What's the best way to inverse sort in scala?

问题

What is the best way to do an inverse sort in scala? I imagine the following is somewhat slow.

list.sortBy(_.size).reverse

Is there a conveinient way of using sortBy but getting a reverse sort? I would rather not need to use sortWith.

回答1:

There may be the obvious way of changing the sign, if you sort by some numeric value

list.sortBy(- _.size)

More generally, sorting may be done by method sorted with an implicit Ordering, which you may make explicit, and Ordering has a reverse (not the list reverse below) You can do

list.sorted(theOrdering.reverse)

If the ordering you want to reverse is the implicit ordering, you can get it by implicitly[Ordering[A]] (A the type you're ordering on) or better Ordering[A]. That would be

list.sorted(Ordering[TheType].reverse)

sortBy is like using Ordering.by, so you can do

list.sorted(Ordering.by(_.size).reverse)

Maybe not the shortest to write (compared to minus) but intent is clear

Update

The last line does not work. To accept the _ in Ordering.by(_.size), the compiler needs to know on which type we are ordering, so that it may type the _. It may seems that would be the type of the element of the list, but this is not so, as the signature of sorted is def sorted[B >: A](ordering: Ordering[B]). The ordering may be on A, but also on any ancestor of A (you might use byHashCode : Ordering[Any] = Ordering.by(_.hashCode)). And indeed, the fact that list is covariant forces this signature. One can do

list.sorted(Ordering.by((_: TheType).size).reverse)

but this is much less pleasant.

回答2:

list.sortBy(_.size)(Ordering[Int].reverse)

回答3:

maybe to shorten it a little more:

def Desc[T : Ordering] = implicitly[Ordering[T]].reverse

List("1","22","4444","333").sortBy( _.size )(Desc)

回答4:

Easy peasy (at least in case of size):

scala> val list = List("abc","a","abcde")
list: List[java.lang.String] = List(abc, a, abcde)

scala> list.sortBy(-_.size)
res0: List[java.lang.String] = List(abcde, abc, a)

scala> list.sortBy(_.size)
res1: List[java.lang.String] = List(a, abc, abcde)

回答5:

sortBy has implicit parameter ord which provides ordering

def sortBy [B] (f: (A) ⇒ B)(implicit ord: Ordering[B]): List[A]

so, we can define own Ordering object

scala> implicit object Comp extends Ordering[Int] {
 | override def compare (x: Int, y: Int): Int = y - x
 | }
defined module Comp

List(3,2,5,1,6).sortBy(x => x)
res5: List[Int] = List(6, 5, 3, 2, 1)

回答6:

val list = List(2, 5, 3, 1)
list.sortWith(_>_) -> res14: List[Int] = List(5, 3, 2, 1)
list.sortWith(_<_) -> res14: List[Int] = List(1, 2, 3, 5)

回答7:

Both sortWith and sortBy have a compact syntax:

case class Foo(time:Long, str:String)

val l = List(Foo(1, "hi"), Foo(2, "a"), Foo(3, "X"))

l.sortWith(_.time > _.time)  // List(Foo(3,X), Foo(2,a), Foo(1,hi))

l.sortBy(- _.time)           // List(Foo(3,X), Foo(2,a), Foo(1,hi))

l.sortBy(_.time)             // List(Foo(1,hi), Foo(2,a), Foo(3,X))

I find the one with sortWith easier to understand.

回答8:

Another possibility in cases where you pass a function that you may not be able to modify directly to an Arraybuffer via sortWith for example:

val buf = collection.mutable.ArrayBuffer[Int]()
buf += 3
buf += 9
buf += 1

// the sort function (may be passed through from elsewhere)
def sortFn = (A:Int, B:Int) => { A < B }

// the two ways to sort below
buf.sortWith(sortFn)                        // 1, 3, 9
buf.sortWith((A,B) => { ! sortFn(A,B) })    // 9, 3, 1

回答9:

this is my code ;)

val wordCounts = logData.flatMap(line => line.split(" "))
                        .map(word => (word, 1))
                        .reduceByKey((a, b) => a + b)

wordCounts.sortBy(- _._2).collect()

来源：https://stackoverflow.com/questions/7802851/whats-the-best-way-to-inverse-sort-in-scala