blocks|key|261317|text|在这种情况下，您还可以使用groupingBy，因为有2种组的可能性(活跃的和不活动的员工)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|261318|Map<Boolean,+List<Employee>>+grouped+=+employees.stream()
++++++++++++++++.collect(Collectors.groupingBy(Employee::isActive));

List<Employee>+activeEmployees+=+grouped.get(true);
List<Employee>+formerEmployees+=+grouped.get(false);|code-block|syntax|javascript|261319|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/util/stream/Collectors.html#groupingBy-java.util.function.Function-^0|D|A|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

You can also use <a href="https://docs.oracle.com/javase/8/docs/api/java/util/stream/Collectors.html#groupingBy-java.util.function.Function-" rel="noreferrer">groupingBy</a> in this case as there are 2 group posibilities (active and inactive employees):

<pre><code>Map&lt;Boolean, List&lt;Employee&gt;&gt; grouped = employees.stream()
 .collect(Collectors.groupingBy(Employee::isActive));

List&lt;Employee&gt; activeEmployees = grouped.get(true);
List&lt;Employee&gt; formerEmployees = grouped.get(false);
</code></pre>

blocks|key|260728|text|如果您愿意使用第三方库，这将使用来自Collectors2.partition的日食收藏。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|260729|PartitionMutableList<Employee>+partition+=
++++++++employees.stream().collect(
++++++++++++++++Collectors2.partition(Employee::isActive,+PartitionFastList::new));

List<Employee>+activeEmployees+=+partition.getSelected();
List<Employee>+formerEmployees+=+partition.getRejected();|code-block|syntax|javascript|260730|您还可以使用ListIterate简化事情。|260731|PartitionMutableList<Employee>+partition+=
++++++++ListIterate.partition(employees,+Employee::isActive);

List<Employee>+activeEmployees+=+partition.getSelected();
List<Employee>+formerEmployees+=+partition.getRejected();|260732|PartitionMutableList是从PartitionIterable扩展而来的类型。PartitionIterable的每个子类型都有一个阳性结果getSelected()和阴性结果getRejected()的集合。|260733|注意:我是Eclipse集合的提交者。|260734|entityMap|0|LINK|mutability|MUTABLE|url|https://www.eclipse.org/collections/javadoc/9.2.0/org/eclipse/collections/impl/collector/Collectors2.html#partition-org.eclipse.collections.api.block.predicate.Predicate-java.util.function.Supplier-|1|https://github.com/eclipse/eclipse-collections|2|https://www.eclipse.org/collections/javadoc/9.2.0/org/eclipse/collections/api/partition/list/PartitionMutableList.html|3|https://www.eclipse.org/collections/javadoc/9.2.0/org/eclipse/collections/api/partition/PartitionIterable.html^0|I|L|I|L|0|14|4|1|0|0|6|B|0|0|0|K|M|H|1B|H|26|D|2O|D|0|K|2|M|H|3|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@$9|17|A|18|B|C]]|D|@$9|19|A|1A|1|1B]|$9|1C|A|1D|1|1E]]|E|$]]|$1|F|3|G|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1G|8|@$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1J|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1K|8|@$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]|$9|1R|A|1S|B|C]|$9|1T|A|1U|B|C]]|D|@$9|1V|A|1W|1|1X]|$9|1Y|A|1Z|1|20]]|E|$]]|$1|Q|3|R|5|6|7|21|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|22|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]|10|$5|V|W|X|E|$Y|11]]|12|$5|V|W|X|E|$Y|13]]|14|$5|V|W|X|E|$Y|15]]]]

If you are open to using a third-party library, this will work using <a href="https://www.eclipse.org/collections/javadoc/9.2.0/org/eclipse/collections/impl/collector/Collectors2.html#partition-org.eclipse.collections.api.block.predicate.Predicate-java.util.function.Supplier-" rel="nofollow noreferrer"><code>Collectors2.partition</code></a> from <a href="https://github.com/eclipse/eclipse-collections" rel="nofollow noreferrer">Eclipse Collections</a>.

