blocks|key|3180376|text|Scala：您可能应该看看Kevin+Wright's+answer，它提供了自Scala2.8.1以来可用的更好的解决方案，具有更少的隐式魔力。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3180377|3180378|您可以从scala.collection.JavaConversions导入隐式转换，然后无缝地创建一个新的Scala集合，例如：|style|CODE|3180379|import+collection.JavaConversions._
println(List()+%2B%2B+javaIterator)|code-block|syntax|javascript|3180380|您的Java迭代器被JavaConversions.asScalaIterator转换为Scala迭代器。带有A类型元素的Scala迭代器实现了TraversableOnce[A]，这是连接集合和%2B%2B所需的参数类型。|3180381|如果您需要其他集合类型，只需将List()更改为您需要的任何类型(例如，IndexedSeq()或collection.mutable.Seq()等)。|3180382|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/6014881/create-an-immutable-list-from-a-java-lang-iterator/6015315#6015315|1|http://www.scala-lang.org/api/current/scala/collection/JavaConversions$.html^0|D|L|0|0|0|4|W|4|W|1|0|0|A|V|1J|1|20|I|2Q|2|0|F|6|10|C|1D|O|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@$A|12|B|13|1|14]]|C|$]]|$1|D|3|-4|5|6|7|15|8|@]|9|@]|C|$]]|$1|E|3|F|5|6|7|16|8|@$A|17|B|18|G|H]]|9|@$A|19|B|1A|1|1B]]|C|$]]|$1|I|3|J|5|K|7|1C|8|@]|9|@]|C|$L|M]]|$1|N|3|O|5|6|7|1D|8|@$A|1E|B|1F|G|H]|$A|1G|B|1H|G|H]|$A|1I|B|1J|G|H]|$A|1K|B|1L|G|H]]|9|@]|C|$]]|$1|P|3|Q|5|6|7|1M|8|@$A|1N|B|1O|G|H]|$A|1P|B|1Q|G|H]|$A|1R|B|1S|G|H]]|9|@]|C|$]]|$1|R|3|-4|5|6|7|1T|8|@]|9|@]|C|$]]]|S|$T|$5|U|V|W|C|$X|Y]]|Z|$5|U|V|W|C|$X|10]]]]

EDIT: You should probably look at <a href="https://stackoverflow.com/questions/6014881/create-an-immutable-list-from-a-java-lang-iterator/6015315#6015315">Kevin Wright's answer</a>, which provides a better solution available since Scala 2.8.1, with less implicit magic.

<hr>

You can import the implicit conversions from <a href="http://www.scala-lang.org/api/current/scala/collection/JavaConversions$.html" rel="nofollow noreferrer"><code>scala.collection.JavaConversions</code></a> and then create a new Scala collection seamlessly, e.g. like this:

<pre><code>import collection.JavaConversions._
println(List() ++ javaIterator)
</code></pre>

Your Java iterator is converted to a Scala iterator by <code>JavaConversions.asScalaIterator</code>. A Scala iterator with elements of type <code>A</code> implements <code>TraversableOnce[A]</code>, which is the argument type needed to concatenate collections with <code>++</code>.

If you need another collection type, just change <code>List()</code> to whatever you need (e.g., <code>IndexedSeq()</code> or <code>collection.mutable.Seq()</code>, etc.).

