问题
Now I have 3 RDDs like this:
rdd1:
1 2
3 4
5 6
7 8
9 10
rdd2:
11 12
13 14
rdd3:
15 16
17 18
19 20
and I want to do this:
rdd1.zip(rdd2.union(rdd3))
and I want the result is like this:
1 2 11 12
3 4 13 14
5 6 15 16
7 8 17 18
9 10 19 20
but I have an exception like this:
Exception in thread "main" java.lang.IllegalArgumentException: Can't zip RDDs with unequal numbers of partitions
someone tell me I can do this without exception:
rdd1.zip(rdd2.union(rdd3).repartition(1))
But it seems like it is a little cost. So I want to know if there is other ways to solve this problem.
回答1:
I'm not sure what you mean by "cost", but you're right to suspect that repartition(1)
is not the right solution. It will repartition the RDD to a single partition.
- If your data does not fit on a single machine, this will fail.
- It only works if
rdd1
has a single partition. When you have more data this will probably no longer hold. repartition
performs a shuffle, so your data can end up ordered differently.
I think the right solution is to give up on using zip
, because you likely cannot ensure that the partitioning will match up. Create a key and use join
instead:
val indexedRDD1 = rdd1.zipWithIndex.map { case (v, i) => i -> v }
val indexedRDD2 = rdd2.zipWithIndex.map { case (v, i) => i -> v }
val offset = rdd2.count
val indexedRDD3 = rdd3.zipWithIndex.map { case (v, i) => (i + offset) -> v }
val combined =
indexedRDD1.leftOuterJoin(indexedRDD2).leftOuterJoin(indexedRDD3).map {
case (i, ((v1, v2Opt), v3Opt)) => i -> (v1, v2Opt.getOrElse(v3Opt.get))
}
This will work no matter the partitioning. If you like, you can sort the result and remove the index at the end:
val unindexed = combined.sortByKey().values
来源:https://stackoverflow.com/questions/29814499/cant-zip-rdds-with-unequal-numbers-of-partitions