java IntStream 逐步迭代
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/40358827/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
IntStream iterate in steps
提问by JasperJ
How do you iterate through a range of numbers (0-100) in steps(3) with IntStream?
如何使用 IntStream 在步骤 (3) 中迭代一系列数字 (0-100)？
I tried iterate, but this never stops executing.
我试过了iterate，但这永远不会停止执行。
IntStream.iterate(0, n -> n + 3).filter(x -> x > 0 && x < 100).forEach(System.out::println)
采纳答案by JasperJ
Actually rangeis ideal for this.
实际上range是理想的。
IntStream.range(0, 100).filter(x -> x % 3 == 0); //107,566 ns/op [Average]
Edit: Holgers's solution is the fastest performing solution.
编辑：Holgers 的解决方案是性能最快的解决方案。
Since the following lines of code
由于以下代码行
IntStream.range(0, 100).filter(x -> x % 3 == 0).forEach((x) -> x = x + 2);
IntStream.range(0, 100 / 3).map(x -> x * 3).forEach((x) -> x = x + 2);
int limit = ( 100 / 3 ) + 1;
IntStream.iterate(0, n -> n + 3).limit(limit).forEach((x) -> x = x + 2);
show these benchmark results
显示这些基准测试结果
Benchmark Mode Cnt Score Error Units
Benchmark.intStreamTest avgt 5 485,473 ± 58,402 ns/op
Benchmark.intStreamTest2 avgt 5 202,135 ± 7,237 ns/op
Benchmark.intStreamTest3 avgt 5 280,307 ± 41,772 ns/op
回答by gil.fernandes
Actually you can also achieve the same results with a combination of peek and allMatch:
实际上，您也可以通过 peek 和 allMatch 的组合获得相同的结果：
IntStream.iterate(0, n -> n + 3).peek(n -> System.out.printf("%d,", n)).allMatch(n -> n < 100 - 3);
This prints
这打印
0,3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99,
0,3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72, 75,78,81,84,87,90,93,96,99,
But nevertheless, this one is faster:
但是，这个速度更快：
IntStream.range(0, 100 / 3 + 1).map(x -> x * 3).forEach((x) -> System.out.printf("%d,", x));
Java 9 Update:
Java 9 更新：
Now the same iteration easier to achieve with Java 9:
现在使用 Java 9 更容易实现相同的迭代：
Stream.iterate(0, i -> i <= 100, i -> 3 + i).forEach(i -> System.out.printf("%d,", i));
回答by davidsheldon
回答by Saravana
limitcan also be used
limit也可以使用
int limit = ( 100 / 3 ) + 1;
IntStream.iterate(0, n -> n + 3).limit(limit).forEach(System.out::println);
回答by Donald Raab
If you are ok adding a library dependency, the IntIntervalclass in Eclipse Collectionshas the step function I think you are looking for. I tried a few different approaches converting IntIntervalto an IntStream, since the original question asked for IntStream. Here are the solutions I came up with using IntIntervaland then converting it to an IntStream.
如果您可以添加库依赖项，那么Eclipse Collections 中的IntInterval类具有我认为您正在寻找的 step 函数。我尝试了几种不同的转换为 an 的方法，因为最初的问题要求. 这是我想出的解决方案，然后将其转换为.IntIntervalIntStreamIntStreamIntIntervalIntStream
IntInterval interval = IntInterval.zeroTo(99).by(3);
interval.each(System.out::print);
IntStream.of(interval.toArray()).forEach(System.out::print);
IntStream.Builder builder = IntStream.builder();
interval.each(builder::add);
builder.build().forEach(System.out::print);
IntStream.generate(interval.intIterator()::next)
.limit(interval.size()).forEach(System.out::print);
IntIntervalis inclusive on the from and to like IntStream.rangeClosed().
IntInterval是包容对 from 和 to 喜欢IntStream.rangeClosed()。
Note: I am a committer for Eclipse Collections
注意：我是 Eclipse Collections 的提交者
回答by Sriram
Elegant Solution:
优雅的解决方案：
IntStream.iterate(0, n -> n < 100, n -> n + 3).forEach(System.out::println)
Stream.iterate() supports a hasNext()predicate(added in Java 9) which can be used to limit the stream in more natural way.
Stream.iterate() 支持hasNext()谓词（在 Java 9 中添加），可用于以更自然的方式限制流。
回答by gstackoverflow
More generic solution:
更通用的解决方案：
LongStream.range(0L, (to - from) / step) // +1 depends on inclusve or exclusive
.mapToObj(i -> (from + i * step))
.collect(Collectors.toList());
