[Solved]-In Scala, how to use Ordering[T] with List.min or List.max and keep code readable-scala

You could always define your own implicit conversion:

implicit def funToOrdering[T,R <% Ordered[R]](f: T => R) = new Ordering[T] {
  def compare(x: T, y: T) = f(x) compare f(y)
}

val list = ("a", 5) :: ("b", 3) :: ("c", 2) :: Nil

list.min { t: (String,Int) => t._2 }  // (c, 2)

EDIT: Per @Dario's comments.

Might be more readable if the conversion wasn't implicit, but using an "on" function:

def on[T,R <% Ordered[R]](f: T => R) = new Ordering[T] {
  def compare(x: T, y: T) = f(x) compare f(y)
}

val list = ("a", 5) :: ("b", 3) :: ("c", 2) :: Nil

list.min( on { t: (String,Int) => t._2 } ) // (c, 2)

list.min(Ordering.fromLessThan[(String, Int)](_._2 < _._2))

Which is still too verbose, of course. I'd probably declare it as a val or object.

The function Ordering#on witnesses the fact that Ordering is a contra-variant functor. Others include Comparator, Function1, Comparable and scalaz.Equal.

Scalaz provides a unified view on these types, so for any of them you can adapt the input with value contramap f, or with symbolic denotation, value ∙ f

scala> import scalaz._
import scalaz._

scala> import Scalaz._
import Scalaz._

scala> val ordering = implicitly[scala.Ordering[Int]] ∙ {x: (_, Int) => x._2}
ordering: scala.math.Ordering[Tuple2[_, Int]] = scala.math.Ordering$$anon$2@34df289d

scala> List(("1", 1), ("2", 2)) min ordering  
res2: (java.lang.String, Int) = (1,1)

Here's the conversion from the Ordering[Int] to Ordering[(_, Int)] in more detail:

scala> scalaz.Scalaz.maContravariantImplicit[Ordering, Int](Ordering.Int).contramap { x: (_, Int) => x._2 }
res8: scala.math.Ordering[Tuple2[_, Int]] = scala.math.Ordering$$anon$2@4fa666bf

One thing you can do is use the more concise standard tuple type syntax instead of using Tuple2:

val min = list.min(new Ordering[(String, Int)] { 
  def compare(x: (String, Int), y: (String, Int)): Int = x._2 compare y._2 
})

Or use reduceLeft to have a more concise solution altogether:

val min = list.reduceLeft((a, b) => (if (a._2 < b._2) a else b))

Or you could sort the list by your criterion and get the first element (or last for the max):

val min = list.sort( (a, b) => a._2 < b._2 ).first

Which can be further shortened using the placeholder syntax:

val min = list.sort( _._2 < _._2 ).first

Which, as you wrote yourself, can be shortened to:

val min = list.sortBy( _._2 ).first

But as you suggested sortBy yourself, I'm not sure if you are looking for something different here.

In Scala 2.9, you can do list minBy { _._2 }.