Question

想象一个像这样返回Future的函数。

def findById(Int id): Future[MyObject]

我收集了一些ID。

val ids = Vector(1,2,3)

迭代集合并为每个值并行调用函数以产生（有效）这个的最惯用的方法是什么？

val result: Vector[Future[MyObject]] = Vector(findById(1), findById(2), findById(3))

Answer 1

我认为简单的map就够了：ids.map(findById)，或者更长ids.map(id => findById(id))

请考虑以下示例代码：

import concurrent.ExecutionContext.Implicits.global
import concurrent.duration._
import concurrent.{Await, Future}


def findById(id: Int): concurrent.Future[String] = concurrent.Future { Thread.sleep(3000); "OK" }
val ids = Vector(1,2,3)
val sequenced = Future.sequence(ids.map(findById)) // we sequence them so we can wait for one future of a collection, not collection of futures
Await.result(sequenced, 10.seconds) // you probably don't want to block normally, but this is for tests it should be resolved in about 3 seconds, so it runs in parallel

您可以在REPL中运行它（只需运行scala并执行:paste）。

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