基於排序機制的Wordcount程序

Java版：

package cn.spark.sparktest;

import org.apache.spark.SparkConf;
import org.apache.spark.SparkContext;
import org.apache.spark.api.java.JavaPairRDD;
import org.apache.spark.api.java.JavaRDD;
import org.apache.spark.api.java.JavaSparkContext;
import org.apache.spark.api.java.function.*;
import scala.Tuple2;

import java.util.Arrays;

public class sortWordcount {

    public static void main(String[] args){

        SparkConf conf = new SparkConf()
                .setMaster("local")
                .setAppName("sortWordcount");

        JavaSparkContext sc = new JavaSparkContext(conf);

        JavaRDD<String> lines = sc.textFile("C://User//Desktop//spark.txt");

        JavaRDD<String> mid = lines.flatMap(new FlatMapFunction<String, String>() {
            @Override
            public Iterable<String> call(String s) throws Exception {
                return Arrays.asList(s.split(" "));
            }
        });

        JavaPairRDD<String ,Integer> pairs = mid.mapToPair(new PairFunction<String, String, Integer>() {
            @Override
            public Tuple2<String, Integer> call(String s) throws Exception {
                return new Tuple2<String, Integer>(s,1);
            }
        });

        JavaPairRDD<String,Integer> redu = pairs.reduceByKey(new Function2<Integer, Integer, Integer>() {
            @Override
            public Integer call(Integer i, Integer h) throws Exception {
                return i + h;
            }
        });


        // 到這裏爲止，就得到了每個單詞出現的次數
        // 但是，問題是，我們的新需求，是要按照每個單詞出現次數的順序，降序排序
        // wordCounts RDD內的元素是什麼？應該是這種格式的吧：(hello, 3) (you, 2)
        // 我們需要將RDD轉換成(3, hello) (2, you)的這種格式，才能根據單詞出現次數進行排序把！

        // 進行key-value的反轉映射

        JavaPairRDD<Integer,String>  tran = redu.mapToPair(new PairFunction<Tuple2<String, Integer>, Integer, String>() {
            @Override
            public Tuple2<Integer, String> call(Tuple2<String, Integer> t) throws Exception {
                return new Tuple2<Integer, String>(t._2,t._1);
            }
        });

        // 按照key進行排序
        JavaPairRDD<Integer , String> sort = tran.sortByKey(false);
        // 再次將value-key進行反轉映射
        JavaPairRDD<String,Integer> tran1 = sort.mapToPair(new PairFunction<Tuple2<Integer, String>, String, Integer>() {
            @Override
            public Tuple2<String, Integer> call(Tuple2<Integer, String> s) throws Exception {
                return new Tuple2<String, Integer>(s._2,s._1);
            }
        });


        // 到此爲止，我們獲得了按照單詞出現次數排序後的單詞計數
        // 打印出來

        tran1.foreach(new VoidFunction<Tuple2<String, Integer>>() {
            @Override
            public void call(Tuple2<String, Integer> stringIntegerTuple2) throws Exception {
                System.out.println(stringIntegerTuple2);
            }
        });

        sc.close();

    }
}

測試：

Scala版：

package cn.spark.study.core

import org.apache.spark.{SparkConf, SparkContext}

object sortWordcount {
  def main(args: Array[String]): Unit = {

    val conf = new SparkConf()
      .setAppName("sortWordcount")
      .setMaster("local")
    val sc = new SparkContext(conf)

    val lines = sc.textFile("C://Users//gaochen//Desktop//spark.txt")

    val flat = lines.flatMap(x => x.split(" "))
    val mp =flat.map(x => (x ,1))
    val red = mp.reduceByKey( _ + _)

    val tran = red.map(x => (x._2,x._1))
    val sortCount = tran.sortByKey(false)
    val tran1 = sortCount.map(x => (x._2,x._1))

    tran1.foreach(x => println(x._1,x._2))

  }
}

測試：