Scala 复杂分词求和（二元组）

package chapter07

object Test18_ComplexWordCount {
  def main(args: Array[String]): Unit = {
    val tupleList: List[(String, Int)] = List(
      ("hello", 1),
      ("hello world", 2),
      ("hello scala", 3),
      ("hello spark from scala", 1),
      ("hello flink from scala", 2)
    )

    // 思路一：直接展开为普通版本
    val newStringList: List[String] = tupleList.map(
      kv => {
        (kv._1.trim + " ") * kv._2
      }
    )
    println(newStringList)

    // 接下来操作与普通版本完全一致
    val wordCountList: List[(String, Int)] = newStringList
      .flatMap(_.split(" "))    // 空格分词
      .groupBy( word => word )     // 按照单词分组
      .map( kv => (kv._1, kv._2.size) )     // 统计出每个单词的个数
      .toList
      .sortBy(_._2)(Ordering[Int].reverse)
      .take(3)

    println(wordCountList)

    println("================================")

    // 思路二：直接基于预统计的结果进行转换
    // 1. 将字符串打散为单词，并结合对应的个数包装成二元组List((hello,1), (hello,2), (world,2), (hello,3), (scala,3), (
    val preCountList: List[(String, Int)] = tupleList.flatMap(
      tuple => {
        val strings: Array[String] = tuple._1.split(" ")
        strings.map( word => (word, tuple._2) )
      }
    )
    println(preCountList)

    // 2. 对二元组按照单词进行分组
    val preCountMap: Map[String, List[(String, Int)]] = preCountList.groupBy( _._1 )
    println(preCountMap)

    // 3. 叠加每个单词预统计的个数值
    val countMap: Map[String, Int] = preCountMap.mapValues(
      tupleList => tupleList.map(_._2).sum
    )
    println(countMap)

    // 4. 转换成list，排序取前3
    val countList = countMap.toList
      .sortWith(_._2 > _._2)
      .take(3)
    println(countList)
  }
}

("hello", 1), 说明“hello”字符串已知出现了两次！