# The problem

Let’s say there are two data sets A and B such that,
A has the fields {id, time} and B has the fields {id, start-time, end-time, points}.

Find the sum of points for a given row in A such that A.id = B.id and A.time is in between B.start-time and B.end-time.

Let’s make it clearer by adding example data -

A
id time
1 10:00
1 10:15
2 10:01
B
id start-time end-time points
1 9:30 10:30 10
1 10:01 10:05 20
1 10:08 10:20 30
1 10:30 10:45 40
2 9:30 10:30 50

Its such a simple question. An equally simple and correct answer would be to write -

Based on our answer, the output should be -

id time sum(points)
1 10:00 30 // 10
1 10:15 40 // 10 + 30
2 10:01 50 // 50

Perhaps, a picture might give you a better intuition:

Here are a few observations:

• The join between A and B is a many-to-many join. It may not seem like much at this scale but it is going to be a nightmare for large datasets.
• Not all the records of the join are useful. For instance, the 4th row in B, with 40 points was part of the output of the JOIN clause, but it was filtered out in the WHERE clause.
• There is an overlap in the aggregation(a by product of the many-to-many join). For instance, The first row of B with 10 points, is part of the aggregation of first and second rows in A

Much of these observations do not matter at smaller scales. But when you are dealing with a data set to a scale of billions or trillions of records, it begins to bother you; More so, if there is a skew in that data. In traditional databases, we got by fine by pushing the condition in WHERE clause to the JOIN clause. Unfortunately for us, because of the way big data systems are designed, non-equi joins cannot be implemented efficiently.

# A solution

Now, you could simply implement the same query as above translated to RDDs or dataframes. As we said before, it is not scalable enough. It might not even finish depending upon the size of the data. What we have is a different approach a time-range join, without a join itself. A senior developer in my team introduced me to this solution.

#### The algorithm

1. Generalize the datasets and combine them.
2. Partition the data and sort them.
3. Aggregate per partition.

Stage 1: Generalize the datasets and combine them:
Let’s consider a dataset G with the fields - {id, time, row-type, points}. Datasets A and B can be generalized into G as follows:

G
id time row-type points
A id time ‘a’ NULL
B+ id start-time ‘b+’ points
B- id end-time ‘b-‘ -1 * points
• Because we are generalizing the datasets, in order to differentiate between the two, row-type is added to G.
• We split the B dataset into two datasets - B+ and B-. B+ has start-time as time in G.
• B+ dataset has points as a positive integer. B- has points as a negative integer(multiplied by -1)

so,
$G = A \cup B_+ \cup B_-$

Repeating it more formally - In dataset G if the row-type = b+, then the value of time field is the start-time in B. The same way, if the row-type = b- then the value of time in G is the start-time in B.

For simplicity’s sake, I will use the terms a records, b+ records or b- records to denote records with row-type = a, row-type=b+ and row-type=b- respectively.

mapped A
id time row-type points
1 10:00 a null
1 10:15 a null
2 10:01 a null
mapped B
id time row-type points
1 9:30 b+ 10
1 10:30 b- -10
1 10:01 b+ 20
1 10:05 b- -20
1 10:08 b+ 30
1 10:20 b- -30
1 10:30 b+ 40
1 10:45 b- -40
2 9:30 b+ 50
2 10:30 b- -50

Stage 2: Partition the data and sort it:
Now that we have a combined dataset of A and B(in terms of B+ and B-), the next step is to partition the data by id and sort by time and row-type within every partition.

The hash for the same key gives the same result everytime. Based on this idea, hash partition the data on id. The records with the same value of id will be part of the same partition[1]. Recall that our simple sql had a join on id - A a LEFT JOIN B b ON (a.id = b.id). Roughly speaking, when you partition the data by id, it is similar to a join on id. For intution, you can think of this partitioning as a JOIN

Sort the data by time and row-type within every partition. This implicity positions the a records in between b+ and b- records, where time is in between start-time and end-time. To get an intution, relate this to the condition in our SQL- start-time <= time and time <= end-time except nothing is filtered yet. Because, we already hash partitioned by id, all values of the same id are part of the same partition. Sorting this partition guarentees that our “filter condition” will be effective. There is also a sort on row-type. This is useful in cases where the times are equal; row-type will be a tie-breaker in those situations. Putting it in terms of our query, it decides if time < end-time or time <= end-time. row-type can be useful on real data sets, if you have some meaningful values like {0,1,2}. We used {a, b+, b-} so that it would be easier to explain.

There is a nice function in spark that does this for us - repartitionAndSortWithinPartitions()

you should really look at this, if you want a detailed explanation on its usage, or simply you can look up spark API. So, in order to use the function, we need to specify how it partition and sort the data.

First, create case classes with for G -

case class GKey (id, time, rowType) extends Ordered[GKey]{
def compare(that) = (this.time, this.rowType)
.compare(that.time, that.rowType)
}

case class GValue (points)


we want a key-value pair for that function to work, so we created a case class for key with {id, time, row-type}. Also, we specified an ordering for the class. Notice that we included only time and row-type. There is no harm in including id in the sorting but we did not, just to illustrate that we need not. If you don’t like the idea of specifying ordering on the case class, you can choose to implement ordering implictly.

