blocks|key|236211|text|您可以先使用SortedSet，因为它上存在subSet方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|236212|您还可以将set的内容添加到List，并对其使用subList方法。但这取决于存储在Set中的数据量，因为您不希望复制大量数据。|236213|否则，你应该在集合上继续迭代，因为它会更有效率。|236214|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/util/SortedSet.html#subSet%2528E,%2520E%2529^0|6|9|M|6|M|6|0|0|E|4|O|7|16|3|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@$9|W|A|X|1|Y]]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|16|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|K|$L|$5|M|N|O|E|$P|Q]]]]

You can either use at first a <code>SortedSet</code> as the <a href="http://docs.oracle.com/javase/7/docs/api/java/util/SortedSet.html#subSet%28E,%20E%29" rel="noreferrer"><code>subSet</code></a> method exists on it.

You can also add the content of your set to a <code>List</code> and use the <code>subList</code> method on it. But it depends on the amount of data stored in your <code>Set</code> as you would not want to duplicate an enormous volume of data.

Otherwise you should stay with the iteration over the Set as it will be more efficient.

blocks|key|453924|text|你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|453925|Set<Integer>+set+=+new+LinkedHashSet<>();
for+(int+i+=+0;+i+<+50;+i%2B%2B)+{
+++set.add(i);
}

List<Integer>+list+=+new+ArrayList<>(set);
Set<Integer>+subSet+=+new+LinkedHashSet<>(list.subList(0,+20));|code-block|syntax|javascript|453926|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|$]]

You could do this:

<pre><code>Set&lt;Integer&gt; set = new LinkedHashSet&lt;&gt;();
for (int i = 0; i &lt; 50; i++) {
 set.add(i);
}

List&lt;Integer&gt; list = new ArrayList&lt;&gt;(set);
Set&lt;Integer&gt; subSet = new LinkedHashSet&lt;&gt;(list.subList(0, 20));
</code></pre>

blocks|key|453957|text|在Guava：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|453958|Set<Integer>+subset+=+ImmutableSet.copyOf(Iterables.limit(set,+20));|code-block|syntax|javascript|453959|请注意，Iterables.limit()是延迟计算的，因此只创建了一个额外的集合。|offset|length|style|CODE|453960|entityMap^0|0|0|4|H|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

In Guava:

<pre><code>Set&lt;Integer&gt; subset = ImmutableSet.copyOf(Iterables.limit(set, 20));
</code></pre>

Note that <code>Iterables.limit()</code> is evaluated lazily, so only one extra collection is created.

blocks|key|236306|text|简单的帮助器方法(您可以将其用于Set或任何其他集合)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|236307|public+static+<T>+List<T>+listOf(final+Collection<T>+set,+final+int+limit)+{
++++final+List<T>+list+=+new+ArrayList<>(limit);

++++final+Iterator<T>+i+=+set.iterator();
++++for+(int+j+=+0;+j+<+limit+&&+i.hasNext();+j%2B%2B)+{
++++++++list.add(i.next());
++++}

++++return+list;
}|code-block|syntax|javascript|236308|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|$]]

Simple helper method (You can use it for Set or any other collection):

<pre><code>public static &lt;T&gt; List&lt;T&gt; listOf(final Collection&lt;T&gt; set, final int limit) {
 final List&lt;T&gt; list = new ArrayList&lt;&gt;(limit);

 final Iterator&lt;T&gt; i = set.iterator();
 for (int j = 0; j &lt; limit &amp;&amp; i.hasNext(); j++) {
 list.add(i.next());
 }

 return list;
}
</code></pre>

blocks|key|454440|text|使用流和收集器的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|454441|Set<Integer>+subSet+=+set.stream()
++++//+.skip(10)+//+Use+this+to+get+elements+later+in+the+stream
++++.limit(20)
++++.collect(toCollection(LinkedHashSet::new));
++++//+You+could+also+collect+to+something+else+
++++//+with+another+collector+like+this:+
++++//+.collect(toList());|code-block|syntax|javascript|454442|这里假设有以下导入：|454443|import+static+java.util.stream.Collectors.toCollection;|454444|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

A solution using streams and collectors:
<pre><code>Set&lt;Integer&gt; subSet = set.stream()
 // .skip(10) // Use this to get elements later in the stream
 .limit(20)
 .collect(toCollection(LinkedHashSet::new));
 // You could also collect to something else 
 // with another collector like this: 
 // .collect(toList());
</code></pre>
This assumes the following import:
<pre><code>import static java.util.stream.Collectors.toCollection;
</code></pre>

blocks|key|236450|text|在Java+8中，您可以执行以下操作|type|unstyled|depth|inlineStyleRanges|entityRanges|data|236451|mySet.stream()
+++.skip(start)+//+the+offset
+++.limit(count)+//+how+many+items+you+want
+++.collect(Collectors.toSet());|code-block|syntax|javascript|236452|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|$]]

In Java 8 you can do

<pre><code>mySet.stream()
 .skip(start) // the offset
 .limit(count) // how many items you want
 .collect(Collectors.toSet());
</code></pre>

I have a <code>LinkedHashSet</code>, i.e an ordered set. I'm trying to find a function to just return a subset of the set, i.e the first 20 elements of the set. I know I can do it by creating a new set and then populating using an iteration of the first set but I was hoping for something more succinct.

Also took a look at Google's Guava libraries, but couldn't see what I wanted.

Creating subset of a Set in Java

Java

我有一个LinkedHashSet，也就是有序集。我正在尝试找到一个函数，它只返回集合的一个子集，即集合的前20个元素。我知道我可以通过创建一个新的集合，然后使用第一个集合的迭代来填充，但我希望有更简洁的东西。我也看了一下Google的Guava库，但看不到我想要什么。

问在Java中创建集合的子集
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