnyRef]()
map3.put("payout", 2.asInstanceOf[AnyRef])
list.add(map3)
val map4:java.util.Map[String, AnyRef] = new java.util.HashMap[String, AnyRef]()
map4.put("payout", 1.asInstanceOf[AnyRef])
list.add(map4)
println(list)
val result = list.asScala
//result Buffer({payout=3}, {payout=2}, {payout=2}, {payout=1})

And i wish: list.asScala.groupBy(_("payout")).toList save its ordering (sort by payout) but .toList.sortBy(_._1) throw error: error: No implicit Ordering defined for java.lang.Object.

val result = list.groupBy(_("payout")).toList.sortBy(_._1)

Upvotes: 2

Views: 1406

Answers (1)

thoredge
thoredge

Reputation: 12601

This gives a result, but I don't know if its what you wanted:

val result = list.asScala.map(_.asScala).groupBy(_("payout")).toList.sortWith(_._1.asInstanceOf[Int] > _._1.asInstanceOf[Int])

I added map(.asScala) in order to convert your java maps to scala maps. The group by value is a java.lang.Object which does not have an ordering; using sortWith(._1.asInstanceOf[Int] > _._1.asInstanceOf[Int]) I cast it to Int in order to sort it. This will of course crash if some other object is used, but there is no way to order an object that you don't know anything about.

Upvotes: 3

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