Java中从有限和无限流中获取每N个元素的方法
Java中从有限和无限流中获取每N个元素的方法
1. 概述
Java Stream API提供了多种操作方法来处理元素序列。然而，如果我们只想处理流的一部分，例如每N个元素，这就不容易了。这在我们处理代表CSV文件或数据库表的原始数据流，并且只想处理特定列时可能会很有用。
**我们将处理两种类型的流：有限的和无限的。**第一种情况可以通过将Stream转换为List来解决，这允许索引访问。另一方面，无限流将需要一种不同的方法。在本教程中，我们将学习如何使用各种技术来解决这个挑战。
2. 测试设置
我们将使用参数化测试来检查我们解决方案的正确性。将有几个案例，分别对应N-th元素和预期结果：
Arguments.of(
Stream.of("Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"),
List.of("Wednesday", "Saturday"), 3),
Arguments.of(
Stream.of("Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"),
List.of("Friday"), 5),
Arguments.of(
Stream.of("Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"),
List.of("Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"), 1)
现在，我们可以深入不同的方法来处理流中的第N个元素。
3. 使用 filter(Predicate)
在第一种方法中，我们可以创建一个单独的流，其中只包含我们想要处理的元素的索引。我们以使用 filter(Predicate) 来创建这样一个数组：
void givenListSkipNthElementInListWithFilterTestShouldFilterNthElement(Stream`````````````````````````<String>````````````````````````` input, List`````````````````````````<String>````````````````````````` expected, int n) {
final List`````````````````````````<String>````````````````````````` sourceList = input.collect(Collectors.toList());
final List`````````````````````````<String>````````````````````````` actual = IntStream.range(0, sourceList.size())
.filter(s -> (s + 1) % n == 0)
.mapToObj(sourceList::get)
.collect(Collectors.toList());
assertEquals(expected, actual);
}
如果我们想要操作允许索引访问的数据结构，例如_List_，这种方法将有效。所需的元素可以收集到一个新的_List_或使用_forEach(Consumer)_进行处理。
4. 使用 iterate()
这种方法与前一种类似，需要一个具有索引访问的数据结构。然而，我们不是过滤掉我们不需要的索引，而是在开始时只生成我们想要使用的索引：
void givenListSkipNthElementInListWithIterateTestShouldFilterNthElement(Stream`````````````````````````<String>````````````````````````` input, List`````````````````````````<String>````````````````````````` expected, int n) {
final List`````````````````````````<String>````````````````````````` sourceList = input.collect(Collectors.toList());
int limit = sourceList.size() / n;
final List`````````````````````````<String>````````````````````````` actual = IntStream.iterate(n - 1, i -> (i + n))
.limit(limit)
.mapToObj(sourceList::get)
.collect(Collectors.toList());
assertEquals(expected, actual);
}
在这种情况下，我们使用_IntStream.iterate(int, IntUnaryOperator)_，它允许我们创建一个步长为_n_的整数序列。
5. 使用 subList()
这种方法使用_Stream.iterate_，类似于前一种，但它创建了一个_List_的流，每个_List_都从_nk-th_索引开始：
void givenListSkipNthElementInListWithSublistTestShouldFilterNthElement(Stream`````````````````````````<String>````````````````````````` input, List`````````````````````````<String>````````````````````````` expected, int n) {
final List`````````````````````````<String>````````````````````````` sourceList = input.collect(Collectors.toList());
int limit = sourceList.size() / n;
final List`````````````````````````<String>````````````````````````` actual = Stream.iterate(sourceList, s -> s.subList(n, s.size()))
.limit(limit)
.map(s -> s.get(n - 1))
.collect(Collectors.toList());
assertEquals(expected, actual);
}
我们应该取每个这些_Lists_的第一个元素来获得所需的结果。
6. 使用自定义 Collector
作为一个更高级和透明的解决方案，我们可以实现一个自定义_Collector_，它只收集所需的元素：
class SkippingCollector {
private static final BinaryOperator`<SkippingCollector>` IGNORE_COMBINE = (a, b) -> a;
private final int skip;
private final List`````````````````````````<String>````````````````````````` list = new ArrayList<>();
private int currentIndex = 0;
private SkippingCollector(int skip) {
this.skip = skip;
}
private void accept(String item) {
int index = ++currentIndex % skip;
if (index == 0) {
list.add(item);
}
}
private List`````````````````````````<String>````````````````````````` getResult() {
return list;
}
public static Collector<String, SkippingCollector, List`````````````````````````<String>`````````````````````````> collector(int skip) {
return Collector.of(() -> new SkippingCollector(skip),
SkippingCollector::accept,
IGNORE_COMBINE,
SkippingCollector::getResult);
}
}
这种方法更复杂，需要一些编码。同时，这个解决方案不允许并行化，并且在技术上甚至可能在顺序流上失败，因为组合是一个实现细节，可能会在未来的版本中改变：
public static List`````````````````````````<String>````````````````````````` skipNthElementInStreamWithCollector(Stream`````````````````````````<String>````````````````````````` sourceStream, int n) {
return sourceStream.collect(SkippingCollector.collector(n));
}
然而，可以使用_Spliterators_使这种方法适用于并行流，但这应该有一个很好的理由。
7. 简单循环
所有前面的解决方案都是可行的，但总的来说，它们过于复杂，而且常常误导。**通常，解决问题的最佳方式是尽可能简单的实现。**这是我们如何使用_for_循环实现相同目标的方法：
void givenListSkipNthElementInListWithForTestShouldFilterNthElement(Stream`````````````````````````<String>````````````````````````` input, List`````````````````````````<String>````````````````````````` expected, int n) {
final List`````````````````````````<String>````````````````````````` sourceList = input.collect(Collectors.toList());
List`````````````````````````<String>````````````````````````` result = new ArrayList<>();
for (int i = n - 1; i < sourceList.size(); i += n) {
result.add(sourceList.get(i));
}
final List`````````````````````````<String>````````````````````````` actual = result;
assertEquals(expected, actual);
}
然而，有时我们需要直接使用_Stream_，这不允许我们直接通过索引访问元素。在这种情况下，我们可以使用带有_while_循环的_Iterator_：
void givenListSkipNthElementInStreamWithIteratorTestShouldFilterNthElement(Stream`````````````````````````<String>````````````````````````` input, List`````````````````````````<String>````````````````````````` expected, int n) {
List`````````````````````````<String>````````````````````````` result = new ArrayList<>();
final Iterator`````````````````````````<String>````````````````````````` iterator = input.iterator();
int count = 0;
while (iterator.hasNext()) {
if (count % n == n - 1) {
result.add(iterator.next());
} else {
iterator.next();
}
++count;
}
final List`````````````````````````<String>````````````````````````` actual = result;
assertEquals(expected, actual);
}
这些解决方案更清洁、更直接易懂，同时解决了相同的问题。
8. 结论
Java Stream API是一个强大的工具，它有助于使代码更加声明性和可读。此外，通过利用参数化，流可以实现更好的性能。然而，想要在任何地方都使用流可能不是使用这个API的最佳方式。
虽然在流不完全适用的情况下应用流操作的精神体操可能很有趣，但这也可能导致“聪明代码”。通常，像循环这样的最简单结构，可以用更少的代码实现相同的结果，并且代码更易于理解。
如常，本文中使用的所有代码都可以在GitHub上找到。