scala – Future.traverse似乎按顺序工作而不是并行工作.这是真
发布时间:2020-12-16 18:21:07 所属栏目:安全 来源:网络整理
导读:我的问题很简单,关于Future.traverse方法. 所以我有一个String-s列表.每个字符串都是网页的URL.然后我有一个类可以获取URL,加载网页并解析一些数据.所有这些都包含在Future {}中,因此异步处理结果. 该类简化如下: class RatingRetriever(context:ExecutionC
我的问题很简单,关于Future.traverse方法.
所以我有一个String-s列表.每个字符串都是网页的URL.然后我有一个类可以获取URL,加载网页并解析一些数据.所有这些都包含在Future {}中,因此异步处理结果. 该类简化如下: class RatingRetriever(context:ExecutionContext) { def resolveFilmToRating(url:String):Future[Option[Double]]={ Future{ //here it creates Selenium web driver,loads the url and parses it. }(context) } } 然后在另一个对象中我有这个: implicit val executionContext = ExecutionContext.fromExecutor(Executors.newFixedThreadPool(2)) ....... val links:List[String] = films.map(film => film.asInstanceOf[WebElement].getAttribute("href")) val ratings: Future[List[Option[Double]]] = Future.traverse(links)(link => new RatingRetriever(executionContext).resolveFilmToRating(link)) 当它工作时我绝对可以看到它按顺序进行收集.如果我将执行上下文从固定大小池更改为单线程池,则行为是相同的. 解决方法
看看traverse的资料来源:
in.foldLeft(successful(cbf(in))) { (fr,a) => //we sequentially traverse Collection val fb = fn(a) //Your function comes here for (r <- fr; b <- fb) yield (r += b) //Just add elem to builder }.map(_.result()) //Getting the collection from builder 那么代码的并行程度取决于你的函数fn,看看两个例子: 1)此代码: import scala.concurrent.Future import scala.concurrent.ExecutionContext.Implicits.global object FutureTraverse extends App{ def log(s: String) = println(s"${Thread.currentThread.getName}: $s") def withDelay(i: Int) = Future{ log(s"withDelay($i)") Thread.sleep(1000) i } val seq = 0 to 10 Future { for(i <- 0 to 5){ log(".") Thread.sleep(1000) } } val resultSeq = Future.traverse(seq)(withDelay(_)) Thread.sleep(6000) } 有这样的输出: ForkJoinPool-1-worker-5: . ForkJoinPool-1-worker-3: withDelay(0) ForkJoinPool-1-worker-1: withDelay(1) ForkJoinPool-1-worker-7: withDelay(2) ForkJoinPool-1-worker-5: . ForkJoinPool-1-worker-3: withDelay(3) ForkJoinPool-1-worker-1: withDelay(4) ForkJoinPool-1-worker-7: withDelay(5) ForkJoinPool-1-worker-5: . ForkJoinPool-1-worker-3: withDelay(6) ForkJoinPool-1-worker-1: withDelay(7) ForkJoinPool-1-worker-7: withDelay(8) ForkJoinPool-1-worker-5: . ForkJoinPool-1-worker-3: withDelay(9) ForkJoinPool-1-worker-1: withDelay(10) ForkJoinPool-1-worker-5: . ForkJoinPool-1-worker-5: . 2)只需更改withDelay函数: def withDelay(i: Int) = { Thread.sleep(1000) Future { log(s"withDelay($i)") i } } 你会得到一个顺序输出: ForkJoinPool-1-worker-7: . ForkJoinPool-1-worker-7: . ForkJoinPool-1-worker-5: withDelay(0) ForkJoinPool-1-worker-7: . ForkJoinPool-1-worker-1: withDelay(1) ForkJoinPool-1-worker-7: . ForkJoinPool-1-worker-1: withDelay(2) ForkJoinPool-1-worker-7: . ForkJoinPool-1-worker-1: withDelay(3) ForkJoinPool-1-worker-7: . ForkJoinPool-1-worker-1: withDelay(4) ForkJoinPool-1-worker-7: withDelay(5) ForkJoinPool-1-worker-1: withDelay(6) ForkJoinPool-1-worker-1: withDelay(7) ForkJoinPool-1-worker-7: withDelay(8) ForkJoinPool-1-worker-7: withDelay(9) ForkJoinPool-1-worker-7: withDelay(10) 所以Future.traverse不一定是并行的,它只是提交任务,它可以按顺序执行,整个并行的东西都在你提交的函数中. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |