计算 Stream 的元素 [英] Counting elements of a Stream
问题描述
我想计算一个流的不同元素，想知道为什么
I want to count the different elements of a stream and am wondering why
Stream<String> stream = Stream.of("a", "b", "a", "c", "c", "a", "a", "d");
Map<String, Integer> counter1 = stream.collect(Collectors.toMap(s -> s, 1, Integer::sum));
不起作用.Eclipse 告诉我
doesn't work. Eclipse tells me
类型Collectors中的toMap(Function, Function, BinaryOperator)方法不适用于参数((s) -> {}, int, Integer::sum)
The method toMap(Function, Function, BinaryOperator) in the type Collectors is not applicable for the arguments (( s) -> {}, int, Integer::sum)
顺便说一句，我知道那个解决方案:
By the way, I know about that solution:
Map<String, Long> counter2 = stream.collect(Collectors.groupingBy(s -> s, Collectors.counting()));
所以我有两个问题:
- 我的第一种方法有什么错误?
- 您将如何实现这样的计数器?
我自己解决了第一个问题:
I solved the first question by myself:
Map<String, Integer> counter1 = stream.collect(Collectors.toMap(s -> s, s -> 1, Integer::sum));
Java 期望一个函数作为第二个参数.
Java is expecting a function as second argument.
推荐答案
确实有几种方法可以做到.你没有提到的是 .collect(groupingBy(x -> x, summingInt(x -> 1)));
There are indeed several ways to do it. The one you haven't mentioned is .collect(groupingBy(x -> x, summingInt(x -> 1)));
在性能上有一些差异.
如果每个存储桶的对象很少，方法 #1 将处于最佳状态.在每个桶只有 1 个对象的理想情况下，您可以立即获得最终映射，而无需修改条目.在有大量重复对象的最坏情况下，它将不得不进行大量装箱/拆箱.
Approach #1 is going to be at its best if there are very few objects per bucket. In the ideal case of only 1 object per bucket, you end up with the final map right away with no need to modify the entries. In the worst case of having a very large number of repeated objects, it will have to do a lot of boxing/unboxing.
方法 #2 依赖于 counting() 收集器，它没有具体说明它应该如何进行计数.当前实现转发到 reducing 但这可能会改变.
Approach #2 relies on counting() collector, which doesn't specify exactly how it should do the counting. The current implementation forwards to reducing but that might change.
summingInt 方法将在 int 而不是 Integer 中累积计数，因此不需要任何装箱/拆箱.如果对象重复很多次，那将是最好的.
The summingInt approach will accumulate the count in int rather than Integer and thus will not require any boxing/unboxing. It will be at its best if objects repeat a very large number of times.
至于选择哪一个，最好把代码写清楚，必要时再优化.对我来说，groupingBy(x->x,counting()) 最清楚地表达了意图，所以这是我喜欢的.
As for which one to choose, it is best to code for clarity and optimize when it becomes necessary. To me, groupingBy(x->x, counting()) expresses the intent most clearly, so that's the one I would favor.
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