词频统计 Java 8

Question

如何统计Java 8中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}

Answer 1

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
            ));

Answer 2

（注意：请参阅下面的编辑）

作为Mounas answer的替代方案，这是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toConcurrentMap和groupingByConcurrent方法，不同的输入列表大小和不同长度的随机单词。 该测试表明toConcurrentMap方法更快。 当考虑这些方法“在幕后”有多么不同时，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toMap 、 groupingBy 、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（至少，对我来说）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 - 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;
    }
}

仅对于具有 10000 个元素和 2 个字母单词的列表的串行情况，时间相似。

值得检查一下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运行另一个详细的基准测试。

Answer 3

使用泛型查找集合中最常见的项目：

private <V> V findMostFrequentItem(final Collection<V> items)
{
  return items.stream()
      .filter(Objects::nonNull)
      .collect(Collectors.groupingBy(Functions.identity(), Collectors.counting()))
      .entrySet()
      .stream()
      .max(Comparator.comparing(Entry::getValue))
      .map(Entry::getKey)
      .orElse(null);
}

计算项目频率：

private <V> Map<V, Long> findFrequencies(final Collection<V> items)
{
  return items.stream()
      .filter(Objects::nonNull)
      .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
}

Answer 4

如果您使用Eclipse Collections ，您只需将List转换为Bag 。

Bag<String> words = 
    Lists.mutable.with("hello", "bye", "ciao", "bye", "ciao").toBag();

Assert.assertEquals(2, words.occurrencesOf("ciao"));
Assert.assertEquals(1, words.occurrencesOf("hello"));
Assert.assertEquals(2, words.occurrencesOf("bye"));

您也可以直接使用Bags工厂类创建Bag 。

Bag<String> words = 
    Bags.mutable.with("hello", "bye", "ciao", "bye", "ciao");

此代码适用于 Java 5+。

注意：我是 Eclipse Collections 的提交者

Answer 5

我将在这里介绍我制作的解决方案（分组的解决方案要好得多:)）。

static private void test0(List<String> input) {
    Set<String> set = input.stream()
            .collect(Collectors.toSet());
    set.stream()
            .collect(Collectors.toMap(Function.identity(),
                    str -> Collections.frequency(input, str)));
}

只是我的 0.02 美元

Answer 6

这是一种使用映射函数创建频率图的方法。

List<String> words = Stream.of("hello", "bye", "ciao", "bye", "ciao").collect(toList());
Map<String, Integer> frequencyMap = new HashMap<>();

words.forEach(word ->
        frequencyMap.merge(word, 1, (v, newV) -> v + newV)
);

System.out.println(frequencyMap); // {ciao=2, hello=1, bye=2}

或者

words.forEach(word ->
       frequencyMap.compute(word, (k, v) -> v != null ? v + 1 : 1)
);

Answer 7

你可以使用 Java 8 Streams

    Arrays.asList(s).stream()
          .collect(Collectors.groupingBy(Function.<String>identity(), 
          Collectors.<String>counting()));

Answer 8

给定一个数组，我的另外 2 美分：

import static java.util.stream.Collectors.*;

String[] str = {"hello", "bye", "ciao", "bye", "ciao"};    
Map<String, Integer> collected 
= Arrays.stream(str)
        .collect(groupingBy(Function.identity(), 
                    collectingAndThen(counting(), Long::intValue)));

Answer 9

public class Main {

    public static void main(String[] args) {


        String testString ="qqwweerrttyyaaaaaasdfasafsdfadsfadsewfywqtedywqtdfewyfdweytfdywfdyrewfdyewrefdyewdyfwhxvsahxvfwytfx"; 
        long java8Case2 = testString.codePoints().filter(ch -> ch =='a').count();
        System.out.println(java8Case2);

        ArrayList<Character> list = new ArrayList<Character>();
        for (char c : testString.toCharArray()) {
          list.add(c);
        }
        Map<Object, Integer> counts = list.parallelStream().
            collect(Collectors.toConcurrentMap(
                w -> w, w -> 1, Integer::sum));
        System.out.println(counts);
    }

}

Answer 10

我认为有一种更具可读性的方式：

var words = List.of("my", "more", "more", "more", "simple", "way");
var count = words.stream().map(x -> Map.entry(x, 1))
                    .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, Integer::sum));

类似于 map-reduce 方法，首先将 map 中的每个单词w到 a ( w , 1)。 然后聚合（减少部分）所有对的计数（ Map.Entry::getValue ），其中它们的键（word w ）相似，（ Map.Entry::getKey ）并计算总和（ Integer::sum ）。

最后的终端操作将返回一个HashMap<String, Integer> ：

{more=3, simple=1, my=1, way=1}

Answer 11

  public static void main(String[] args) {
    String str = "Hi Hello Hi";
    List<String> s = Arrays.asList(str.split(" "));
    Map<String, Long> hm = 
              s.stream().collect(Collectors.groupingBy(Function.identity(), 
              Collectors.counting()));

              hm.entrySet().forEach(entry -> {

             System.out.println(entry.getKey() + " " + entry.getValue());
              });

}