如何在Java 8中统计列表中各单词的出现频率?
List <String> wordsList = Lists.newArrayList("hello", "bye", "ciao", "bye", "ciao");
结果必须是:
{ciao=2, hello=1, bye=2}
如何在Java 8中统计列表中各单词的出现频率?
List <String> wordsList = Lists.newArrayList("hello", "bye", "ciao", "bye", "ciao");
结果必须是:
{ciao=2, hello=1, bye=2}
我希望分享我找到的解决方案,因为一开始我预计要使用map和reduce方法,但实际上略有不同。
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:请参见下面的编辑内容)
作为Mounas答案的替代方案,这里提供了一种并行计算单词数的方法:
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:根据评论,我使用JMH进行了小型测试,比较了
toConcurrentMap
和groupingByConcurrent
方法,使用不同大小的输入列表和不同长度的随机单词。这个测试表明,toConcurrentMap
方法更快。考虑到这两种方法在“内部”有多大的不同,很难预测这样的结果。进一步扩展,基于更多的评论,我将测试扩展到涵盖了
toMap
、groupingBy
、串行和并行四种组合。结果仍然是
toMap
方法更快,但出乎意料(至少对我来说),在两种情况下,“concurrent”版本都比串行版本更慢...
(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
我在JMH方面经验不太丰富,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;
}
}
在拥有10000个元素和2个字母的列表的串行情况下,时间相似。
值得检查的是,在更大的列表大小下,并发版本是否最终会胜过串行版本,但目前没有时间对所有这些配置进行另一个详细的基准测试运行。
使用泛型查找集合中出现频率最高的项:
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()));
}
List
转换为Bag
即可。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"));
Bags
工厂类直接创建一个Bag
。Bag<String> words =
Bags.mutable.with("hello", "bye", "ciao", "bye", "ciao");
这段代码适用于Java 5或以上版本。
注意: 我是Eclipse Collections的贡献者。
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)));
}
我只是提供一些意见。
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)
);
List list = Arrays.asList("hello", "bye", "ciao", "bye", "ciao")
- mancocapac您可以使用Java 8 Streams来实现此功能。
Arrays.asList(s).stream()
.collect(Collectors.groupingBy(Function.<String>identity(),
Collectors.<String>counting()));
我再提供两分钱的意见,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 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());
});
}
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));
类似于map-reduce方法,首先将每个单词w映射到(w, 1)。然后聚合(reduce部分)所有键(单词w)相似的对的计数(Map.Entry::getValue
),(Map.Entry::getKey
)并计算总和(Integer::sum
)。
最终的终端操作将返回一个HashMap<String, Integer>
:
{more=3, simple=1, my=1, way=1}
Collectors.groupingByConcurrent(w->w, Collectors.counting())
会更有效率。 - HolgergroupingBy
方法也可以并行工作,扩展基准测试以查看所有变体(groupingBy
vstoMap
)×(普通 vs.Concurrent
)之间的差异将是很棒的... - Holger