Java 8中如何统计List的单词频率？
List <String> wordsList = Lists.newArrayList("hello", "bye", "ciao", "bye", "ciao");
结果必须是：
{ciao=2, hello=1, bye=2}
我想分享我找到的解决方案，因为一开始我希望使用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
));
（注意：请参阅下面的编辑）
作为 Mounas 答案的替代方案，这里有一种并行进行字数统计的方法：
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);
}
}
编辑为了回应评论，我用 JMH 进行了一个小测试，比较了
和toConcurrentMap方法，使用不同的输入列表大小和不同长度的随机单词。该测试表明groupingByConcurrent方法更快。当考虑到这些方法“在幕后”有多么不同时，很难预测这样的事情。toConcurrentMap作为进一步的扩展，根据进一步的评论，我将测试扩展为涵盖
、toMap、串行和并行的所有四种组合。groupingBy结果仍然是
方法更快，但出乎意料的是（至少对我来说）两种情况下的“并发”版本都比串行版本慢......：toMap
(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 不太有经验，也许我在这里做错了什么 - 欢迎建议和更正：
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;
}
}
仅对于包含 10000 个元素和 2 个字母单词的列表的串行情况，时间相似。
可能值得检查一下，对于更大的列表大小，并发版本最终是否优于串行版本，但目前没有时间对所有这些配置进行另一次详细的基准测试。
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()));
}
如果您使用 Eclipse Collections，您只需将
ListBagBag<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"));
您还可以直接使用
BagBagsBag<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
我将在这里展示我所做的解决方案（带有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$
这是一种使用地图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)
);
您可以使用 Java 8 Streams
Arrays.asList(s).stream()
.collect(Collectors.groupingBy(Function.<String>identity(),
Collectors.<String>counting()));
我的另外 2 美分，给定一个数组：
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)));
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);
}
}
我认为有一种更易读的方式：
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));
与映射归约方法类似，首先将每个单词 w 映射到 (w, 1)。然后聚合（减少部分）所有键（单词
w）相似的对的计数（
Map.Entry::getValue），（Map.Entry::getKeyInteger::sum最终的终端操作会返回一个
HashMap<String, Integer>{more=3, simple=1, my=1, way=1}
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());
});
}
Map.getOrDefaultMap<String, Integer> res = new HashMap<>();
wordsList.forEach(w -> res.put(w, res.getOrDefault(w, 0) + 1));
default V getOrDefault(Object key, V defaultValue)