blocks|key|4001901|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4001902|entities.stream()
++++++++.collect(toMap(YourEntityType::getId,+
+++++++++++++++++++++++YourEntityType::getProperty,
+++++++++++++++++++++++(x,+y)+->+"A".equals(x)+?+x+:+y));|code-block|syntax|javascript|4001903|更新：|4001904|我刚刚注意到@Holger在他的评论中发布了完全相同的东西，而我正在编辑我的答案。不过，这里有两个小区别：|4001905|4001906|我更喜欢使用方法引用而不是lambda(这是一个品味问题，但我发现它们比lambdas)|unordered-list-item|4001907|"A".equals(x)更不容易出错，比更安全。如果x为null.|offset|length|style|CODE|4001908|，则NPE将导致后者失败|4001909|更新2：|4001910|既然您想要列表形式的结果，那么此选项应该适用于您：|4001911|entities.stream()
++++++++.collect(collectingAndThen(
++++++++++++++++++++toMap(YourEntityType::getId,+
++++++++++++++++++++++++++identity(),
++++++++++++++++++++++++++(x,+y)+->+"A".equals(x.getProperty())+?+x+:+y,
++++++++++++++++++++m+->+new+ArrayList<YourEntityType>(m.values()));|4001912|entityMap^0|0|0|0|0|0|0|0|D|R|1|T|4|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|16|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|17|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|18|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|19|8|@]|9|@]|A|$]]|$1|O|3|P|5|N|7|1A|8|@$Q|1B|R|1C|S|T]|$Q|1D|R|1E|S|T]|$Q|1F|R|1G|S|T]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1H|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1I|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1J|8|@]|9|@]|A|$]]|$1|10|3|11|5|D|7|1K|8|@]|9|@]|A|$E|F]]|$1|12|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]]|13|$]]

Try this:
<pre><code>entities.stream()
 .collect(toMap(YourEntityType::getId, 
 YourEntityType::getProperty,
 (x, y) -&gt; &quot;A&quot;.equals(x) ? x : y));
</code></pre>
<h3>Update:</h3>
I've just noticed that @Holger posted exactly the same thing in his comment, while I was editing my answer. There are two minor differences though:
<ul>
<li>I prefer using method references instead of lambdas (It's a matter of taste, but I find them less error prone than lambdas)</li>
<li><code>&quot;A&quot;.equals(x)</code> is safer than <code>x.equals(&quot;A&quot;)</code>. The latter will fail with NPE in case <code>x</code> is <code>null</code>.</li>
</ul>
<h3>Update 2:</h3>
Since you want the result in form of a list, then this option should work for you:
<pre><code>entities.stream()
 .collect(collectingAndThen(
 toMap(YourEntityType::getId, 
 identity(),
 (x, y) -&gt; &quot;A&quot;.equals(x.getProperty()) ? x : y,
 m -&gt; new ArrayList&lt;YourEntityType&gt;(m.values()));
</code></pre>

blocks|key|4001942|text|如果您坚持使用分组收集器，则?+another+collector+function+?将如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4001943|++++public+static+void+main(String[]+args)+{
++++++++record+C(String+id,+String+property)+{}
++++++++List<C>+list+=+List.of(new+C("FOO","A"),new+C("FOO","Z"),new+C("JOHN","A"),new+C("DOE","Z"),new+C("XXX","X"));
++++++++System.out.println(list);
++++++++Object+result+=+list.stream()
++++++++++++++++.collect(Collectors.groupingBy(o+->+o.id,+Collectors.collectingAndThen(Collectors.reducing((l,r)+->+"A".equals(l.property)+?+l+:+r),+Optional::get)));
++++++++System.out.println(result);+++++
++++}|code-block|syntax|javascript|4001944|，这一想法建议纠正几乎@ETO的解决方案。|4001945|(抱歉，我有记录，我懒得使用getter。)|4001946|entityMap^0|E|U|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|V|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

If you insisted on using grouping collector, the <code>? another collector function ?</code> would look like this:
<pre class="lang-java prettyprint-override"><code> public static void main(String[] args) {
 record C(String id, String property) {}
 List&lt;C&gt; list = List.of(new C(&quot;FOO&quot;,&quot;A&quot;),new C(&quot;FOO&quot;,&quot;Z&quot;),new C(&quot;JOHN&quot;,&quot;A&quot;),new C(&quot;DOE&quot;,&quot;Z&quot;),new C(&quot;XXX&quot;,&quot;X&quot;));
 System.out.println(list);
 Object result = list.stream()
 .collect(Collectors.groupingBy(o -&gt; o.id, Collectors.collectingAndThen(Collectors.reducing((l,r) -&gt; &quot;A&quot;.equals(l.property) ? l : r), Optional::get)));
 System.out.println(result); 
 }
</code></pre>
, which Idea suggest to correct to almost @ETO's solution.
(Apologize, having records I am lazy to use getters.)

blocks|key|4001954|text|//+Utility+class+with+methods+you+can+reuse+for+other+purposes+too
public+final+class+StreamUtil+{

++++private+StreamUtil()+{}

++++public+static+<T,+U>+Predicate<T>+distinctBy(Function<T,+U>+keyFunction)+{
++++++++final+Set+uniqueKeys+=+synchronizedSet(new+HashSet<>());
++++++++return+item+->+uniqueKeys.add(keyFunction.apply(item));
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4001955|和您的问题的解决方案|unstyled|4001956|entities.stream()
+.sorting(Comparator.comparing(Entity::getProperty))+//+order+entities+according+to+priorities
+.filter(StreamUtil.distinctBy(Entity::getId))+//+take+first+entity+only+per+key
+...+//+here+you+have+stream+as+per+you+requirements.+You+can+collect()+it+or+continue+mapping/filtering|4001957|由于排序的原因，此解决方案比此处建议的其他解决方案要慢。然而，它有它的优势-更容易阅读，而速度慢只会在大型集合上发挥作用。|4001958|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>// Utility class with methods you can reuse for other purposes too
public final class StreamUtil {

 private StreamUtil() {}

 public static &lt;T, U&gt; Predicate&lt;T&gt; distinctBy(Function&lt;T, U&gt; keyFunction) {
 final Set&lt;U&gt; uniqueKeys = synchronizedSet(new HashSet&lt;&gt;());
 return item -&gt; uniqueKeys.add(keyFunction.apply(item));
 }
}
</code></pre>
and solution for your question
<pre><code>entities.stream()
 .sorting(Comparator.comparing(Entity::getProperty)) // order entities according to priorities
 .filter(StreamUtil.distinctBy(Entity::getId)) // take first entity only per key
 ... // here you have stream as per you requirements. You can collect() it or continue mapping/filtering
</code></pre>
This solution is slower than other suggested here because of sorting.
However, it has its advantage - easier to read, while slowness will kick in on large collections only.

I have the following entities:
<pre><code>ID | PROPERTY

FOO | A
FOO | Z
JOHN | A
DOE | Z
</code></pre>
Now I want to group these entities by ID and the take the one with property 'A' if one or both valuesare present , otherwise Z.
Thus the output should be:
<pre><code>FOO | A
JOHN | A
DOE | Z
</code></pre>
I was thinking about the groupingBy stream method, but I don't know how what collectors to use neither if is the correct solution:
<pre><code>list.stream()
 .collect(Collectors.groupingBy(o -&gt; o.getId(), ? another collector function ?)
 .collect(Collectors.toList()));
</code></pre>

java 11 lambda, aggregate by same property and the filter with a certain condition

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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