hadoop mapreduce无序元组作为映射键

Question

根据Hadoop - 权威指南中的wordcount示例，我开发了一个mapreduce工作来计算字符串无序元组的出现次数。输入看起来像这样（只是更大）：

  

a b
  c c
  d d
  b a   a d
  d d

运行mapreduce我希望输出（对于这个例子）：

  

c c 1
  d d 1
  a b 2
  a d 1
  d d 1

这意味着，我希望元组a，b和b，a被认为是相同的。这里已经提出了这个问题：Hadoop MapReduce: Two values as key in Mapper-Reducer并且可能在这里解决了https://developer.yahoo.com/hadoop/tutorial/module5.html#keytypes。

对于大输入文件，我得到这样的输出，第一列是resp的hashCode。键：

  

151757761 a a 62822
  153322274 a b 62516
  154886787 a c 62248
  156451300 a d 62495
  153322274 b a 62334
  154902916 b b 62232
  158064200 b d 62759
  154886787 c a 62200
  156483558 c b 124966
  158080329 c c 62347
  159677100 d c 125047
  156451300 d a 62653
  158064200 d b 62603
  161290000 d d 62778

可以看出，有些键是重复的，如a，b和b的153322274，a。对于其他人，例如c，b（和b，c）和c，d（和d，c），计数是正确的。因为测试数据是随机抽取的，所以数量大大超过其他数量。

我已经在一段时间内一直在寻找这个问题，现在已经没有想法，为什么在减少阶段之后仍然会有重复的重复。

以下是我使用的代码：

首先是我的自定义WritableComparable的代码

import org.apache.hadoop.io.WritableComparable;
import org.apache.hadoop.io.WritableUtils;

import java.io.DataInput;
import java.io.DataOutput;
import java.io.IOException;
import java.math.BigInteger;

public class Pair implements WritableComparable<Pair> {

    private String first;
    private String second;

    public Pair(String first, String second) {
        this.first = first;
        this.second = second;
    }

    public Pair() {
       this("", "");
    }

    @Override
    public String toString() {
        return this.hashCode() + "\t" + first + "\t" + second;
    }

    @Override
    public void write(DataOutput out) throws IOException {
        WritableUtils.writeString(out, first);
        WritableUtils.writeString(out, second);
    }

    @Override
    public void readFields(DataInput in) throws IOException {
        first = WritableUtils.readString(in);
        second = WritableUtils.readString(in);
    }

    @Override
    public int hashCode() {
        BigInteger bA = BigInteger.ZERO;
        BigInteger bB = BigInteger.ZERO;

        for(int i = 0; i < first.length(); i++) {
            bA = bA.add(BigInteger.valueOf(127L).pow(i+1).multiply(BigInteger.valueOf(first.codePointAt(i))));
        }

        for(int i = 0; i < second.length(); i++) {
            bB = bB.add(BigInteger.valueOf(127L).pow(i+1).multiply(BigInteger.valueOf(second.codePointAt(i))));
        }

        return bA.multiply(bB).intValue();
    }

    @Override
    public boolean equals(Object o) {
        if (o instanceof Pair) {
            Pair other = (Pair) o;

            boolean result = ( first.compareTo(other.first) == 0 && second.compareTo(other.second) == 0 )
                    || ( first.compareTo(other.second) == 0 && second.compareTo(other.first) == 0 );

            return result;
        }
        return false;
    }

    @Override
    public int compareTo(Pair other) {
        if (( first.compareTo(other.first) == 0 && second.compareTo(other.second) == 0 )
                || ( first.compareTo(other.second) == 0 && second.compareTo(other.first) == 0 ) ) {
            return 0;
        } else {
            int cmp = first.compareTo( other.first );

            if (cmp != 0) {
                return cmp;
            }

            return second.compareTo( other.second );
        }
    }
}

其余的：

import java.io.IOException;
import java.util.StringTokenizer;
import org.apache.hadoop.conf.Configuration;
import org.apache.hadoop.fs.Path;
import org.apache.hadoop.io.IntWritable;
import org.apache.hadoop.io.Text;
import org.apache.hadoop.mapreduce.Job;
import org.apache.hadoop.mapreduce.Mapper;
import org.apache.hadoop.mapreduce.Reducer;
import org.apache.hadoop.mapreduce.lib.input.FileInputFormat;
import org.apache.hadoop.mapreduce.lib.output.FileOutputFormat;
import org.apache.hadoop.util.GenericOptionsParser;

public class PairCount {

    public static void main(String[] args) throws Exception {
        Configuration conf = new Configuration();
        String[] otherArgs = new GenericOptionsParser(conf, args).getRemainingArgs();

        if (otherArgs.length < 2) {
            System.err.println("Usage: paircount <in-dir> <out-dir>");
            System.exit(2);
        }

