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
你可以重写代码如下:
List<Long> sourceLongList = Arrays.asList(1L, 10L, 50L, 80L, 100L, 120L, 133L, 333L);
List<Long> targetLongList = sourceLongList.stream().filter(l -> l > 100).collect(Collectors.toList());
其他回答
在Java 16中有一个新的方法Stream.toList():
List<Long> targetLongList = sourceLongList
.stream()
.filter(l -> l > 100)
.toList();
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));
collect(Collectors.toList());
这个调用可以用来将任何流转换为列表。
更具体地说:
List<String> myList = stream.collect(Collectors.toList());
来自:
https://www.geeksforgeeks.org/collectors-tolist-method-in-java-with-examples/
一个更有效的方法(避免创建源列表和由过滤器自动开箱):
List<Long> targetLongList = LongStream.of(1L, 10L, 50L, 80L, 100L, 120L, 133L, 333L)
.filter(l -> l > 100)
.boxed()
.collect(Collectors.toList());
