scala sortBy and sortWith

sortBy: sortBy[B](f: (A) ⇒ B)(implicit ord: math.Ordering[B]): List[A] 按照应用函数f之后产生的元素进行排序

sorted： sorted[B >: A](implicit ord: math.Ordering[B]): List[A] 按照元素自身进行排序

sortWith： sortWith(lt: (A, A) ⇒ Boolean): List[A] 使用自定义的比较函数进行排序，比较函数boolean

用法

<code class="hljs coffeescript has-numbering">val nums = List(<span class="hljs-number">1</span>,<span class="hljs-number">3</span>,<span class="hljs-number">2</span>,<span class="hljs-number">4</span>)
val sorted = nums.sorted  <span class="hljs-regexp">//</span>List(<span class="hljs-number">1</span>,<span class="hljs-number">2</span>,<span class="hljs-number">3</span>,<span class="hljs-number">4</span>)

val users = List((<span class="hljs-string">"HomeWay"</span>,<span class="hljs-number">25</span>),(<span class="hljs-string">"XSDYM"</span>,<span class="hljs-number">23</span>))
val sortedByAge = users.sortBy{<span class="hljs-reserved">case</span><span class="hljs-function"><span class="hljs-params">(user,age)</span> =></span> age}  <span class="hljs-regexp">//</span>List((<span class="hljs-string">"XSDYM"</span>,<span class="hljs-number">23</span>),(<span class="hljs-string">"HomeWay"</span>,<span class="hljs-number">25</span>))
val sortedWith = users.sortWith{<span class="hljs-reserved">case</span><span class="hljs-function"><span class="hljs-params">(user1,user2)</span> =></span> user1._2 < user2._2} <span class="hljs-regexp">//</span>List((<span class="hljs-string">"XSDYM"</span>,<span class="hljs-number">23</span>),(<span class="hljs-string">"HomeWay"</span>,<span class="hljs-number">25</span>))</code>

How to sort a Scala Map by key or value (sortBy, sortWith)

By Alvin Alexander. Last updated: Jun 6, 2015

This is an excerpt from the Scala Cookbook (partially modified for the internet). This is Recipe 11.23, “How to Sort an Existing Map by Key or Value”

Problem

You have an unsorted map and want to sort the elements in the map by the key or value.

Solution

Given a basic, immutable Map:

scala> val grades = Map("Kim" -> 90,
     |     "Al" -> 85,
     |     "Melissa" -> 95,
     |     "Emily" -> 91,
     |     "Hannah" -> 92
     | )
grades: scala.collection.immutable.Map[String,Int] = Map(Hannah -> 92, Melissa -> 95, Kim -> 90, Emily -> 91, Al -> 85)

You can sort the map by key, from low to high, using sortBy:

scala> import scala.collection.immutable.ListMap
import scala.collection.immutable.ListMap

scala> ListMap(grades.toSeq.sortBy(_._1):_*)
res0: scala.collection.immutable.ListMap[String,Int] = Map(Al -> 85, Emily -> 91, Hannah -> 92, Kim -> 90, Melissa -> 95)

You can also sort the keys in ascending or descending order using sortWith:

// low to high
scala> ListMap(grades.toSeq.sortWith(_._1 < _._1):_*)
res0: scala.collection.immutable.ListMap[String,Int] = Map(Al -> 85, Emily -> 91, Hannah -> 92, Kim -> 90, Melissa -> 95)

// high to low
scala> ListMap(grades.toSeq.sortWith(_._1 > _._1):_*)
res1: scala.collection.immutable.ListMap[String,Int] = Map(Melissa -> 95, Kim -> 90, Hannah -> 92, Emily -> 91, Al -> 85)

You can sort the map by value using sortBy:

scala> ListMap(grades.toSeq.sortBy(_._2):_*)
res0: scala.collection.immutable.ListMap[String,Int] = Map(Al -> 85, Kim -> 90, Emily -> 91, Hannah -> 92, Melissa -> 95)

You can also sort by value in ascending or descending order using sortWith:

// low to high
scala> ListMap(grades.toSeq.sortWith(_._2 < _._2):_*)
res0: scala.collection.immutable.ListMap[String,Int] = Map(Al -> 85, Kim -> 90, Emily -> 91, Hannah -> 92, Melissa -> 95)

// high to low
scala> ListMap(grades.toSeq.sortWith(_._2 > _._2):_*)
res1: scala.collection.immutable.ListMap[String,Int] = Map(Melissa -> 95, Hannah -> 92, Emily -> 91, Kim -> 90, Al -> 85)

In all of these examples, you’re not sorting the existing map; the sort methods result in a new sorted map, so the output of the result needs to be assigned to a new variable.

Also, you can use either a ListMap or a LinkedHashMap in these recipes. This example shows how to use a LinkedHashMap and assign the result to a new variable:

scala> val x = collection.mutable.LinkedHashMap(grades.toSeq.sortBy(_._1):_*)
x: scala.collection.mutable.LinkedHashMap[String,Int] =
    Map(Al -> 85, Emily -> 91, Hannah -> 92, Kim -> 90, Melissa -> 95)

scala> x.foreach(println)
(Al,85)
(Emily,91)
(Hannah,92)
(Kim,90)
(Melissa,95)

转：https://blog.csdn.net/mengxing87/article/details/51636080

有关 sortWith 底层，请看这里：https://www.cnblogs.com/nucdy/p/9270594.html