斯卡拉相当于java.util.ArrayList中的 [英] Scala equivalent of java.util.ArrayList

问题描述

我做的斯卡拉项目，但我相当新的语言，有一个Java的背景。我看到Scala没有ArrayList中，所以我想知道什么Scala的相当于Java的ArrayList的叫了，如果有是Java和Scala版本之间的重要区别。

I am doing a project in Scala, but am fairly new to the language and have a Java background. I see that Scala doesn't have ArrayList, so I am wondering what Scala's equivalent of Java's ArrayList is called, and if there are any important differences between the Java and Scala versions.

编辑：我不是在寻找一个特定的行为，而更像是一种内部重新presentation（存储在数组中的数据，但整个数组是不可见的，只有一部分您使用）。

I'm not looking for a specific behavior so much as an internal representation (data stored in an array, but the whole array isn't visible, only the part you use).

推荐答案

的ArrayList 是一个被广泛误用默认列表在Java中。它具有当元素添加/删除经常表现糟糕，但工程pretty以及为阵列的替代品。所以，当你问什么是Scala等价，我能想到的三种不同的实际的问题：

ArrayList is a widely misused "default" List in Java. It has terrible performance when elements are added/removed frequently, but works pretty well as a replacement for Array. So, when you ask what is the equivalent in Scala, I can think of three different actual questions:

• 什么是Scala的默认集合？


• 什么斯卡拉收藏有类似的ArrayList 特点？


• 什么是一个很好的替代品斯卡拉阵列？

• What is Scala's default collection?
• What Scala collection has characteristics similar to ArrayList?
• What's a good replacement for Array in Scala?

因此​​，这里有答案了这些：

So here are the answers for these:

Scala的相当于Java中的列表接口是 SEQ 。一个更普遍的接口中也存在，这是 GenSeq - 主要的区别是，一个 GenSeq 可能有操作处理串行或并行，根据实现

Scala's equivalent of Java's List interface is the Seq. A more general interface exists as well, which is the GenSeq -- the main difference being that a GenSeq may have operations processed serially or in parallel, depending on the implementation.

由于斯卡拉允许程序员使用 SEQ 作为工厂，他们不经常，除非他们关心它定义一个特定的实现麻烦。当他们这样做，他们通常会挑选要么Scala的列表或矢量。它们都是不可变的，而矢量具有良好的索引访问性能。在另一方面，列表做得非常好它做得很好的操作。

Because Scala allows programmers to use Seq as a factory, they don't often bother with defining a particular implementation unless they care about it. When they do, they'll usually pick either Scala's List or Vector. They are both immutable, and Vector has good indexed access performance. On the other hand, List does very well the operations it does well.

这是 scala.collection.mutable.ArrayBuffer 。

好了，好消息是，你可以使用阵列 Scala中！在Java中，阵列通常避免，因为它与一般的仿制药不兼容。它是共变体集合，而泛型是不变的，它是可变的 - 这使得它的协方差的危险时，它接受原语，其中泛型不这样做，并且它有一个pretty有限集的方法

Well, the good news is, you can just use Array in Scala! In Java, Array is often avoided because of its general incompatibility with generics. It is a co-variant collection, whereas generics is invariant, it is mutable -- which makes it's co-variance a danger, it accepts primitives where generics don't, and it has a pretty limited set of methods.

在Scala中，阵列 - 这仍然是相同的阵列在Java中 - 是不变的，这使大多数问题消失。斯卡拉接受 AnyVal （图元的等值）作为类型的仿制药，即使它会做自动装箱。并通过丰富了我的图书馆模式，所有 SEQ 方法可用于阵列。

In Scala, Array -- which is still the same Array as in Java -- is invariant, which makes most problems go away. Scala accepts AnyVal (the equivalent of primitives) as types for its "generics", even though it will do auto-boxing. And through the "enrich my library" pattern, ALL of Seq methods are available to Array.

所以，如果你想有一个更强大的阵列，只使用阵列。

So, if you want a more powerful Array, just use an Array.

适用于所有生产的所有集合默认的方法新的的集合。例如，如果我这样做：

The default methods available to all collections all produce new collections. For example, if I do this:

val ys = xs filter (x => x % 2 == 0)

然后 YS 将是一个的新的的集合，而 XS 仍然会是与此相同命令之前。这是真实的，不管是什么 XS 是：阵列，列表等。

Then ys will be a new collection, while xs will still be the same as before this command. This is true no matter what xs was: Array, List, etc.

当然，这是有代价的 - 毕竟，你的是的产生一个新的集合。 Scala的不可变的集合是在处理这个成本要好得多，因为他们的永久的，而是取决于执行什么操作。

Naturally, this has a cost -- after all, you are producing a new collection. Scala's immutable collections are much better at handling this cost because they are persistent, but it depends on what operation is executed.

没有收集可以做很多关于过滤，但列表对由$生成一个新的集合优异性能p $ ppending一个元件或取出头 - 堆的基本操作，事实上。 矢量有一堆作业性能好，但如果集合不小，只支付。对于，比如说收藏，百达元素，整体成本可能会超过收益。

No collection can do much about filter, but a List has excellent performance on generating a new collection by prepending an element or removing the head -- the basic operations of a stack, as a matter of fact. Vector has good performance on a bunch of operations, but it only pays if the collection isn't small. For collections of, say, up to a hundred elements, the overall cost might exceed the gains.

所以，实际上你可以添加或删除元素的阵列，和Scala会产生的新的的阵列为你，但是当你做，你将支付的完整副本的成本。

So you can actually add or remove elements to an Array, and Scala will produce a new Array for you, but you'll pay the cost of a full copy when you do that.

Scala的可变集合添加一些其他的方法。特别是，可以增加或减少大小集合 - 而不会产生一个新的集合 - 贯彻落实<一个href=\"http://www.scala-lang.org/archives/downloads/distrib/files/nightly/docs/library/index.html#scala.collection.generic.Growable\"><$c$c>Growable和<一个href=\"http://www.scala-lang.org/archives/downloads/distrib/files/nightly/docs/library/index.html#scala.collection.generic.Shrinkable\"><$c$c>Shrinkable特征。他们不保证良好的性能上的这些操作，虽然，但他们会点你要检查出的集合。

Scala mutable collections add a few other methods. In particular, the collections that can increase or decrease size -- without producing a new collection -- implement the Growable and Shrinkable traits. They don't guarantee good performance on these operations, though, but they'll point you to the collections you want to check out.

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