英文:
Is map applied on all the list before filter findAny?
问题
我想筛选一个列表中具有非空属性的元素，并返回该属性：
list.stream.map(a -> StringUtils.trimToEmpty(a.getProp())).filter(p -> StringUtils.isNotEmpty(p)).findAny().orElse("");
上面的代码首先是否映射了所有元素？出于效率原因，我想逐个处理元素。
英文:
I want to filter a list for elements having a non null property and returning that property:
list.stream.map(a -> StringUtils.trimToEmpty(a.getProp())).filter( p -> StringUtils.isNotEmpty(p)).findAny().orElse("");
Do the code above first map all the elements? For efficiency reasons I'd like to process element by element.
答案1
得分: 5
Stream::findAny 是一个短路终端操作，简而言之，如果存在一个项目，它将终止流。
也就是说，如果第一个元素符合 Stream::filter 管道并通过它，Stream::findAny 会立即返回它，而不会处理原始集合中的其他元素。
有一个使用 Stream::peek 来理解 Stream API 如何工作的好例子（假设一个类 Foo 只有一个 final 字符串参数 prop）：
List<Foo> list = List.of(
new Foo(""), // 第一个是空的，不会通过 Stream::filter
new Foo("one"), // 这个是合格的
new Foo("two")); // 这个也是合格的
list.stream()
.peek(item -> System.out.println("# Streamed " + item.getProp())) // 流式处理
.map(a -> StringUtils.trimToEmpty(a.getProp()))
.peek(item -> System.out.println("# Mapped " + item)) // 映射
.filter(StringUtils::isNotEmpty)
.peek(item -> System.out.println("# Filtered " + item)) // 过滤
.findAny()
.orElse("");
输出显示第一个元素不会通过 Stream::filter（为空），并在 Stream::map 之前结束。第二个元素通过了 Stream::filter 并且随后的 Stream::map，最终到达了 Stream::findAny。只要 Stream::findAny 是一个具有存在结果的短路终端操作，它就会终止流。
# Streamed
# Mapped
# Streamed one
# Mapped one
# Filtered one
英文:
Stream::findAny is a short-circuiting terminal operation, which in a nutshell means that if an item is present, it terminates the Stream.
Ie. if the first element is qualified for the Stream::filter pipe and passes it through, the Stream::findAny returns it immediately without processing further elements in the original collection.
There is a nice proof using Stream::peek to understand how Stream API works (let's say a class Foo has only one final String parameter prop):
List<Foo> list = List.of(
new Foo(""), // the first is empty and doesn't pass through Stream::filter
new Foo("one"), // this is qualified
new Foo("two")); // this is qualified
list.stream()
.peek(item -> System.out.println("# Streamed " + item.getProp())) // streamed
.map(a -> StringUtils.trimToEmpty(a.getProp()))
.peek(item -> System.out.println("# Mapped " + item)) // mapped
.filter(StringUtils::isNotEmpty)
.peek(item -> System.out.println("# Filtered " + item)) // filtered
.findAny()
.orElse("");
The output shows the first element doesn't pass through Stream::filter (is empty) and ended before Stream::map. The second element passed through Stream::filter and the subsequent Stream::map and reached finally Stream::findAny. As long as Stream::findAny is a short-circuiting and terminal operation with a present result, it terminates the Stream.
# Streamed
# Mapped
# Streamed one
# Mapped one
# Filtered one
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