blocks|key|79649|text|我认为您只是在寻找overload，它需要另一个Collector来指定如何处理每个组……然后使用Collectors.counting()进行计数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|79650|import+java.util.*;
import+java.util.stream.*;

class+Test+{
++++public+static+void+main(String[]+args)+{
++++++++List<String>+list+=+new+ArrayList<>();

++++++++list.add("Hello");
++++++++list.add("Hello");
++++++++list.add("World");

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

++++++++System.out.println(counted);
++++}
}|code-block|syntax|javascript|79651|结果：|79652|{Hello=2,+World=1}|79653|(也有可能使用groupingByConcurrent来提高效率。对于您的真实代码，如果它在您的上下文中是安全的，请牢记这一点。)|79654|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/8/docs/api/java/util/stream/Collectors.html#groupingBy-java.util.function.Function-java.util.stream.Collector-^0|O|9|1D|L|9|8|0|0|0|0|0|7|K|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

I think you're just looking for the <a href="http://docs.oracle.com/javase/8/docs/api/java/util/stream/Collectors.html#groupingBy-java.util.function.Function-java.util.stream.Collector-" rel="noreferrer">overload</a> which takes another <code>Collector</code> to specify what to do with each group... and then <code>Collectors.counting()</code> to do the counting:

<pre><code>import java.util.*;
import java.util.stream.*;

class Test {
 public static void main(String[] args) {
 List&lt;String&gt; list = new ArrayList&lt;&gt;();

 list.add("Hello");
 list.add("Hello");
 list.add("World");

 Map&lt;String, Long&gt; counted = list.stream()
 .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

 System.out.println(counted);
 }
}
</code></pre>

Result:

<pre><code>{Hello=2, World=1}
</code></pre>

(There's also the possibility of using <code>groupingByConcurrent</code> for more efficiency. Something to bear in mind for your real code, if it would be safe in your context.)

blocks|key|79665|text|List<String>+list+=+new+ArrayList<>();

list.add("Hello");
list.add("Hello");
list.add("World");

Map<String,+List<String>>+collect+=+list.stream()
++++++++++++++++++++++++++++++++++++++++.collect(Collectors.groupingBy(o+->+o));
collect.entrySet()
+++++++.forEach(e+->+System.out.println(e.getKey()+%2B+"+-+"+%2B+e.getValue().size()));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|79666|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>List&lt;String&gt; list = new ArrayList&lt;&gt;();

list.add("Hello");
list.add("Hello");
list.add("World");

Map&lt;String, List&lt;String&gt;&gt; collect = list.stream()
 .collect(Collectors.groupingBy(o -&gt; o));
collect.entrySet()
 .forEach(e -&gt; System.out.println(e.getKey() + " - " + e.getValue().size()));
</code></pre>

blocks|key|4586398|text|下面是由StreamEx提供的简单解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4586399|StreamEx.of(list).groupingBy(Function.identity(),+MoreCollectors.countingInt());|code-block|syntax|javascript|4586400|这样做的好处是减少了Java+stream样板代码：collect(Collectors.|style|CODE|4586401|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/amaembo/streamex^0|4|8|0|0|0|Q|J|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@$A|10|B|11|K|L]]|9|@]|C|$]]|$1|M|3|-4|5|6|7|12|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

Here is the simple solution by <a href="https://github.com/amaembo/streamex" rel="nofollow noreferrer">StreamEx</a>:
<pre class="lang-java prettyprint-override"><code>StreamEx.of(list).groupingBy(Function.identity(), MoreCollectors.countingInt());
</code></pre>
This has the advantage of reducing the Java stream boilerplate code: <code>collect(Collectors.</code>

blocks|key|79749|text|以下是对象列表示例|type|unstyled|depth|inlineStyleRanges|entityRanges|data|79750|Map<String,+Long>+requirementCountMap+=+requirements.stream().collect(Collectors.groupingBy(Requirement::getRequirementType,+Collectors.counting()));|code-block|syntax|javascript|79751|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is example for list of Objects

<pre><code>Map&lt;String, Long&gt; requirementCountMap = requirements.stream().collect(Collectors.groupingBy(Requirement::getRequirementType, Collectors.counting()));
</code></pre>

blocks|key|79807|text|这里有一些稍微不同的选项来完成手头的任务。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|79808|使用toMap|offset|length|style|CODE|79809|list.stream()
++++.collect(Collectors.toMap(Function.identity(),+e+->+1,+Math::addExact));|code-block|syntax|javascript|79810|使用Map::merge|79811|Map<String,+Integer>+accumulator+=+new+HashMap<>();
list.forEach(s+->+accumulator.merge(s,+1,+Math::addExact));|79812|entityMap^0|0|2|5|0|0|2|A|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|10|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

Here are slightly different options to accomplish the task at hand.

using <code>toMap</code>:

<pre><code>list.stream()
 .collect(Collectors.toMap(Function.identity(), e -&gt; 1, Math::addExact));
</code></pre>

using <code>Map::merge</code>:

<pre><code>Map&lt;String, Integer&gt; accumulator = new HashMap&lt;&gt;();
list.forEach(s -&gt; accumulator.merge(s, 1, Math::addExact));
</code></pre>

I want to collect the items in a stream into a map which groups equal objects together, and maps to the number of occurrences.
<pre><code>List&lt;String&gt; list = Arrays.asList(&quot;Hello&quot;, &quot;Hello&quot;, &quot;World&quot;);
Map&lt;String, Long&gt; wordToFrequency = // what goes here?
</code></pre>
So in this case, I would like the map to consist of these entries:
<pre><code>Hello -&gt; 2
World -&gt; 1
</code></pre>
How can I do that?

How can I count occurrences with groupBy?

Java

我想将流中的项收集到一个映射中，该映射将相等的对象分组在一起，并映射到出现的次数。List<String> list = Arrays.asList("Hello", "Hello", "World");Map<String, Long> wordToFrequency = // what goes here?因此，在...

问我如何使用groupBy来统计出现的次数？
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