Next, we define a hash partitioner to use only id from the case class to partition.

class IdPartitioner extends HashPartitioner(numPartitions) {
override def numPartitions = numPartitions
override def getPartition(key) = {
val id = key.asInstanceOf[GKey].id
super.getPartition(id)
}
}


Now you can use that function -

// I'm sure you if you are reading this, you know how to convert
// the data to an RDD and a key-value pair.
val gRddKV = gRdd.map(toKeyValuePair)
val partitioner = new IdPartitioner(numPartitions)
val gPartitioned = gRddKV.repartitionAndSortWithinPartitions(partitioner)

Partition - 1
id time row-type points
1 9:30 b+ 10
1 10:00 a null
1 10:01 b+ 20
1 10:05 b- -20
1 10:08 b+ 30
1 10:15 a null
1 10:20 b- -30
1 10:30 b+ 40
1 10:30 b- -10
1 10:45 b- -40
Partition - 2
id time row-type points
2 9:30 b+ 50
2 10:01 a null
2 10:30 b- -50

[1] : In reality, pigeon hole principle applies. While it is true that the same key belongs to the same partition, it is also true that other keys might also be part of this partition.

Stage 3: Aggregate per partition:
The data by this stage is generalized, partitioned and sorted within that partition.

Now, follow these steps for every partition, separately:

1. Maintain a map M such that key = id and value = points. Initially, M is empty.
2. Iterate through all the time ascending sorted records.
3. For every record,
• if it is a b+ or b- record, merge it with M.
• if it is an a record, take the value of current record’s id in M and attach it to the current record.

merge. When I said merge it with M, I meant that if M already has the id from current record, add the points from the current record with the value of that id in M. If the current record is a b+ record, points is added to M but when the current record is a b- record, points is taken away from M(because it has a negative value).
If M doesn’t already have the id from current record, add a new entry in M.

merge(currentRecord: (String, Long), M: mutable.HashMap[String,Long]) = {
//extract the id and points from currentRecord
currId = currentRecord._1
//note that currPoints exists only for b+ and  b- records
//it is null otherwise
currPoints = currentRecord._2
M.get(currId) match {
case None =>
//add a new entry in the map
M += currId -> currPoints
case Some(existingValue) =>
//sum the value in map with currentRecord's value
M += currId -> (currPoints + existingValue)
}
}


The code per partition will look like -

perPartition(itr: Iterator[(GKey, GValue)]) : Option[(GKey, Long)] = {
//define map M
M = new mutable.HashMap[String,Long]()
// define an iterator that will aggregate points
// when it is a b- or b+ record, and extract those aggregated
// points if it is an a record.
new Iterator: [Option[(GKey, Long)]] {
override def hasNext = itr.hasNext

override def next() = {
val row = itr.next
val key = row._1
val value = row._2

// if current record is an a record
if (key.rowType.equals("a")) {
//get the value of id from M
val aggregatedPoints = M.getOrElse(key.id, 0)
Some((GKey,aggregatedPoints))
}
else {
// otherwise it is a b+ or b- record, merge it
// with the current record
merge((key.id, value.points), M)
//go on until you find an a record
if (itr.hasNext) {
this.next
}
else {
None
}
}
}
}


That’s it! What you get after these 3 stages is an RDD[Option[(GKey, Long)]]. What you choose to from here on its up to you.

One last note before we close, I want to talk a little about why we used an iterator.

1. perPartition() function will be called by RDD’s mapPartitions(). mapPartitions expects an iterator in return.
2. This is a perfect use case for an iterator. Once, you have final output (Gkey, Long), you don’t need it again for the records that are read after that. So, there is no need to keep track of the final output records. You could very well store the output records in a list and return an iterator of that list. But that would be a waste of memory; And you would have to deal with memory issues, JVM tuning, etc.
Partition - 1 Intermediate
id time row-type points in M
1 9:30 b+ 10 M = {1 -> 10}
1 10:00 a null copy 10 points
1 10:01 b+ 20 M = {1 -> 30}
1 10:05 b- -20 M = {1 -> 10}
1 10:08 b+ 30 M = {1 -> 40}
1 10:15 a null copy 40 points
1 10:20 b- -30 M = {1 -> 10}
1 10:30 b+ 40 M = {1 -> 50}
1 10:30 b- -10 M = {1 -> 40}
1 10:45 b- -40 M = {1 -> 0}
Partition - 1 Final
id time row-type points
1 10:00 a 10
1 10:15 a 40
Partition - 2 Intermediate
id time row-type points
2 9:30 b+ 50 M = {1 -> 50}
2 10:01 a null copy 50 points
2 10:30 b- -50 M = {1 -> 0}
Partition - 2 Final
id time row-type points
2 10:01 a 50

#### Conclusion

What we have illustrated in this post is a very simple example. In many cases, its a lot more complicated. But the if you have understood the base idea, the rest shouldn’t be too hard, just make a variant of this solution.

For example, if the points need be be grouped, rather than being aggregated, you could make a change to merge() in our code from (currPoints + existingValue) to using a HashSet/LinkedList data structure that adds points to the set/list if it is b+ and remove from the set/list if it is b- record. Or maybe you could implement merge with a function parameter with which you can control how you merge at every step.

So. Sigh! Let me know what you think of this post. Write a comment, ask a question or point out what is wrong. :)