        Job job = new Job(conf, "word count");
        job.setJarByClass(PairCount.class);

        job.setMapperClass(TokenizerMapper.class);
        job.setReducerClass(IntSumReducer.class);

        job.setMapOutputKeyClass(Pair.class);
        job.setMapOutputValueClass(IntWritable.class);

        job.setOutputKeyClass(Pair.class);
        job.setOutputValueClass(IntWritable.class);

        for (int i = 0; i < otherArgs.length - 1; ++i) {
            FileInputFormat.addInputPath(job, new Path(otherArgs[i]));
        }

        FileOutputFormat.setOutputPath(job, new Path(otherArgs[otherArgs.length - 1]));

        System.exit(job.waitForCompletion(true) ? 0 : 1);
    }

    public static class TokenizerMapper extends Mapper<Object, Text, Pair, IntWritable> {

        private final static IntWritable one = new IntWritable(1);

        public void map(Object key, Text value, Context context) throws IOException, InterruptedException {
            StringTokenizer itr = new StringTokenizer(value.toString());

            while (itr.hasMoreTokens()) {
                context.write(new Pair(itr.nextToken(), itr.nextToken()), one);
            }
        }
    }

    public static class IntSumReducer extends Reducer<Pair, IntWritable, Pair, IntWritable> {
        private IntWritable result = new IntWritable();

        public void reduce(Pair key, Iterable<IntWritable> values, Context context) throws IOException, InterruptedException {
            int sum = 0;

            for (IntWritable val : values) {
                sum += val.get();
            }

            result.set(sum);
            context.write( key, result);
        }
    }
}

编辑：我为hashCode（）和compareTo（）函数添加了单元测试。他们工作得很好。

import org.junit.Test;
import static org.junit.Assert.assertEquals;
import static org.junit.Assert.assertNotEquals;
import static org.junit.Assert.assertTrue;

public class Tests  {
    @Test
    public void testPairComparison() {
        assertTrue( 0 == new Pair("a", "a").compareTo(new Pair("a", "a")) );
        assertTrue( 0 == new Pair("a", "b").compareTo(new Pair("b", "a")) );
        assertTrue( 0 == new Pair("a", "c").compareTo(new Pair("c", "a")) );
        assertTrue( 0 == new Pair("a", "d").compareTo(new Pair("d", "a")) );

        assertTrue( 0 == new Pair("b", "b").compareTo(new Pair("b", "b")) );
        assertTrue( 0 == new Pair("b", "c").compareTo(new Pair("c", "b")) );
        assertTrue( 0 == new Pair("b", "d").compareTo(new Pair("d", "b")) );

        assertTrue( 0 == new Pair("c", "c").compareTo(new Pair("c", "c")) );
        assertTrue( 0 == new Pair("c", "d").compareTo(new Pair("d", "c")) );

        assertTrue( 0 == new Pair("d", "d").compareTo(new Pair("d", "d")) );

        assertTrue( 0 > new Pair("a", "a").compareTo(new Pair("b", "b")) );
        assertTrue( 0 > new Pair("a", "a").compareTo(new Pair("c", "b")) );
        assertTrue( 0 < new Pair("d", "d").compareTo(new Pair("c", "b")) );
        assertTrue( 0 < new Pair("c", "d").compareTo(new Pair("c", "a")) );
    }

    @Test
    public void testPairHashcode(){
        assertTrue( 0 != new Pair("a", "a").hashCode());
        assertTrue( 0 != new Pair("a", "b").hashCode());
        assertTrue( 0 != new Pair("a", "c").hashCode());
        assertTrue( 0 != new Pair("a", "d").hashCode());

        assertTrue( 0 != new Pair("b", "b").hashCode());
        assertTrue( 0 != new Pair("b", "c").hashCode());
        assertTrue( 0 != new Pair("b", "d").hashCode());

        assertTrue( 0 != new Pair("c", "c").hashCode());
        assertTrue( 0 != new Pair("c", "d").hashCode());

        assertTrue( 0 != new Pair("d", "d").hashCode());

        assertEquals( new Pair("a", "a").hashCode(), new Pair("a", "a").hashCode() );
        assertEquals( new Pair("a", "b").hashCode(), new Pair("b", "a").hashCode() );
        assertEquals( new Pair("a", "c").hashCode(), new Pair("c", "a").hashCode() );
        assertEquals( new Pair("a", "d").hashCode(), new Pair("d", "a").hashCode() );

