如何在列表中找到匹配的元素并将其映射为 Scala API 方法？
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原文地址: http://stackoverflow.com/questions/5479227/
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How to find a matching element in a list and map it in as an Scala API method?
提问by agilefall
Is there a method to do the following without doing both methods: findand map?
有没有一种方法可以在不执行这两种方法的情况下执行以下操作：find和map？
val l = 0 to 3
l.find(_ * 33 % 2 == 0).map(_ * 33) // returns Some(66)
回答by Wilfred Springer
How about using collect?
使用收集怎么样？
// Returns List(66)
List(1, 2, 3) collect { case i if (i * 33 % 2 == 0) => i * 33 }
However that will return allmatches and not just the first one.
但是，这将返回所有匹配项，而不仅仅是第一个匹配项。
The better answer would have been, based on Scala 2.9:
更好的答案是，基于 Scala 2.9：
// Returns Some(66)
List(1, 2, 3) collectFirst { case i if (i * 33 % 2 == 0) => i * 33 }
The solution suggested in the comments to append a headto get a Scala 2.8 version of that is not very efficient, I'm afraid. Perhaps in that case I would stick to your own code. In any case, in order to make sure it returns an option, you should not call head, but headOption.
head恐怕在评论中建议的解决方案附加 a以获得 Scala 2.8 版本的效率不是很高。也许在那种情况下，我会坚持使用您自己的代码。在任何情况下，为了确保它返回一个选项，你不应该调用head, 而是headOption。
// Returns Some(66)
List(1, 2, 3) collect { case i if (i * 33 % 2 == 0) => i * 33 } headOption
回答by Garrett Hall
If you don't want to do your map()operation multiple times (for instance if it's an expensive DB lookup) you can do this:
如果您不想map()多次执行操作（例如，如果它是一个昂贵的数据库查找），您可以这样做：
l.view.map(_ * 33).find(_ % 2 == 0)
l.view.map(_ * 33).find(_ % 2 == 0)
The viewmakes the collection lazy, so the number of map()operations is minimized.
这view使得集合变得懒惰，因此map()操作的数量被最小化。
回答by Alex Cruise
Hey look, it's my little buddy findMapagain!
嘿看，又是我的小伙伴findMap！
/**
* Finds the first element in the list that satisfies the partial function, then
* maps it through the function.
*/
def findMap[A,B](in: Traversable[A])(f: PartialFunction[A,B]): Option[B] = {
in.find(f.isDefinedAt(_)).map(f(_))
}
Note that, unlike in the accepted answer, but like the collectFirstmethod mentioned in one of its comments, this guy stops as soon as he finds a matching element.
请注意，与已接受的答案不同，但与collectFirst其中一条评论中提到的方法一样，这个人一旦找到匹配的元素就会停止。
回答by thoredge
This can do it, but it would be easier if you tell what you're really trying to achieve:
这可以做到，但如果你说出你真正想要实现的目标会更容易：
l.flatMap(n => if (n * 33 % 2 == 0) Some(n * 33) else None).headOption
