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.
Action count() returns the number of elements in RDD. For example, RDD has values {1, 2, 2, 3, 4, 5, 5, 6} in this RDD “rdd. count()” will give the result 8.
Take the first num elements of the RDD. It works by first scanning one partition, and use the results from that partition to estimate the number of additional partitions needed to satisfy the limit.
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
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