Question

假设我有一个枚举器

val e：Enumerator [String] = Enumerator（＆＃34; a＆＃34;，＆＃34; b＆＃34;，＆＃34; c＆＃34;）

和另一个自然数的调查员

val count = Enumerator.enumerate（Iterator.from（1））

我怎样才能生成一个拉链它们的枚举器。或者用索引拉链e。所以在这种情况下输出是： val zipped：Enumerator [（String，Int）] = Enumerator（（＆＃34; a＆＃34;，1），（＆＃34; b＆＃34;，2），（＆＃34; c＆＃34;， 3））

Answer 1

I'm somewhat new to Iteratees but since there are no responses yet, I will share my less than expert knowledge.

It seems like there should be a zip operator that does what you describe. The closest is interleave, but based on my understanding of its behavior, I don't think it will fit you use case.

If you just need an index for each element you could use something like this (I acknowledge the ugly mutable state)

scala> import scala.concurrent.ExecutionContext.Implicits.global
import scala.concurrent.ExecutionContext.Implicits.global

scala> import play.api.libs.iteratee._
import play.api.libs.iteratee._

scala> var counter = -1
counter: Int = -1

scala> Enumerator("a", "b", "c").map[(String, Int)]{ s => counter+=1; (s, counter) }
res1: play.api.libs.iteratee.Enumerator[(String, Int)] = play.api.libs.iteratee.Enumerator$$anon$3@3c0c6967

scala> res1 run Iteratee.foreach( println _ )
(a,0)
(b,1)
res2: scala.concurrent.Future[Unit] = scala.concurrent.impl.Promise$DefaultPromise@7814bb7d
(c,2)

scala>

If you want something that doesn't close around a var (as you should!) you could look into implementing this functionality on the side of the iteratee and pass the counter around as state in the Step process.

Play Framework Enumerator：如何使用Index压缩枚举器

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