无需迭代即可获取番石榴多重集中元素的实例数
我在番石榴中有一个多重集，我想检索给定元素的实例数，而不迭代这个多重集（我不想迭代，因为我认为迭代需要相当长的时间，因为它会查看所有集合）。
为此，我首先考虑使用多重集的entryset()方法，来获取包含单个实例及其相应计数的集合。然后，将此集合转换为哈希图（其中键是集合的元素，值是它们的实例计数）。因为这样我就可以使用 hashmap 的方法直接从其键中检索值 - 完成！但是，只有当我可以快速地将集合转换为哈希图（无需迭代所有元素）时，这才有意义：可能吗？
（正如我所说，我预计这个问题在多个方面都有缺陷，如果您能阐明我在这里可能犯的概念性错误，我会很高兴。谢谢！）
I have a multiset in guava and I would like to retrieve the number of instances of a given element without iterating over this multiset (I don't want to iterate because I assume that iterating takes quite some time, as it looks through all the collection).
To do that, I was thinking first to use the entryset() method of multiset, to obtain a set with single instances and their corresponding count. Then, transform this set into a hashmap (where keys are the elements of my set, and values are their instance count). Because then I can use the methods of hashmap to directly retrieve a value from its key - done! But this makes sense only if I can transform the set into the hashmap in a quick way (without iterating trhough all elements): is it possible?
(as I said I expect this question to be flawed on multiple counts, I'd be happy if you could shed light on the conceptual mistakes I probably make here. Thx!)
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只需调用
count(element)在你的多重集上 - 瞧！Simply invoke
count(element)on your multiset -- voila!你可能知道在 Guava 中
Multiset是一个接口，而不是一个类。如果您只想知道某个元素的重复次数，请调用
Multiset.count(Object element)。请忘记我的以下声明：
然后，如果您使用流行的实现
HashMultiset，则场景下已经有一个HashMap在工作。也就是说，当
HashMultiset迭代时，HashMap 也会迭代。无需转变成另一个HashMap。You may know in Guava
Multisetis an interface, not a class.If you just want to know the repeated number of an element, call
Multiset.count(Object element).Please forget my following statement:
Then if you are using a popular implementation
HashMultiset, there is already aHashMap<E, AtomicInteger>working under the scene.That is, when the
HashMultisetiterates, also a HashMap iterates. No need to transform into another HashMap.