词频统计 Java 8

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Java 8如何统计List的词频?

 List <String> wordsList = Lists.newArrayList("hello", "bye", "ciao", "bye", "ciao");

结果必须是:

 {ciao=2, hello=1, bye=2}

原文由 Mouna 发布,翻译遵循 CC BY-SA 4.0 许可协议

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2 个回答

我想分享我找到的解决方案,因为起初我希望使用 map-and-reduce 方法,但它有点不同。

 Map<String,Long> collect = wordsList.stream()
    .collect( Collectors.groupingBy( Function.identity(), Collectors.counting() ));

或者对于整数值:

 Map<String,Integer> collect = wordsList.stream()
     .collect( Collectors.groupingBy( Function.identity(), Collectors.summingInt(e -> 1) ));

编辑

我添加了如何按值对地图进行排序:

 LinkedHashMap<String, Long> countByWordSorted = collect.entrySet()
            .stream()
            .sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
            .collect(Collectors.toMap(
                    Map.Entry::getKey,
                    Map.Entry::getValue,
                    (v1, v2) -> {
                        throw new IllegalStateException();
                    },
                    LinkedHashMap::new
            ));

原文由 Mouna 发布,翻译遵循 CC BY-SA 4.0 许可协议

注意:请参阅下面的编辑

作为 Mounas answer 的替代方法,这里有一种并行计算字数的方法:

 import java.util.Arrays;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

public class ParallelWordCount
{
    public static void main(String[] args)
    {
        List<String> list = Arrays.asList(
            "hello", "bye", "ciao", "bye", "ciao");
        Map<String, Integer> counts = list.parallelStream().
            collect(Collectors.toConcurrentMap(
                w -> w, w -> 1, Integer::sum));
        System.out.println(counts);
    }
}

编辑 作为对评论的回应,我用 JMH 进行了一个小测试,比较了 toConcurrentMapgroupingByConcurrent 方法,具有不同的输入列表大小和不同长度的随机词。该测试表明 toConcurrentMap 方法更快。在考虑这些方法“在幕后”有何不同时,很难预测这样的事情。

作为进一步的扩展,基于进一步的评论,我扩展了测试以涵盖 toMapgroupingBy 、串行和并行的所有四种组合。

结果仍然是 toMap 方法更快,但出乎意料的是(至少对我而言)两种情况下的“并发”版本都比串行版本慢……:

              (method)  (count) (wordLength)  Mode  Cnt     Score    Error  Units
      toConcurrentMap     1000            2  avgt   50   146,636 ±  0,880  us/op
      toConcurrentMap     1000            5  avgt   50   272,762 ±  1,232  us/op
      toConcurrentMap     1000           10  avgt   50   271,121 ±  1,125  us/op
                toMap     1000            2  avgt   50    44,396 ±  0,541  us/op
                toMap     1000            5  avgt   50    46,938 ±  0,872  us/op
                toMap     1000           10  avgt   50    46,180 ±  0,557  us/op
           groupingBy     1000            2  avgt   50    46,797 ±  1,181  us/op
           groupingBy     1000            5  avgt   50    68,992 ±  1,537  us/op
           groupingBy     1000           10  avgt   50    68,636 ±  1,349  us/op
 groupingByConcurrent     1000            2  avgt   50   231,458 ±  0,658  us/op
 groupingByConcurrent     1000            5  avgt   50   438,975 ±  1,591  us/op
 groupingByConcurrent     1000           10  avgt   50   437,765 ±  1,139  us/op
      toConcurrentMap    10000            2  avgt   50   712,113 ±  6,340  us/op
      toConcurrentMap    10000            5  avgt   50  1809,356 ±  9,344  us/op
      toConcurrentMap    10000           10  avgt   50  1813,814 ± 16,190  us/op
                toMap    10000            2  avgt   50   341,004 ± 16,074  us/op
                toMap    10000            5  avgt   50   535,122 ± 24,674  us/op
                toMap    10000           10  avgt   50   511,186 ±  3,444  us/op
           groupingBy    10000            2  avgt   50   340,984 ±  6,235  us/op
           groupingBy    10000            5  avgt   50   708,553 ±  6,369  us/op
           groupingBy    10000           10  avgt   50   712,858 ± 10,248  us/op
 groupingByConcurrent    10000            2  avgt   50   901,842 ±  8,685  us/op
 groupingByConcurrent    10000            5  avgt   50  3762,478 ± 21,408  us/op
 groupingByConcurrent    10000           10  avgt   50  3795,530 ± 32,096  us/op

我对 JMH 不是很有经验,也许我在这里做错了什么——欢迎提出建议和更正:

 import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.Random;
import java.util.concurrent.TimeUnit;
import java.util.function.Function;
import java.util.stream.Collectors;

import org.openjdk.jmh.annotations.Benchmark;
import org.openjdk.jmh.annotations.BenchmarkMode;
import org.openjdk.jmh.annotations.Mode;
import org.openjdk.jmh.annotations.OutputTimeUnit;
import org.openjdk.jmh.annotations.Param;
import org.openjdk.jmh.annotations.Scope;
import org.openjdk.jmh.annotations.Setup;
import org.openjdk.jmh.annotations.State;
import org.openjdk.jmh.infra.Blackhole;

@State(Scope.Thread)
public class ParallelWordCount
{

    @Param({"toConcurrentMap", "toMap", "groupingBy", "groupingByConcurrent"})
    public String method;

    @Param({"2", "5", "10"})
    public int wordLength;

    @Param({"1000", "10000" })
    public int count;

    private List<String> list;

    @Setup
    public void initList()
    {
         list = createRandomStrings(count, wordLength, new Random(0));
    }

    @Benchmark
    @BenchmarkMode(Mode.AverageTime)
    @OutputTimeUnit(TimeUnit.MICROSECONDS)
    public void testMethod(Blackhole bh)
    {

        if (method.equals("toMap"))
        {
            Map<String, Integer> counts =
                list.stream().collect(
                    Collectors.toMap(
                        w -> w, w -> 1, Integer::sum));
            bh.consume(counts);
        }
        else if (method.equals("toConcurrentMap"))
        {
            Map<String, Integer> counts =
                list.parallelStream().collect(
                    Collectors.toConcurrentMap(
                        w -> w, w -> 1, Integer::sum));
            bh.consume(counts);
        }
        else if (method.equals("groupingBy"))
        {
            Map<String, Long> counts =
                list.stream().collect(
                    Collectors.groupingBy(
                        Function.identity(), Collectors.<String>counting()));
            bh.consume(counts);
        }
        else if (method.equals("groupingByConcurrent"))
        {
            Map<String, Long> counts =
                list.parallelStream().collect(
                    Collectors.groupingByConcurrent(
                        Function.identity(), Collectors.<String> counting()));
            bh.consume(counts);
        }
    }

    private static String createRandomString(int length, Random random)
    {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < length; i++)
        {
            int c = random.nextInt(26);
            sb.append((char) (c + 'a'));
        }
        return sb.toString();
    }

    private static List<String> createRandomStrings(
        int count, int length, Random random)
    {
        List<String> list = new ArrayList<String>(count);
        for (int i = 0; i < count; i++)
        {
            list.add(createRandomString(length, random));
        }
        return list;
    }
}

对于具有 10000 个元素和 2 个字母的单词的列表的连续情况,时间仅相似。

可能值得检查对于更大的列表大小,并发版本最终是否优于串行版本,但目前没有时间使用所有这些配置进行另一次详细的基准测试。

原文由 Marco13 发布,翻译遵循 CC BY-SA 3.0 许可协议

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