24
votes

I'd like to select a range of elements in a Spark RDD. For example, I have an RDD with a hundred elements, and I need to select elements from 60 to 80. How do I do that?

I see that RDD has a take(i: int) method, which returns the first i elements. But there is no corresponding method to take the last i elements, or i elements from the middle starting at a certain index.

4
I think the most natural way to do this as of 1.0.0 is to register your RDD as a table and use Spark SQL to add a ROW_NUMBER() or RANK() to your dataset and then SELECT the desired rows. For a small RDD this is overkill, but this approach should work efficiently for very large RDDs.Nick Chammas
@NickChammas Checkout updated answeraaronman

4 Answers

12
votes

I don't think there is an efficient method to do this yet. But the easy way is using filter(), lets say you have an RDD, pairs with key value pairs and you only want elements from 60 to 80 inclusive just do.

val 60to80 = pairs.filter {
    _ match {
        case (k,v) => k >= 60 && k <= 80
        case _ => false //incase of invalid input
    }
}

I think it's possible that this could be done more efficiently in the future, by using sortByKey and saving information about the range of values mapped to each partition. Keep in mind this approach would only save anything if you were planning to query the range multiple times because the sort is obviously expensive.

From looking at the spark source it would definitely be possible to do efficient range queries using RangePartitioner:

// An array of upper bounds for the first (partitions - 1) partitions
  private val rangeBounds: Array[K] = {

This is a private member of RangePartitioner with the knowledge of all the upper bounds of the partitions, it would be easy to only query the necessary partitions. It looks like this is something spark users may see in the future: SPARK-911

UPDATE: Way better answer, based on pull request I'm writing for SPARK-911. It will run efficiently if the RDD is sorted and you query it multiple times.

val sorted = sc.parallelize((1 to 100).map(x => (x, x))).sortByKey().cache()
val p: RangePartitioner[Int, Int] = sorted.partitioner.get.asInstanceOf[RangePartitioner[Int, Int]];
val (lower, upper) = (10, 20)
val range = p.getPartition(lower) to p.getPartition(upper)
println(range)
val rangeFilter = (i: Int, iter: Iterator[(Int, Int)]) => {
  if (range.contains(i))
    for ((k, v) <- iter if k >= lower && k <= upper) yield (k, v)
  else
    Iterator.empty
}
for((k,v) <- sorted.mapPartitionsWithIndex(rangeFilter, preservesPartitioning = true).collect()) println(s"$k, $v")

If having the whole partition in memory is acceptable you could even do something like this.
val glommedAndCached = sorted.glom()cache(); glommedAndCached.map(a => a.slice(a.search(lower),a.search(upper)+1)).collect()

search is not a member BTW I just made an implicit class that has a binary search function, not shown here

8
votes

How big is your data set? You might be able to do what you need with:

data.take(80).drop(59)

This seems inefficient, but for small to medium-sized data, should work.

Is it possible to solve this in another way? What's the case for picking exactly a certain range out of the middle of your data? Would takeSample serve you better?

5
votes

Following should be able to get the range. Note the cache will save you some overhead, because internally zipWithIndex need to scan the RDD partition to get the number of elements in each partition.

scala>val r1 = sc.parallelize(List("a", "b", "c", "d", "e", "f", "g"), 3).cache
scala>val r2 = r1.zipWithIndex
scala>val r3 = r2.filter(x=> {x._2>2 && x._2 < 4}).map(x=>x._1)
scala>r3.foreach(println)
d
1
votes

For those who stumble on this question looking for Spark 2.x-compatible answer, you can use filterByRange