英文:
Java stream get single element from collection
问题
我正在使用非流式方法从集合中获取单个元素。
List<MyCustomClass> list = OtherObject.getMyList();if (list.size() != 1) {throw new RuntimeException();}MyCustomClass customClass = list.get(0);
除了这种多行方法，是否有一些通过流（streams）实现这个的方式？
英文:
I am using a non stream way to get single element from collection.
List<MyCustomClass> list = OtherObject.getMyList();if (list.size() != 1) {throw new RuntimeException();}MyCustomClass customClass = list.get(0);
Instead of this multi liner approach, is there some way to achieve this via streams?
答案1
得分: 2
你可以使用 reduce(accumulator) 和 orElseThrow(exceptionSupplier) 来确保流产生且仅产生一个结果。
MyCustomClass customClass = list.stream().reduce((a,b) -> { throw new RuntimeException("存在过多的值"); }).orElseThrow(() -> { throw new RuntimeException("不存在值"); });
英文:
You can use reduce(accumulator) and orElseThrow(exceptionSupplier) to ensure the stream produces exactly one result.
MyCustomClass customClass = list.stream().reduce((a,b) -> { throw new RuntimeException("Too many values present"); }).orElseThrow(() -> { throw new RuntimeException("No value present"); });
答案2
得分: 0
你可以尝试从findFirst()或findAny()返回一个可选项。
List<String> strings = new ArrayList<>();Optional<String> maybeFirst = strings.stream().findFirst();// 现在我们有了一个可选项，让我们强制获取一个值String value = maybeFirst.orElseThrow(IllegalArgumentException::new);// 如果没有值，我们会抛出非法参数异常。
这可以简化为以下形式。
String value = strings.stream().findFirst().orElseThrow(() -> new IllegalArgumentException("至少必须有一个字符串。"));
希望对您有所帮助。
英文:
You could try returning an optional from findFirst() or findAny().
List<String> strings = new ArrayList<>();Optional<String> maybeFirst = strings.stream().findFirst();// we now have an optional, lets force a valueString value = maybeFirst.orElseThrow(IllegalArgumentException::new);// if there isn't a value, we'll throw an illegal argument exception.
This can collapsed into the following.
String value = strings.stream().findFirst().orElseThrow(() -> new IllegalArgumentException("There must be at least one string."));
Hope that helps.
答案3
得分: 0
我曾经寻找过一个只包含单个collect语句的版本，尽管结果并不像Andreas的解决方案那样简洁优雅。它使用了一个Collector的实现，将累积到一个单元素列表中，而组合器在有多于一个元素时会引发异常；完成器会在列表为空时引发异常。
list.stream().collect(Collector.of(ArrayList::new,(a, t) -> {if (!a.isEmpty())throw new RuntimeException();a.add(t);},(a, b) -> {throw new RuntimeException();},a -> {if (a.isEmpty())throw new RuntimeException();return a.get(0);}));
英文:
I was looking for a version with a single collect statement, although it turned out not as concise or elegant as the solution by Andreas. It uses an implementation of Collector that accumulates to a one-element list, while the combiner raises an exception if we have more than one element; the finisher raises an exception when the list is empty.
list.stream().collect(Collector.of( ArrayList::new,(a, t) -> { if (!a.isEmpty())throw new RuntimeException();a.add(t); },(a, b) -> { throw new RuntimeException(); },a -> { if( a.isEmpty() )throw new RuntimeException();return a.get(0);} );
专注分享java语言的经验与见解，让所有开发者获益！
评论