如何拉平内RDD列表? [英] How to flatten list inside RDD?
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问题描述
是否有可能拉平内RDD列表?例如转换:
VAL XXX:org.apache.spark.rdd.RDD [列表[美孚]
到
VAL YYY:org.apache.spark.rdd.RDD [美孚]
如何做到这一点?
解决方案
VAL RDD = sc.parallelize(阵列(名单(1,2,3),列表(4,5, 6),列表(7,8,9),表(10,11,12)))
// org.apache.spark.rdd.RDD [列表[INT]] = ParallelCollectionRDD ...VAL RDDI = rdd.flatMap(名单=>清单)
// RDDI:org.apache.spark.rdd.RDD [INT] = FlatMappedRDD ...//这是相同rdd.flatMap(身份)
//身份是preDEF对象中定义的方法。
// DEF身份[A](X:A):一个rddi.collect()
// RES2:数组[INT] =阵列(1,2,3,4,5,6,7,8,9,10,11,12)
Is it possible to flatten list inside RDD? For example convert:
val xxx: org.apache.spark.rdd.RDD[List[Foo]]
to:
val yyy: org.apache.spark.rdd.RDD[Foo]
How to do this?
解决方案
val rdd = sc.parallelize(Array(List(1,2,3), List(4,5,6), List(7,8,9), List(10, 11, 12)))
// org.apache.spark.rdd.RDD[List[Int]] = ParallelCollectionRDD ...
val rddi = rdd.flatMap(list => list)
// rddi: org.apache.spark.rdd.RDD[Int] = FlatMappedRDD ...
// which is same as rdd.flatMap(identity)
// identity is a method defined in Predef object.
// def identity[A](x: A): A
rddi.collect()
// res2: Array[Int] = Array(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)
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