如何将Java8流的元素添加到现有List中 [英] How to add elements of a Java8 stream into an existing List

问题描述

收藏家的Javadoc 节目如何将流的元素收集到新的List中。是否有单行将结果添加到现有的ArrayList中？

Javadoc of Collector shows how to collect elements of a stream into a new List. Is there an one-liner that adds the results into an existing ArrayList?

推荐答案

注意： nosid的答案显示如何使用 forEachOrdered（）。这是一种用于改变现有集合的有用且有效的技术。我的答案解决了为什么你不应该使用收集器来改变现有的集合。

NOTE: nosid's answer shows how to add to an existing collection using forEachOrdered(). This is a useful and effective technique for mutating existing collections. My answer addresses why you shouldn't use a Collector to mutate an existing collection.

简短回答是否，至少在一般情况下，您不应该使用收集器来修改现有的集合。

The short answer is no, at least, not in general, you shouldn't use a Collector to modify an existing collection.

原因是收集器旨在支持并行性，即使是非线程安全的集合也是如此。他们这样做的方法是让每个线程独立地在其自己的中间结果集合上运行。每个线程获取自己的集合的方式是调用 Collector.supplier（），这是每次返回 new 集合所必需的。

The reason is that collectors are designed to support parallelism, even over collections that aren't thread-safe. The way they do this is to have each thread operate independently on its own collection of intermediate results. The way each thread gets its own collection is to call the Collector.supplier() which is required to return a new collection each time.

然后以线程限制的方式合并这些中间结果集合，直到有一个结果集合。这是 collect（）操作的最终结果。

These collections of intermediate results are then merged, again in a thread-confined fashion, until there is a single result collection. This is the final result of the collect() operation.

来自 Balder 和 assylias 建议使用 Collectors.toCollection（）然后传递返回现有列表而不是新列表的供应商。这违反了供应商的要求，即每次都会返回一个新的空集合。

A couple answers from Balder and assylias have suggested using Collectors.toCollection() and then passing a supplier that returns an existing list instead of a new list. This violates the requirement on the supplier, which is that it return a new, empty collection each time.

这将适用于简单的案例，如答案中的示例所示。但是，它会失败，特别是如果流并行运行。 （未来版本的库可能会以某种无法预料的方式发生变化，导致它失败，即使在连续的情况下也是如此。）

This will work for simple cases, as the examples in their answers demonstrate. However, it will fail, particularly if the stream is run in parallel. (A future version of the library might change in some unforeseen way that will cause it to fail, even in the sequential case.)

让我们举一个简单的例子：

Let's take a simple example:

List<String> destList = new ArrayList<>(Arrays.asList("foo"));
List<String> newList = Arrays.asList("0", "1", "2", "3", "4", "5");
newList.parallelStream()
       .collect(Collectors.toCollection(() -> destList));
System.out.println(destList);

当我运行这个程序时，我经常得到一个 ArrayIndexOutOfBoundsException 。这是因为多个线程在 ArrayList 上运行，这是一个线程不安全的数据结构。好的，让它同步：

When I run this program, I often get an ArrayIndexOutOfBoundsException. This is because multiple threads are operating on ArrayList, a thread-unsafe data structure. OK, let's make it synchronized:

List<String> destList =
    Collections.synchronizedList(new ArrayList<>(Arrays.asList("foo")));

这将不会再出现异常。但不是预期的结果：

This will no longer fail with an exception. But instead of the expected result:

[foo, 0, 1, 2, 3]

它给出了奇怪的结果：

[foo, 2, 3, foo, 2, 3, 1, 0, foo, 2, 3, foo, 2, 3, 1, 0, foo, 2, 3, foo, 2, 3, 1, 0, foo, 2, 3, foo, 2, 3, 1, 0]

这是结果我在上面描述的线程限制的累积/合并操作。使用并行流，每个线程调用供应商以获得其自己的集合以进行中间累积。如果您通过了返回相同集合的供应商，则每个线程会将其结果附加到该集合。由于线程之间没有排序，因此结果将以任意顺序附加。

This is the result of the thread-confined accumulation/merging operations I described above. With a parallel stream, each thread calls the supplier to get its own collection for intermediate accumulation. If you pass a supplier that returns the same collection, each thread appends its results to that collection. Since there is no ordering among the threads, results will be appended in some arbitrary order.

然后，当合并这些中间集合时，这基本上将列表与其自身合并。使用 List.addAll（）合并列表，如果在操作期间修改了源集合，则表明结果未定义。在这种情况下， ArrayList.addAll（）执行数组复制操作，因此它最终会复制自身，这是人们所期望的类型，我猜。 （请注意，其他List实现可能具有完全不同的行为。）无论如何，这解释了目标中的奇怪结果和重复元素。

Then, when these intermediate collections are merged, this basically merges the list with itself. Lists are merged using List.addAll(), which says that the results are undefined if the source collection is modified during the operation. In this case, ArrayList.addAll() does an array-copy operation, so it ends up duplicating itself, which is sort-of what one would expect, I guess. (Note that other List implementations might have completely different behavior.) Anyway, this explains the weird results and duplicated elements in the destination.

您可能会说，我会只需确保按顺序运行我的流并继续编写这样的代码

You might say, "I'll just make sure to run my stream sequentially" and go ahead and write code like this

stream.collect(Collectors.toCollection(() -> existingList))

无论如何。我建议不要这样做。如果您控制流，当然，您可以保证它不会并行运行。我希望在流式传输而不是集合的情况下会出现一种编程风格。如果有人给你一个流并且你使用这个代码，如果流恰好是并行的，它将会失败。更糟糕的是，有人可能会给你一个顺序流，这段代码可以正常工作一段时间，通过所有测试等等。然后，一些任意的时间后，系统中其他地方的代码可能会改变为使用并行流将导致你的代码要打破。

anyway. I'd recommend against doing this. If you control the stream, sure, you can guarantee that it won't run in parallel. I expect that a style of programming will emerge where streams get handed around instead of collections. If somebody hands you a stream and you use this code, it'll fail if the stream happens to be parallel. Worse, somebody might hand you a sequential stream and this code will work fine for a while, pass all tests, etc. Then, some arbitrary amount of time later, code elsewhere in the system might change to use parallel streams which will cause your code to break.

好的，那就确保记得打电话给 sequential（）在使用此代码之前的任何流上：

OK, then just make sure to remember to call sequential() on any stream before you use this code:

stream.sequential().collect(Collectors.toCollection(() -> existingList))

当然，你会记得每次都这样做，对吧？ :-)假设你这样做。然后，性能团队将会想知道为什么他们所有精心设计的并行实现都没有提供任何加速。再一次，他们会将其追溯到你的代码，这会强制整个流顺序运行。

Of course, you'll remember to do this every time, right? :-) Let's say you do. Then, the performance team will be wondering why all their carefully crafted parallel implementations aren't providing any speedup. And once again they'll trace it down to your code which is forcing the entire stream to run sequentially.

不要这样做。

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