Scala中做简易wordCount

使用foldLeft函数，实现简易的wordCount

import scala.collection.mutable

object Demo_019 {
  def main(args: Array[String]): Unit = {
    val list = List("bigdata han hello ", "bigdata han aaa aaa aaa ccc ddd uuu")
    val map01 = mutable.Map[String, Int]()
    list.foldLeft(map01)(count2)
    println(map01)
  }

  def count2(map2: mutable.Map[String, Int], str: String): mutable.Map[String, Int] = {
    val arr: Array[String] = str.split(" ")
    for (elem <- arr) {
      map2 += (elem -> (map2.getOrElse(elem, 0) + 1))
    }
    map2
  }

运行结果为：

 

当然这种方式，还是显得太罗嗦了，还有更为简洁的方式，而且还可以结果进行正序或逆序排序

简单一句就是：

 val result = list.flatMap(_.split(" ")).map((_,1)).groupBy(_._1).map(s => (s._1,s._2.size))

显然不容易懂，下面是详细说明

object Demo_019_01 {
  def main(args: Array[String]): Unit = {
    val list = List("bigdata han hello ", "bigdata han aaa aaa aaa ccc ddd uuu")
//  val result2 = list.flatMap((x: String) => x.split(" "))
    val result2 = list.flatMap(_.split(" "))
    println("result2："+result2)
//  val result3 = result2.map((x: String) => (x, 1))
    val result3 = result2.map((_,1))
    println("result3："+result3)
//  val result4 = result3.groupBy((x: (String, Int)) => x._1)
    val result4 = result3.groupBy(_._1)
    println("result4："+result4)
//  val result5 = result4.map((s: (String, List[(String, Int)])) => (s._1, s._2.size))
    val result5= result4.map(s => (s._1,s._2.size))
    println("result5："+result5)
//  val result6 = result5.toList.sortBy((x: (String, Int)) => x._2).reverse
    val result6 = result5.toList.sortBy(_._2)
    println("result6："+result6)
  }
}

　　输出结果为

 

上面使用了参数类型推断，关于参数类型推断，介绍如下