Java Streams，仅过滤第一个 'N' 匹配项

Question

有没有办法使用 java 流仅过滤前“n”个匹配项？

例如，如果我们有以下代码:

List<String> words = Arrays.asList("zero","one","two","three","four","five","one","one");

List<String> filteredWords = words.stream()
                .filter(word->!word.equals("one"))//filter all "one" strings..
                .collect(Collectors.toList());

System.out.println(filteredWords);

这将从单词流中过滤掉所有“one”字符串。

那么，如何仅过滤前“n”个匹配项并保持流的其余部分完好无损？

换句话说，如果n=1，那么程序应该输出

"zero","two","three","four","five","one","one"

如果 n=2 则

“零”、“二”、“三”、“四”、“五”、“一”

Answer 1

您可以创建一个类来为您进行过滤

class LimitedFilter<T> implements Predicate<T> {
    int matches = 0;
    final int limit;
    private Predicate<T> delegate;
    public LimitedFilter<T>(Predicate<T> p, int limit) { 
        delegate = p; this.limit = limit;
    }
    public boolean test(T toTest) {
        if (matches > limit) return true;
        boolean result = delegate.test(toTest);
        if (result) matches++;
        return result;
    }
}

然后用它来过滤

Predicate<String> limited = new LimitedFilter<>(w -> !"one".equals(w), 5);
List<String> filteredWords = words.stream()
            .filter(limited) //filter first five "one" strings..
            .collect(Collectors.toList());