如何将 Java 8 IntStream 转换为列表？
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How do I convert a Java 8 IntStream to a List?
提问by Eric Wilson
I'm looking at the docs for the IntStream, and I see an toArraymethod, but no way to go directly to a List<Integer>
我正在查看 的文档IntStream，我看到了一种toArray方法，但无法直接转到List<Integer>
Surely there is a way to convert a Streamto a List?
当然有一种方法可以将 a 转换Stream为 aList吗？
采纳答案by Ian Roberts
IntStream::boxed
IntStream::boxed
IntStream::boxedturns an IntStreaminto a Stream<Integer>, which you can then collectinto a List:
IntStream::boxed将 anIntStream变成 a Stream<Integer>，然后您可以将其collect变成 a List：
theIntStream.boxed().collect(Collectors.toList())
The boxedmethod converts the intprimitive values of an IntStreaminto a stream of Integerobjects. The word "boxing"names the int? Integerconversion process. See Oracle Tutorial.
该boxed方法将 a 的int原始值转换IntStream为Integer对象流。“拳击”这个词命名为int？Integer转换过程。请参阅Oracle 教程。
回答by Nikhil Nanivadekar
You can use primitive collections available in Eclipse Collectionsand avoid boxing.
您可以使用Eclipse Collections 中提供的原始集合并避免装箱。
MutableIntList list =
IntStream.range(1, 5)
.collect(IntArrayList::new, MutableIntList::add, MutableIntList::addAll);
Note: I am a contributor to Eclipse Collections.
注意：我是 Eclipse Collections 的贡献者。
回答by Ida Buci?
You could also use mapToObj() on a Stream, which takes an IntFunction and returns an object-valued Stream consisting of the results of applying the given function to the elements of this stream.
您还可以在 Stream 上使用 mapToObj()，它接受一个 IntFunction 并返回一个对象值 Stream，该对象值 Stream 由将给定函数应用于此流的元素的结果组成。
List<Integer> intList = myIntStream.mapToObj(i->i).collect(Collectors.toList());
回答by Vikash
Find the folowing example of finding square of each int element using Java 8 :-
查找以下使用 Java 8 查找每个 int 元素的平方的示例：-
IntStream ints = Arrays.stream(new int[] {1,2,3,4,5});
List<Integer> intsList = ints.map(x-> x*x)
.collect(ArrayList<Integer>::new, ArrayList::add, ArrayList::addAll);
回答by Harry Jones
You can use the collect method:
您可以使用 collect 方法：
IntStream.of(1, 2, 3).collect(ArrayList::new, List::add, List::addAll);
In fact, this is almost exactly what Java is doing when you call .collect(Collectors.toList()) on an object stream:
事实上，当你在对象流上调用 .collect(Collectors.toList()) 时，这几乎就是 Java 所做的：
public static <T> Collector<T, ?, List<T>> toList() {
return new Collectors.CollectorImpl(ArrayList::new, List::add, (var0, var1) -> {
var0.addAll(var1);
return var0;
}, CH_ID);
}
Note: The third parameter is only required if you want to run parallel collection; for sequential collection providing just the first two will suffice.
注意：第三个参数只有在运行并行收集时才需要；对于顺序收集，仅提供前两个就足够了。
