如何计算Java 8中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}
答案 0 :(得分：74)
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(groupingBy(Function.identity(), counting()));
或者对于整数值：
Map<String, Integer> collect =
wordsList.stream().collect(groupingBy(Function.identity(), summingInt(e -> 1)));
修改
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
));
答案 1 :(得分：25)
（注意：请参阅下面的编辑）
作为Mounas answer的替代方法，这里有一种方法可以并行执行单词计数：
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);
}
}
编辑在回复评论时，我使用JMH进行了一项小测试，比较
toConcurrentMap和groupingByConcurrent方法，使用不同的输入列表大小和不同长度的随机字。该测试表明toConcurrentMap方法更快。在考虑这些方法的不同之处时，我们很难预测到这样的事情。作为进一步的扩展，基于进一步的评论，我将测试扩展到涵盖
toMap，groupingBy，串行和并行的所有四种组合。结果仍然是
toMap方法更快，但意外地（至少对我而言）＆＃34;并发＆＃34;两种情况下的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不太熟悉，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;
}
}
时间仅与包含10000个元素和2个字母单词的列表的序列大小写相似。
检查更大的列表大小是否值得，并发版本最终优于串行版本，但目前还没有时间进行所有这些配置的另一个详细基准测试。
答案 2 :(得分：6)
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()));
}
答案 3 :(得分：3)
如果您使用Eclipse Collections，则只需将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"));
此代码适用于Java 5 - 8。
注意：我是Eclipse Collections的提交者
答案 4 :(得分：2)
我将在这里提出我所做的解决方案（分组更好:) :)。
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
答案 5 :(得分：2)
这是一种使用地图功能创建频率图的方法。
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)
);
答案 6 :(得分：1)
您可以使用Java 8流
aux2 = [
(lambda _fns: lambda x, args:
sum( fn(x, *arg) for fn, arg in zip(_fns, args) )
)(fns)
for fns in aux1
]
答案 7 :(得分：0)
我的另外2分，给出了一个阵列：
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)));
答案 8 :(得分：0)
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);
}
}