java 如何使用java流添加所有内容？
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How to do add all with java streams?
提问by user3407267
How do I do add all with java 8 ?
如何使用 java 8 添加所有内容？
processeditemList is a Map<Integer, Map<Item, Boolean>>
As for now I am doing :
至于现在我在做什么：
List<Item> itemList = Lists.newLinkedList();
for (Map<Item, Boolean> entry : processeditemList.values()) {
itemList.addAll(entry.keySet());
}
return itemList;
回答by Beri
You can use flatMap. It's used to combine multiple streams into a single one. So here you need to create a stream of collections, and then create a stream from each of them:
您可以使用flatMap。它用于将多个流合并为一个流。所以在这里你需要创建一个集合流，然后从它们中的每一个创建一个流：
processeditemList.values().stream()
.map(Map::keySet) // Stream<Set<Item>>
.flatMap(Set::stream) // convert each set to a stream
.collect(toList()); // convert to list, this will be an ArrayList implmentation by default
If you want to change default List implementation, then you can use below collector:
如果要更改默认列表实现，则可以使用以下收集器：
Collectors.toCollection(LinkedList::new)
LinkedList would be good if you do not know the final size of the list, and you do more insert operations than read.
如果您不知道列表的最终大小，并且您执行的插入操作多于读取操作，则 LinkedList 会很好。
ArrayList is the oposite: more you read, and less add/remove. Because ArrayList under the hood holds an array, which must be rescaled when adding new elements, but never get's reduced, when you remove elements.
ArrayList 正好相反：您阅读的内容越多，添加/删除的内容就越少。因为引擎盖下的 ArrayList 包含一个数组，在添加新元素时必须重新缩放该数组，但在删除元素时永远不会减少。
回答by Jacob G.
I'm on my phone at the moment so I can't promise that this will be syntactically perfect, but here's how you can do it with a Stream:
我现在正在使用手机，所以我不能保证这在语法上是完美的，但是您可以使用以下方法来做到这一点Stream：
processeditemList.values().stream()
.flatMap(e -> e.keySet().stream())
.collect(Collectors.toCollection(LinkedList::new));
I'm unable to test it currently, but I'll look over the Javadocs and change anything if it's incorrect.
我目前无法测试它，但我会查看 Javadocs 并更改任何不正确的内容。
Edit: I think everything is good. If it doesn't matter which Listimplementation is used, you can change
编辑：我认为一切都很好。如果List使用哪种实现无关紧要，您可以更改
Collectors.toCollection(LinkedList::new)
to
到
Collectors.toList()
