blocks|key|408852|text|使用Guava+library|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|408853|int+size+=+Iterators.size(iterator);|code-block|syntax|javascript|408854|在内部，它只是迭代所有元素，所以这只是为了方便。|408855|entityMap|0|LINK|mutability|MUTABLE|url|https://guava.dev/releases/snapshot/api/docs/com/google/common/collect/Iterators.html#size(java.util.Iterator))^0|2|D|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

Using <a href="https://guava.dev/releases/snapshot/api/docs/com/google/common/collect/Iterators.html#size(java.util.Iterator)" rel="nofollow noreferrer">Guava library</a>:
<pre><code>int size = Iterators.size(iterator);
</code></pre>
Internally it just iterates over all elements so its just for convenience.

blocks|key|409006|text|您将始终需要迭代。但是，您可以使用Java+8、9来执行计数，而无需显式地循环：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|409007|Iterable<Integer>+newIterable+=+()+->+iter;
long+count+=+StreamSupport.stream(newIterable.spliterator(),+false).count();|code-block|syntax|javascript|409008|下面是一个测试：|409009|public+static+void+main(String[]+args)+throws+IOException+{
++++Iterator<Integer>+iter+=+Arrays.asList(1,+2,+3,+4,+5).iterator();
++++Iterable<Integer>+newIterable+=+()+->+iter;
++++long+count+=+StreamSupport.stream(newIterable.spliterator(),+false).count();
++++System.out.println(count);
}|409010|这将打印：|409011|5|409012|有趣的是，您可以通过更改此调用的parallel标志来并行化count操作：|offset|length|style|CODE|409013|long+count+=+StreamSupport.stream(newIterable.spliterator(),+*true*).count();|409014|entityMap^0|0|0|0|0|0|0|G|8|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@$Q|15|R|16|S|T]]|9|@]|A|$]]|$1|U|3|V|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

You will always have to iterate. Yet you can use Java 8, 9 to do the counting without looping explicitely:

<pre><code>Iterable&lt;Integer&gt; newIterable = () -&gt; iter;
long count = StreamSupport.stream(newIterable.spliterator(), false).count();
</code></pre>

Here is a test:

<pre><code>public static void main(String[] args) throws IOException {
 Iterator&lt;Integer&gt; iter = Arrays.asList(1, 2, 3, 4, 5).iterator();
 Iterable&lt;Integer&gt; newIterable = () -&gt; iter;
 long count = StreamSupport.stream(newIterable.spliterator(), false).count();
 System.out.println(count);
}
</code></pre>

This prints:

<pre><code>5
</code></pre>

Interesting enough you can parallelize the count operation here by changing the <code>parallel</code> flag on this call:

<pre><code>long count = StreamSupport.stream(newIterable.spliterator(), *true*).count();
</code></pre>

blocks|key|624108|text|使用Guava+library时，另一种选择是将Iterable转换为List。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|624109|List+list+=+Lists.newArrayList(some_iterator);
int+count+=+list.size();|code-block|syntax|javascript|624110|如果您还需要在获取迭代器的大小后访问其元素，请使用此选项。通过使用Iterators.size()，您不再能够访问迭代的元素。|624111|entityMap|0|LINK|mutability|MUTABLE|url|http://google.github.io/guava/releases/snapshot/api/docs/com/google/common/collect/Iterators.html^0|O|8|Z|4|2|D|0|0|0|X|G|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Using <a href="http://google.github.io/guava/releases/snapshot/api/docs/com/google/common/collect/Iterators.html" rel="noreferrer">Guava library</a>, another option is to convert the <code>Iterable</code> to a <code>List</code>.

<pre><code>List list = Lists.newArrayList(some_iterator);
int count = list.size();
</code></pre>

Use this if you need also to access the elements of the iterator after getting its size. By using <code>Iterators.size()</code> you no longer can access the iterated elements.

blocks|key|408778|text|如果你所拥有的只是迭代器，那么不，没有更好的方法。如果迭代器来自一个集合，那么大小也是如此。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|408779|请记住，Iterator只是一个用于遍历不同值的接口，您很可能会有这样的代码|408780|++++new+Iterator<Long>()+{
++++++++final+Random+r+=+new+Random();
++++++++@Override
++++++++public+boolean+hasNext()+{
++++++++++++return+true;
++++++++}

++++++++@Override
++++++++public+Long+next()+{
++++++++++++return+r.nextLong();
++++++++}

++++++++@Override
++++++++public+void+remove()+{
++++++++++++throw+new+IllegalArgumentException("Not+implemented");
++++++++}
++++};|code-block|syntax|javascript|408781|或|408782|++++new+Iterator<BigInteger>()+{
++++++++BigInteger+next+=+BigInteger.ZERO;

++++++++@Override
++++++++public+boolean+hasNext()+{
++++++++++++return+true;
++++++++}

++++++++@Override
++++++++public+BigInteger+next()+{
++++++++++++BigInteger+current+=+next;
++++++++++++next+=+next.add(BigInteger.ONE);
++++++++++++return+current;
++++++++}

++++++++@Override
++++++++public+void+remove()+{
++++++++++++throw+new+IllegalArgumentException("Not+implemented");
++++++++}
++++};+|408783|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If all you have is the iterator, then no, there is no "better" way. If the iterator comes from a collection you could as that for size.

Keep in mind that Iterator is just an interface for traversing distinct values, you would very well have code such as this

<pre><code> new Iterator&lt;Long&gt;() {
 final Random r = new Random();
 @Override
 public boolean hasNext() {
 return true;
 }

 @Override
 public Long next() {
 return r.nextLong();
 }

 @Override
 public void remove() {
 throw new IllegalArgumentException("Not implemented");
 }
 };
</code></pre>

or

<pre><code> new Iterator&lt;BigInteger&gt;() {
 BigInteger next = BigInteger.ZERO;

 @Override
 public boolean hasNext() {
 return true;
 }

 @Override
 public BigInteger next() {
 BigInteger current = next;
 next = next.add(BigInteger.ONE);
 return current;
 }

 @Override
 public void remove() {
 throw new IllegalArgumentException("Not implemented");
 }
 }; 
</code></pre>

blocks|key|623959|text|如果你只有迭代器，那就没有更有效的方法了。如果迭代器只能使用一次，那么在获取迭代器的内容之前获取计数是...有问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|623960|解决方案是要么更改您的应用程序，使其不需要计数，要么通过一些其他方法获得计数。(例如，传递Collection而不是Iterator+...)|offset|length|style|CODE|623961|entityMap^0|0|19|A|1M|8|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

There is no more efficient way, if all you have is the iterator. And if the iterator can only be used once, then getting the count before you get the iterator's contents is ... problematic.

The solution is either to change your application so that it doesn't need the count, or to obtain the count by some other means. (For example, pass a <code>Collection</code> rather than <code>Iterator</code> ...)

Is there a "computationally" quick way to get the count of an iterator?

<pre><code>int i = 0;
for ( ; some_iterator.hasNext() ; ++i ) some_iterator.next();
</code></pre>

... seems like a waste of CPU cycles.

What is the best way to get the count/length/size of an iterator?

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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有没有一种“计算性”的快速方法来获得迭代器的计数？int i = 0;for ( ; some_iterator.hasNext() ; ++i ) some_iterator.next();..。这似乎是对CPU周期的浪费。

问获取迭代器的计数/长度/大小的最佳方法是什么？
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回答 5

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问获取迭代器的计数/长度/大小的最佳方法是什么？EN

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问获取迭代器的计数/长度/大小的最佳方法是什么？
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