我有一个 RDD[(String, (Iterable[Int], Iterable[Coordinate]))]
我想做的是将 Iterable[Int]
分解为元组,每个元组都像 (String,Int,Iterable[Coordinate])
举个例子,我想转换:
('a',<1,2,3>,<(45.34,32.33),(45.36,32.34)>)
('b',<1>,<(46.64,32.66),(46.67,32.71)>)
到
('a',1,<(45.34,32.33),(45.36,32.34)>)
('a',2,<(45.34,32.33),(45.36,32.34)>)
('a',3,<(45.34,32.33),(45.36,32.34)>)
('b',1,<(46.64,32.66),(46.67,32.71)>)
Scala 是怎么做的?
最佳答案
尝试使用flatMap:
rdd.flatMap {case (v, i1, i2) => i1.map(i=>(v, i, i2)}
关于scala - 将 RDD 中的元组分解为两个元组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34649761/