我need to use SparkContext instead of JavaSparkContext for the accumulableCollection (if you don't agree check out the linked question并回答它,请)Java在Spark中的Scala Seq?
澄清问题:SparkContext有Java,但希望Scala的序列。我如何让它开心 - 在Java中?
我有这段代码做一个简单的jsc.parallelize
我用JavaSparkContext,但SparkContext想要一个Scala集合。我在这里想到了我正在构建一个Scala范围并将其转换为Java列表,不知道如何让核心范围成为Scala Seq,这就是parallelize from SparkContext is asking for。
// The JavaSparkContext way, was trying to get around MAXINT limit, not the issue here
// setup bogus Lists of size M and N for parallelize
//List<Integer> rangeM = rangeClosed(startM, endM).boxed().collect(Collectors.toList());
//List<Integer> rangeN = rangeClosed(startN, endN).boxed().collect(Collectors.toList());
钱线是下一个,我如何创建一个Java中的Scala Seq来并行化?
// these lists above need to be scala objects now that we switched to SparkContext
scala.collection.Seq<Integer> rangeMscala = scala.collection.immutable.List(startM to endM);
// setup sparkConf and create SparkContext
... SparkConf setup
SparkContext jsc = new SparkContext(sparkConf);
RDD<Integer> dataSetMscala = jsc.parallelize(rangeMscala);
我正在看[JavaConversions](http://docs.scala-lang.org/overviews/collections/conversions-between-java-and-scala-collections)对象,看起来像它在两个方向工作,在Java中还是在斯卡拉? – JimLohse
我怀疑你可以在Java中创建一个'Seq',因为它是'trait',并且在Java中没有等价物。我认为在Scala中使用'JavaConversions'是正确的方法。 – davidshen84
我想也许我正在复制[this](http://stackoverflow.com/questions/35988315/convert-java-list-to-scala-seq?rq=1),我会尝试解决方案,并会发布一个回答如果它的工作,似乎像JavaConversion可以在Java中使用,如果我正确读取此:http://stackoverflow.com/questions/35988315/convert-java-list-to-scala-seq?rq=1 – JimLohse