问题描述
我编写了这段代码，将一个单词列表缩减为多少单词以'A'开头。我只是写它来学习Java 8，所以我想更好地理解它 [免责声明：我意识到这可能不是编写这段代码的最好方法;这只是为了练习！] 。
I wrote this code to reduce a list of words to a long count of how many words start with an 'A'. I'm just writing it to learn Java 8, so I'd like to understand it a little better [Disclaimer: I realize this is probably not the best way to write this code; it's just for practice!].
Long countOfAWords = results.stream().reduce(
0L,
(a, b) -> b.charAt(0) == 'A' ? a + 1 : a,
Long::sum);
中间参数/ lambda（称为累加器）似乎能够减少完整列表而不用最后的'Combiner'参数。事实上，Javadoc实际上说：
The middle parameter/lambda (called the accumulator) would seem to be capable of reducing the full list without the final 'Combiner' parameter. In fact, the Javadoc actually says:
- 以下陈述是错误的，所以不要让它混淆你;我只是把它放在这里，所以我不会破坏答案的原始背景。
- The following statement is wrong, so don't let it confuse you; I'm just keeping it here so I don't ruin the original context of the answers.
无论如何，我可以推断累加器必须只输出1和0组合器结合了。我没有从文档中发现这一点特别明显。
Anyway, I can infer that the accumulator must just be outputting 1's and 0's which the combiner combines. I didn't find this particularly obvious from the documentation though.
我的问题
有没有办法在组合器执行之前看到输出是什么，所以我可以看到组合器组合的1和0的列表？这将有助于调试更复杂的情况，我相信我最终会遇到这种情况。
Is there a way to see what the output would be before the combiner executes so I can see the list of 1's and 0's that the combiner combines? This would be helpful in debugging more complex situations which I'm sure I'll come across eventually.
推荐答案
组合器没有减少0和1的列表。当流不是并行运行时，在这种情况下不使用它，因此以下循环是等效的：
The combiner does not reduce a list of 0's and 1's. When the stream is not run in parallel it's not used in this case so that the following loop is equivalent:
U result = identity;
for (T element : this stream)
result = accumulator.apply(result, element)
return result;
当您并行运行流时，任务将跨越多个线程。因此，例如，管道中的数据被划分为块，其独立地评估和产生结果。然后组合器用于合并该结果。
When you run the stream in parallel, the task is spanned into multiple threads. So for example the data in the pipeline is partitioned into chunks that evaluate and produce a result independently. Then the combiner is used to merge this results.
因此，您不会看到减少的列表，而是2个值，即身份值或由任务计算的另一个值求和。例如，如果在合并器中添加打印语句
So you won't see a list that is reduced, but rather 2 values either the identity value or with another value computed by a task that are summed. For example if you add a print statement in the combiner
(i1, i2) -> {System.out.println("Merging: "+i1+"-"+i2); return i1+i2;});
你可以看到这样的东西：
you could see something like this:
Merging: 0-0
Merging: 0-0
Merging: 1-0
Merging: 1-0
Merging: 1-1
更一般地说，如果你想在途中看到管道上的数据你可以使用 peek （或者调试器也可以帮助）。因此适用于您的示例：
More generally if you want to see the data on the pipeline on the go you can use peek (or the debugger could also help). So applied to your example:
long countOfAWords = result.stream().map(s -> s.charAt(0) == 'A' ? 1 : 0).peek(System.out::print).mapToLong(l -> l).sum();
可以输出：
100100
实现任务的惯用方法是过滤器流然后只需使用 count ：
The idiomatic way to achieve your task would be to filter the stream and then simply use count:
long countOfAWords = result.stream().filter(s -> s.charAt(0) == 'A').count();
希望有所帮助！ ：）
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