<pre><code>PartitionMutableList&lt;Employee&gt; partition =
 employees.stream().collect(
 Collectors2.partition(Employee::isActive, PartitionFastList::new));

List&lt;Employee&gt; activeEmployees = partition.getSelected();
List&lt;Employee&gt; formerEmployees = partition.getRejected();
</code></pre>

You can also simplify things using <code>ListIterate</code>.

<pre><code>PartitionMutableList&lt;Employee&gt; partition =
 ListIterate.partition(employees, Employee::isActive);

List&lt;Employee&gt; activeEmployees = partition.getSelected();
List&lt;Employee&gt; formerEmployees = partition.getRejected();
</code></pre>

<a href="https://www.eclipse.org/collections/javadoc/9.2.0/org/eclipse/collections/api/partition/list/PartitionMutableList.html" rel="nofollow noreferrer"><code>PartitionMutableList</code></a> is a type that extends from <a href="https://www.eclipse.org/collections/javadoc/9.2.0/org/eclipse/collections/api/partition/PartitionIterable.html" rel="nofollow noreferrer"><code>PartitionIterable</code></a>. Every subtype of <code>PartitionIterable</code> has a collection for positive results <code>getSelected()</code> and negative results <code>getRejected()</code>. 

Note: I am a committer for Eclipse Collections.

blocks|key|2147865|text|最复杂的方法是什么？|type|blockquote|depth|inlineStyleRanges|entityRanges|data|2147866|Java12当然有新的Collectors::teeing|unstyled|offset|length|style|CODE|2147867|List<List<Employee>>+divided+=+employees.stream().collect(
++++++Collectors.teeing(
++++++++++++++Collectors.filtering(Employee::isActive,+Collectors.toList()),
++++++++++++++Collectors.filtering(Predicate.not(Employee::isActive),+Collectors.toList()),
++++++++++++++List::of
++++++));

System.out.println(divided.get(0));++//active
System.out.println(divided.get(1));++//inactive|code-block|syntax|javascript|2147868|entityMap^0|0|B|I|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Q|8|@$E|R|F|S|G|H]]|9|@]|A|$]]|$1|I|3|J|5|K|7|T|8|@]|9|@]|A|$L|M]]|$1|N|3|-4|5|D|7|U|8|@]|9|@]|A|$]]]|O|$]]

<blockquote>
What is the most sophisticated way?
</blockquote>
Java 12 of course with new <code>Collectors::teeing</code>
<pre><code>List&lt;List&lt;Employee&gt;&gt; divided = employees.stream().collect(
 Collectors.teeing(
 Collectors.filtering(Employee::isActive, Collectors.toList()),
 Collectors.filtering(Predicate.not(Employee::isActive), Collectors.toList()),
 List::of
 ));

System.out.println(divided.get(0)); //active
System.out.println(divided.get(1)); //inactive
</code></pre>

blocks|key|2147879|text|一种更简单、更简洁的方法是使用stream.filter()和collect()，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2147880|activeEmployees+=+employees.stream().filter(Employee::isActive).collect(Collectors.toList());
formerEmployees+=+employees.stream().filter(employee+->+!employee.isActive()).collect(Collectors.toList());|code-block|syntax|javascript|2147881|entityMap^0|F|F|V|9|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

An easier, cleaner way is using <code>stream.filter()</code> and <code>collect()</code> like this:
<pre class="lang-java prettyprint-override"><code>activeEmployees = employees.stream().filter(Employee::isActive).collect(Collectors.toList());
formerEmployees = employees.stream().filter(employee -&gt; !employee.isActive()).collect(Collectors.toList());
</code></pre>

I have a list of <code>employees</code>. They have <code>isActive</code> boolean field. I would like to divide <code>employees</code> into two lists: <code>activeEmployees</code> and <code>formerEmployees</code>. Is it possible to do using Stream API? What is the most sophisticated way?

Java Stream: divide into two lists by boolean predicate

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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我有一个employees的列表。他们有isActive布尔字段。我想把employees分成两个列表：activeEmployees和formerEmployees。是否可以使用Stream？最复杂的方法是什么？

问Java :按布尔谓词划分为两个列表
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问Java :按布尔谓词划分为两个列表EN