How do I sort a collection of Lists in lexicographic order in Scala?

sorting scala html-lists lexicographic
Scott Morrison picture Scott Morrison · Jun 29, 2010 · Viewed 8k times · Source

If A has the Ordered[A] trait, I'd like to be able to have code that works like this

val collection: List[List[A]] = ... // construct a list of lists of As
val sorted = collection sort { _ < _ }

and get something where the lists have been sorted in lexicographic order. Of course, just because A has the trait Ordered[A] doesn't mean that List[A] has the trait Ordered[List[A]]. Presumably, however, the 'scala way' to do this is with an implicit def.

How do I implicitly convert a List[A] to a Ordered[List[A]], assuming A has the trait Ordered[A] (so that the code above just works)?

I have in mind using the lexicographic ordering on List[A] objects, but I'd like code that can be adapted to others orderings.

Answer

Carl Witty picture Carl Witty · Mar 30, 2011

Inspired by Ben Lings' answer, I managed to work out what seems like the simplest way to sort lists lexicographically: add the line

import scala.math.Ordering.Implicits._

before doing your List[Int] comparison, to ensure that the implicit function infixOrderingOps is in scope.

Related questions

How do I sort an array in Scala?

I can see there's a sorting object, Sorting, with a quicksort method, quickSort, on it. What would be a code example of using it, sorting an array of object of arbitrary type? It looks like I need to pass in …

Read More
Easy idiomatic way to define Ordering for a simple case class

I have a list of simple scala case class instances and I want to print them in predictable, lexicographical order using list.sorted, but receive "No implicit Ordering defined for ...". Is there exist an implicit that provides lexicographical ordering for …

Read More
How to sort a list in Scala by two fields?

how to sort a list in Scala by two fields, in this example I will sort by lastName and firstName? case class Row(var firstName: String, var lastName: String, var city: String) var rows = List(new Row("Oscar", "Wilde", "London"), …

Read More