我一直在使用Java 8 lambdas来轻松地过滤集合。但是我没有找到一种简洁的方法来检索结果,将其作为同一语句中的新列表。以下是我目前为止最简洁的方法:

List<Long> sourceLongList = Arrays.asList(1L, 10L, 50L, 80L, 100L, 120L, 133L, 333L);
List<Long> targetLongList = new ArrayList<>();
sourceLongList.stream().filter(l -> l > 100).forEach(targetLongList::add);

网上的示例没有回答我的问题,因为它们没有生成新的结果列表就停止了。一定有更简洁的方法。我本来希望,Stream类有toList(), toSet(),…

是否有一种方法,变量targetLongList可以直接由第三行分配?


当前回答

LongStream类和提供了收集方法的另一种变体 IntStream和DoubleStream类也类似。

<R> R collect(Supplier<R> supplier,
              ObjLongConsumer<R> accumulator,
              BiConsumer<R,R> combiner)

对此流的元素执行可变缩减操作。可变约简是指被约简的值是一个可变的结果容器,例如ArrayList,元素是通过更新结果的状态而不是替换结果来合并的。这样产生的结果相当于:

R result = supplier.get();
  for (long element : this stream)
       accumulator.accept(result, element);
  return result;

像reduce(long, LongBinaryOperator)一样,collect操作可以并行化,而不需要额外的同步。 这是一个终端操作。

用这种收集方法回答你的问题如下:

    LongStream.of(1L, 2L, 3L, 3L).filter(i -> i > 2)
    .collect(ArrayList::new, (list, value) -> list.add(value)
    , (list1, list2) -> list1.addAll(list2));

下面是方法引用变体,它非常聪明,但有些难以理解:

     LongStream.of(1L, 2L, 3L, 3L).filter(i -> i > 2)
    .collect(ArrayList::new, List::add , List::addAll);

下面是HashSet的变体:

     LongStream.of(1L, 2L, 3L, 3).filter(i -> i > 2)
     .collect(HashSet::new, HashSet::add, HashSet::addAll);

类似地,LinkedList的变体是这样的:

     LongStream.of(1L, 2L, 3L, 3L)
     .filter(i -> i > 2)
     .collect(LinkedList::new, LinkedList::add, LinkedList::addAll);

其他回答

这里是常用的算盘代码

LongStream.of(1, 10, 50, 80, 100, 120, 133, 333).filter(e -> e > 100).toList();

披露:我是abacus-common的开发者。

在Java 16中有一个新的方法Stream.toList():

List<Long> targetLongList = sourceLongList
         .stream()
         .filter(l -> l > 100)
         .toList();

如果有人(比如我)正在寻找处理对象而不是基本类型的方法,那么就使用mapToObj()

String ss = "An alternative way is to insert the following VM option before "
        + "the -vmargs option in the Eclipse shortcut properties(edit the "
        + "field Target inside the Shortcut tab):";

List<Character> ll = ss
                        .chars()
                        .mapToObj(c -> new Character((char) c))
                        .collect(Collectors.toList());

System.out.println("List type: " + ll.getClass());
System.out.println("Elem type: " + ll.get(0).getClass());
ll.stream().limit(50).forEach(System.out::print);

打印:

List type: class java.util.ArrayList
Elem type: class java.lang.Character
An alternative way is to insert the following VM o

在一个可变列表中收集:

targetList = sourceList.stream()
                       .filter(i -> i > 100) //apply filter
                       .collect(Collectors.toList());

在一个不可变列表中收集:

targetList = sourceList.stream()
                       .filter(i -> i > 100) //apply filter
                       .collect(Collectors.toUnmodifiableList());

从JavaDoc中collect的解释:

Performs a mutable reduction operation on the elements of this stream using a Collector. A Collector encapsulates the functions used as arguments to collect(Supplier, BiConsumer, BiConsumer), allowing for reuse of collection strategies and composition of collect operations such as multiple-level grouping or partitioning. If the stream is parallel, and the Collector is concurrent, and either the stream is unordered or the collector is unordered, then a concurrent reduction will be performed (see Collector for details on concurrent reduction.) This is a terminal operation. When executed in parallel, multiple intermediate results may be instantiated, populated, and merged so as to maintain isolation of mutable data structures. Therefore, even when executed in parallel with non-thread-safe data structures (such as ArrayList), no additional synchronization is needed for a parallel reduction.

更新:

另一种方法是使用collections . tolist:

targetLongList = 
    sourceLongList.stream().
    filter(l -> l > 100).
    collect(Collectors.toList());

之前的解决方案:

另一种方法是使用collections . tocollection:

targetLongList = 
    sourceLongList.stream().
    filter(l -> l > 100).
    collect(Collectors.toCollection(ArrayList::new));