        assertEquals( new Pair("b", "b").hashCode(), new Pair("b", "b").hashCode() );
        assertEquals( new Pair("b", "c").hashCode(), new Pair("c", "b").hashCode() );
        assertEquals( new Pair("b", "d").hashCode(), new Pair("d", "b").hashCode() );

        assertEquals( new Pair("c", "c").hashCode(), new Pair("c", "c").hashCode() );
        assertEquals( new Pair("c", "d").hashCode(), new Pair("d", "c").hashCode() );

        assertEquals( new Pair("d", "d").hashCode(), new Pair("d", "d").hashCode() );

        assertNotEquals( new Pair("a", "a").hashCode(), new Pair("b", "b").hashCode() );
        assertNotEquals( new Pair("a", "b").hashCode(), new Pair("b", "d").hashCode() );
        assertNotEquals( new Pair("a", "c").hashCode(), new Pair("d", "a").hashCode() );
        assertNotEquals( new Pair("a", "d").hashCode(), new Pair("a", "a").hashCode() );
    }
}

但是我意识到，将compareTo（）更改为始终返回0将导致每个对被认为相同导致输出：

  

156483558 c b 1000000

虽然将hashCode（）更改为始终返回0（对于与上面相同的输入数据）将导致与上面相同的结果，只是键为零。

  

0 a a 62822
  0 a b 62516
  0 a c 62248
  0 a d 62495
  0 b a 62334
  0 b b 62232
  0 b d 62759
  0 c a 62200
  0 c b 124966
  0 c c 62347
  0 d c 125047
  0 d a 62653
  0 d b 62603
  0 d d 62778

编辑：

我进一步调查，使compareTo（）打印出比较的内容。这表明，像a，b和b，a这样的某些键永远不会相互比较，因此不能分组。

如果不是所有的密钥都相互比较，那么如何分组是可能的（除了使用hashCode（）之外它不是什么？

我想我遗失了一些微小的东西。 我很高兴有任何想法！非常感谢你提前。

最好的问候

Answer 1

鉴于{a，b} =：= {b，a}的初始要求，在构造函数中排序元组元素会不容易？

public Pair(String first, String second) {
    boolean swap = first.compareTo(second) > 0;
    this.first = swap ? second : first;
    this.second = swap ? first : second;
}

这将简化compareTo和equals等方法，并且不需要实现分区程序。

Answer 2

我想我在这里看到了这个问题。您尚未实现分区程序。

当你说你遇到大数据集的问题时，我假设你正在使用多个reducer。如果您使用的是单个减速器，则代码将起作用。但是在多个减速器的情况下，你需要一个分区来告诉frameowrk ab＆amp; amp; ba基本上是相同的键，应该使用相同的减速器。

以下是解释性链接：LINK

Answer 3

问题出在compareTo（）函数中。 首先检查a是否相等，b等于b，a。如果不是这种情况，首先要比较对的较小值，如果它们匹配，则比较resp的较大值。对。这解决了这个问题。

这就是我现在实现它的方式：

@Override
public int compareTo(Pair other){
    int cmpFirstFirst = first.compareTo(other.first);
    int cmpSecondSecond =  second.compareTo(other.second);
    int cmpFirstSecond = first.compareTo(other.second);
    int cmpSecondFirst =  second.compareTo(other.first);

    if ( cmpFirstFirst == 0 && cmpSecondSecond == 0 || cmpFirstSecond == 0 && cmpSecondFirst == 0) {
        return 0;
    }

    String thisSmaller;
    String otherSmaller;

    String thisBigger;
    String otherBigger;

    if ( this.first.compareTo(this.second) < 0 ) {
        thisSmaller = this.first;
        thisBigger = this.second;
    } else {
        thisSmaller = this.second;
        thisBigger = this.first;
    }

    if ( other.first.compareTo(other.second) < 0 ) {
        otherSmaller = other.first;
        otherBigger = other.second;
    } else {
        otherSmaller = other.second;
        otherBigger = other.first;
    }

    int cmpThisSmallerOtherSmaller = thisSmaller.compareTo(otherSmaller);
    int cmpThisBiggerOtherBigger = thisBigger.compareTo(otherBigger);

    if (cmpThisSmallerOtherSmaller == 0) {
        return cmpThisBiggerOtherBigger;
    } else {
        return cmpThisSmallerOtherSmaller;
    }
}

这意味着，与我的假设相反，地图输出的分组是使用传递关系而不是键的叉积来完成的。密钥的稳定顺序是必要的。一旦你了解并理解它，这就完全有道理。