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。

我试过了：

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>>，比如[ [a1, a2], [b1, b2], [c1, c2], [d1, d2] ]

但我不知道怎么去：["a1 - a2", "b1 - b2", "c1 - c2", "d1 - d2"]

感谢任何帮助！

Answer 1

你快到了。你已经有了这些对，你现在要做的就是用中间的“-”将它们粉碎成一个字符串。

试试这个：

Arrays.stream(exampleArray)
                .filter(s -> s.length() > 0)   // gets rid of blanks
                .filter(s -> !s.contains("junk"))
                .collect(Collectors.groupingBy(it -> counter.getAndIncrement() / 2))
                .values()
                .stream()                        //stream the pairs
                .map(l -> String.join("-", l))   //and put a "-" between them & into a string
                .collect(Collectors.toList())    //collect all your joined String

Answer 2

Collectors.groupingBy 有一个重载方法，您可以在其中将结果传递给下游收集器。事实上，默认情况下它使用了 toList 收集器，所以你得到了 List。您可以使用 joinBy 来连接字符串。

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

结果是

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

代码

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);
}

Answer 3

实现此输出的简单方法是：

String a[] = {"a1", "junk", "a2", "b1", "junk", "b2", "c1", "junk", "junk", "junk", "c2", "d1", "junk", "d2", "junk-n"};

    Arrays.parallelSort(a); //sorting the array from a to z
    
    List<String> output = new ArrayList<>();
    
    for (int i = 0; i < a.length - 1; i++) {
        
        if(a[i].matches(".*\\d.*")) //verifying if have a number in string
           output.add(a[i] + " - " + a[++i]);
        
    }
    
    System.out.println(output); //["a1 - a2", "b1 - b2", "c1 - c2", "d1 - d2"]