Sau*_*mar 4 java java-8 java-stream
我有一个如下列表.
List<Model> models = ....
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到现在为止,我这样做了
List<String> anotherlist = models.parallelStream().map(Model::getName).collect(Collectors.toList());
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现在,我创建了另一张地图
Map<String, Model> modelsMap = new ConcurrentHashMap<String, Model>();
for (final Model model : models) {
modelsMap.put(model.getId(), model);
}
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是否有可能当我在models
列表上创建并行流时,我可以modelsMap
并排准备我而不必models
再次遍历列表?并行流只能绑定到一种操作吗?
看来你想要的只是:
ConcurrentMap<String, Model> modelsMap =
models.parallelStream()
.collect(toConcurrentMap(Model::getId, identity()));
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这ConcurrentMap
通过收集模型从模型列表中创建Collectors.toConcurrentMap
.键映射器是返回模型名称的函数,值映射器是标识函数.
请注意,如果出现重复的ID,则会引发异常.
如果要在同一个Stream管道中创建名称列表以及此Map,可以使用pairing
此答案中编写的收集器并将结果保存在自定义Pair
类中.Map将是该对中的第一个元素,名称列表将是第二个元素.
Pair<Map<String, Model>, List<String>> pair =
models.parallelStream()
.collect(pairing(
toConcurrentMap(Model::getId, identity()),
mapping(Model::getName, toList()),
Pair::new)
);
static <T, A1, A2, R1, R2, R> Collector<T, ?, R> pairing(Collector<T, A1, R1> c1,
Collector<T, A2, R2> c2, BiFunction<R1, R2, R> finisher) {
EnumSet<Characteristics> c = EnumSet.noneOf(Characteristics.class);
c.addAll(c1.characteristics());
c.retainAll(c2.characteristics());
c.remove(Characteristics.IDENTITY_FINISH);
return Collector.of(() -> new Object[] {c1.supplier().get(), c2.supplier().get()},
(acc, v) -> {
c1.accumulator().accept((A1)acc[0], v);
c2.accumulator().accept((A2)acc[1], v);
},
(acc1, acc2) -> {
acc1[0] = c1.combiner().apply((A1)acc1[0], (A1)acc2[0]);
acc1[1] = c2.combiner().apply((A2)acc1[1], (A2)acc2[1]);
return acc1;
},
acc -> {
R1 r1 = c1.finisher().apply((A1)acc[0]);
R2 r2 = c2.finisher().apply((A2)acc[1]);
return finisher.apply(r1, r2);
}, c.toArray(new Characteristics[c.size()]));
}
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使用以下Pair
课程:
public class Pair<T, U> {
private final T first;
private final U second;
public Pair(T first, U second) {
this.first = first;
this.second = second;
}
public T getFirst() {
return first;
}
public U getSecond() {
return second;
}
}
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