blocks|key|3180402|text|您可能希望重新考虑对List的需求，尽管在Java语言中使用List非常熟悉，而且List是不可变Seq的默认实现，但它通常不是集合的最佳选择。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3180403|list最适合的操作是那些已经通过迭代器可用的操作(基本上是获取连续的head元素和前置元素)。如果迭代器还没有给出你需要的东西，那么我可以保证列表不会是你最好的选择--向量会更合适。|3180404|3180405|把这事解决了……推荐的在Java和Scala集合之间转换的技术(从Scala2.8.1开始)是通过scala.collection.JavaConverters。这为您提供了比JavaConversions更多的控制权，并避免了一些可能的隐式冲突。|3180406|这样就不会有直接的隐式转换。相反，您可以将asScala和asJava方法绑定到集合上，从而允许您显式地执行转换。|3180407|要将Java迭代器转换为Scala迭代器：|3180408|javaIterator.asScala|code-block|syntax|javascript|3180409|要将Java迭代器转换为Scala列表(通过scala迭代器)：|3180410|javaIterator.asScala.toList|3180411|3180412|您可能还想考虑转换toSeq而不是toList。在迭代器的情况下，这将返回一个Stream+-允许您在更丰富的Seq接口中保留迭代器的懒惰行为。|3180413|编辑：没有toVector方法，但是(正如Daniel指出的)有一个toIndexedSeq方法将返回一个Vector作为默认的IndexedSeq子类(就像List是默认的Seq一样)。|BOLD|3180414|javaIterator.asScala.toIndexedSeq|3180415|entityMap|0|LINK|mutability|MUTABLE|url|http://www.scala-lang.org/api/current/scala/collection/JavaConverters$.html^0|A|4|1D|3|0|0|0|1D|V|2G|F|1D|V|0|0|L|7|T|6|0|0|0|0|0|0|9|5|H|6|13|6|1J|3|0|0|3|5|8|Y|C|1H|6|1S|A|27|4|2F|3|0|0^^$0|@$1|2|3|4|5|6|7|1D|8|@$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1I|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]|$1|I|3|J|5|6|7|1K|8|@$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]]|D|@$9|1P|A|1Q|1|1R]]|E|$]]|$1|K|3|L|5|6|7|1S|8|@$9|1T|A|1U|B|C]|$9|1V|A|1W|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1X|8|@]|D|@]|E|$]]|$1|O|3|P|5|Q|7|1Y|8|@]|D|@]|E|$R|S]]|$1|T|3|U|5|6|7|1Z|8|@]|D|@]|E|$]]|$1|V|3|W|5|Q|7|20|8|@]|D|@]|E|$R|S]]|$1|X|3|-4|5|6|7|21|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|22|8|@$9|23|A|24|B|C]|$9|25|A|26|B|C]|$9|27|A|28|B|C]|$9|29|A|2A|B|C]]|D|@]|E|$]]|$1|10|3|11|5|6|7|2B|8|@$9|2C|A|2D|B|12]|$9|2E|A|2F|B|C]|$9|2G|A|2H|B|C]|$9|2I|A|2J|B|C]|$9|2K|A|2L|B|C]|$9|2M|A|2N|B|C]|$9|2O|A|2P|B|C]]|D|@]|E|$]]|$1|13|3|14|5|Q|7|2Q|8|@]|D|@]|E|$R|S]]|$1|15|3|-4|5|6|7|2R|8|@]|D|@]|E|$]]]|16|$17|$5|18|19|1A|E|$1B|1C]]]]

You might want to rethink the need for a <code>List</code>, although it feels very familiar when coming from Java, and List is the default implementation of an immutable <code>Seq</code>, it often isn't the best choice of collection.

The operations that list is optimal for are those already available via an iterator (basically taking consecutive head elements and prepending elements). If an iterator doesn't already give you what you need, then I can pretty much guarantee that a List won't be your best choice - a vector would be more appropriate.

<hr>

Having got that out the way... The recommended technique to convert between Java and Scala collections (since Scala 2.8.1) is via <a href="http://www.scala-lang.org/api/current/scala/collection/JavaConverters$.html" rel="noreferrer"><code>scala.collection.JavaConverters</code></a>. This gives you more control than <code>JavaConversions</code> and avoids some possible implicit conflicts.

You won't have a direct implicit conversion this way. Instead, you get <code>asScala</code> and <code>asJava</code> methods pimped onto collections, allowing you to perform the conversions explicitly.

To convert a Java iterator to a Scala iterator:

<pre><code>javaIterator.asScala
</code></pre>

To convert a Java iterator to a Scala List (via the scala iterator):

<pre><code>javaIterator.asScala.toList
</code></pre>

<hr>

You may also want to consider converting <code>toSeq</code> instead of <code>toList</code>. In the case of iterators, this'll return a <code>Stream</code> - allowing you to retain the lazy behaviour of iterators within the richer <code>Seq</code> interface.

EDIT:
There's no <code>toVector</code> method, but (as Daniel pointed out) there's a <code>toIndexedSeq</code> method that will return a <code>Vector</code> as the default <code>IndexedSeq</code> subclass (just as <code>List</code> is the default <code>Seq</code>).

<pre><code>javaIterator.asScala.toIndexedSeq
</code></pre>

I'm using a library (<a href="http://commons.apache.org/jxpath">JXPath</a>) to query a graph of beans in order to extract matching elements. However, JXPath returns groups of matching elements as an instance of java.lang.Iterator and I'd rather like to convert it into an immutable scala list. Is there any simpler way of doing than iterating over the iterator and creating a new immutable list at each iteration step ?

Create an immutable list from a java.lang.Iterator

Java

我使用一个库()来查询bean的图形，以便提取匹配的元素。但是，JXPath将一组匹配的元素作为java.lang.Iterator的实例返回，我更愿意将其转换为不可变的scala列表。有没有比迭代遍历迭代器并在每个迭代步骤创建一个新的不可变列表更简单的方法呢？

问从java.lang.Iterator创建不可变列表
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