从包含满足流条件的第 n 个元素的 Java 列表中获取子列表
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Getting sublist from a Java list with nth elements that fulfills a condition with streams
提问by user3483969
I have a very simple use case
我有一个非常简单的用例
Given a list of letters with A and Bs, I want to get the sublist that contains the first N Bs, for example:
给定带有 A 和 B 的字母列表，我想获取包含前 N 个 B 的子列表，例如：
- f(3, [A A A B A B A B A A]) = [A A A B A B A B]
- f(2, [A A A B A B A B A A]) = [A A A B A B]
- f(1, [A A B A B A]) = [A A B]
- f(0, [A A B A B]) = []
- f(3, [AAABABABA]) = [AAABABAB]
- f(2, [AAABABABA]) = [AAABAB]
- f(1, [AABABA]) = [AAB]
- f(0, [AABAB]) = []
By following an imperative approach, this is relatively easy, count until we find N Bs, and then get the sublist until that position.
通过遵循命令式方法，这相对容易，计数直到我们找到 N 个 B，然后获取直到该位置的子列表。
However, I couldn't find any functional solution with lambdas, since the operation on every node seems to be independent from the others (which I guess make sense for parallelization).
但是，我找不到 lambda 的任何功能解决方案，因为每个节点上的操作似乎都独立于其他节点（我认为这对并行化很有意义）。
采纳答案by Tagir Valeev
If your input is a Listwith fast random access, you can solve your problem using the stream of indices:
如果您的输入是List具有快速随机访问的，您可以使用索引流解决您的问题：
public static List<String> f(int n, List<String> input) {
int fence = IntStream.range(0, input.size())
.filter(idx -> input.get(idx).equals("B")) // leave only B's
.skip(n-1)
.findFirst() // an index of n-th B
.getAsInt(); // with throw NoSuchElementException if not enough B's
return input.subList(0, fence+1);
}
Usage example:
用法示例：
System.out.println(f(3, Arrays.asList("A", "A", "A", "B", "A", "B", "A", "B", "A", "A")));
System.out.println(f(2, Arrays.asList("A", "A", "A", "B", "A", "B", "A", "B", "A", "A")));
System.out.println(f(1, Arrays.asList("A", "A", "A", "B", "A", "B", "A", "B", "A", "A")));
However despite I love Stream API I would solve this problem imperatively.
然而，尽管我喜欢 Stream API，但我还是会迫切地解决这个问题。
回答by Karol Król
This code snippet do the job.
这段代码片段完成了这项工作。
public static void main(String[] args) {
final String input = "A A A B A B A B A A";
final String[] array = input.split(" ");
final List<String> list = Arrays.asList(array);
final List<String> result = getSubList(list, 3);
System.out.println("Result=" + result);
}
public static List<String> getSubList(final List<String> list, final int limit) {
final AtomicInteger counter = new AtomicInteger();
return list.stream()
.filter(ch -> {
if (ch.equals("B") && counter.incrementAndGet() == limit) return true;
return counter.get() < limit;
})
.collect(Collectors.toList());
}
回答by user_3380739
Here is code by AbacusUtil
这是AbacusUtil 的代码
List<String> input = N.asList("A", "A", "A", "B", "A", "B", "A", "B", "A", "A");
List<String> list = Stream.of(input).takeWhile(MutableInt.of(3), (e, cnt) -> cnt.getAndAdd(e.equals("B") ? -1 : 0) > 0).toList();
N.println(list);
Declaration： I'm the developer of AbacusUtil.
声明： 我是AbacusUtil的开发者。
