如何使用 Java 8 流将列表元素映射到它们的索引？
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How to map elements of the list to their indices using Java 8 streams?
提问by Jaroslaw Pawlak
Having a list of strings, I need to construct a list of objects which are effectively pairs (string, its position in the list). Currently I have such code using google collections:
有一个字符串列表，我需要构造一个有效配对的对象列表(string, its position in the list)。目前我使用谷歌集合有这样的代码：
public Robots(List<String> names) {
ImmutableList.Builder<Robot> builder = ImmutableList.builder();
for (int i = 0; i < names.size(); i++) {
builder.add(new Robot(i, names.get(i)));
}
this.list = builder.build();
}
I would like to do this using Java 8 streams. If there was no index, I could just do:
我想使用 Java 8 流来做到这一点。如果没有索引，我可以这样做：
public Robots(List<String> names) {
this.list = names.stream()
.map(Robot::new) // no index here
.collect(collectingAndThen(
Collectors.toList(),
Collections::unmodifiableList
));
}
To get the index, I would have to do something like this:
要获得索引，我必须执行以下操作：
public Robots(List<String> names) {
AtomicInteger integer = new AtomicInteger(0);
this.list = names.stream()
.map(string -> new Robot(integer.getAndIncrement(), string))
.collect(collectingAndThen(
Collectors.toList(),
Collections::unmodifiableList
));
}
However, the documentation says that mapping function should be stateless, but the AtomicIntegeris effectively its state.
然而，文档说映射函数应该是无状态的，但AtomicInteger实际上是它的状态。
Is there a way to map elements of the sequential stream to their positions in the stream?
有没有办法将顺序流的元素映射到它们在流中的位置？
采纳答案by Alexis C.
You could do something like this:
你可以这样做：
public Robots(List<String> names) {
this.list = IntStream.range(0, names.size())
.mapToObj(i -> new Robot(i, names.get(i)))
.collect(collectingAndThen(toList(), Collections::unmodifiableList));
}
However it may not be as efficient depending on the underlying implementation of the list. You could grab an iterator from the IntStream; then calling next()in the mapToObj.
但是，根据列表的底层实现，它可能效率不高。你可以从IntStream; 中获取一个迭代器。然后调用next()的mapToObj。
As an alternative, the proton-packlibrary defines the zipWithIndexfunctionality for streams:
作为替代方案，proton-pack库定义了zipWithIndex流的功能：
this.list = StreamUtils.zipWithIndex(names.stream())
.map(i -> new Robot(i.getIndex(), i.getValue()))
.collect(collectingAndThen(toList(), Collections::unmodifiableList));
回答by assylias
The easiest way is to stream indices:
最简单的方法是流式传输索引：
List<Robot> robots = IntStream.range(0, names.size())
.mapToObj(i -> new Robot(i, names.get(i))
.collect(toList());
