java - Java - 连接流中的连续元素

Question

我正在尝试连接数组中的两个连续元素。我可以迭代地执行此操作，但我正在尝试学习 Java Streams，并认为这将是一个很好的练习。

如果我有一个字符串数组： String exampleArray[] = ["a1", "junk", "a2", "b1", "junk", "b2", "c1", "junk", "junk", "junk", "c2", "d1", "junk", "d2", "junk-n"]

我想得到： ["a1 - a2", "b1 - b2", "c1 - c2", "d1 - d2"]作为输出。

我试过这个：

Arrays.asList(exampleArray)
    .stream()
    .filter(s -> s.length() > 0)   // gets rid of blanks
    .filter(s -> !s.contains("junk"))
    .collect(Collectors.groupingBy(it -> counter.getAndIncrement() / 2))
    .values();

这会返回一个Collection<List<String>>赞[ [a1, a2], [b1, b2], [c1, c2], [d1, d2] ]

但我不确定如何到达：["a1 - a2", "b1 - b2", "c1 - c2", "d1 - d2"]

任何帮助表示赞赏！

标签： javajava-streamconcatenation

Answer 1

Collectors.groupingBy has a overloaded method where you can pass the result to a downstream collector. In fact, by default it has used toList collector and so you got List. You can use joiningBy to concatenating strings.

.collect(Collectors.groupingBy(it -> counter.getAndIncrement() / 2
       , Collectors.joining(" - ")))

Result is

[a1-a2, b1-b2, c1-c2, d1-d2]

Code

public static void main(String[] args) {
    String exampleArray[] = new String[] {"a1", "junk", "a2", "b1", "junk", "b2", "c1", "junk", "junk", "junk", "c2", "d1", "junk", "d2", "junk-n"};
    AtomicInteger counter = new AtomicInteger(0);
    Collection<String> ans = (Arrays.asList(exampleArray)
            .stream()
            .filter(s -> s.length() > 0)   // gets rid of blanks
            .filter(s -> !s.contains("junk"))
            .collect(Collectors.groupingBy(it -> counter.getAndIncrement() / 2, Collectors.joining(" - ")))
            .values());

    System.out.println(ans);
}