如何有效地在星火每一列找到不同的值

问题描述

为了从阵列中的每一列找到不同的值我试过

RDD[Array[String]].map(_.map(Set(_))).reduce { 
(a, b) => (a.zip(b)).map { case (x, y) => x ++ y}}

成功地执行。不过，我想知道是否有这样做的比我的样本code以上的更有效的方式。谢谢你。

which executes successfully. However, I'd like to know if there was a more efficient way of doing this than my sample code above. Thank you.

推荐答案

总结节省了一步，可能会或可能不会是更有效的。

Aggregate saves a step, might or might not be more efficient

val z = Array.fill(5)(Set[String]()) // or whatever the length is
val d= lists.aggregate(z)({(a, b) => (a.zip(b)).map { case (x, y) => x + y}}, 
                          {(a, b) => (a.zip(b)).map { case (x, y) => x ++ y}})

您也可以尝试使用可变套和修改，而不是产生在每一步一个新的（这是明确允许星火）：

You could also try using mutable sets and modifying rather than producing a new one at each step (which is explicitly allowed by Spark):

val z = Array.fill(5)(scala.collection.mutable.Set[String]())
val d= lists.aggregate(z)({(a, b) => (a.zip(b)).foreach { case (x, y) => x+= y };a},
                          {(a, b) => (a.zip(b)).foreach { case (x, y) => x ++= y};a})

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