Spark column wise word count

Question

We are trying to generate column wise statistics of our dataset in spark. In addition to using the summary function from statistics library. We are using the following procedure:

1. We determine the columns with string values

2. Generate key value pair for the whole dataset, using the column number as key and value of column as value

3. generate a new map of format

   (K,V) ->((K,V),1)

Then we use reduceByKey to find the sum of all unique value in all the columns. We cache this output to reduce further computation time.

In the next step we cycle through the columns using a for loop to find the statistics for all the columns.

We are trying to reduce the for loop by again utilizing the map reduce way but we are unable to find some way to achieve it. Doing so will allow us to generate column statistics for all columns in one execution. The for loop method is running sequentially making it very slow.

Code:

//drops the header

    def dropHeader(data: RDD[String]): RDD[String] = {
         data.mapPartitionsWithIndex((idx, lines) => {
           if (idx == 0) {
             lines.drop(1)
           }
           lines
         })
       }

    def retAtrTuple(x: String) = {
       val newX = x.split(",")
       for (h <- 0 until newX.length) 
          yield (h,newX(h))
    }



    val line = sc.textFile("hdfs://.../myfile.csv")

    val withoutHeader: RDD[String] = dropHeader(line)

    val kvPairs = withoutHeader.flatMap(retAtrTuple) //generates a key-value pair where key is the column number and value is column's value


    var bool_numeric_col = kvPairs.map{case (x,y) => (x,isNumeric(y))}.reduceByKey(_&&_).sortByKey()    //this contains column indexes as key and boolean as value (true for numeric and false for string type)

    var str_cols = bool_numeric_col.filter{case (x,y) => y == false}.map{case (x,y) => x}
    var num_cols = bool_numeric_col.filter{case (x,y) => y == true}.map{case (x,y) => x}

    var str_col = str_cols.toArray   //array consisting the string col
    var num_col = num_cols.toArray   //array consisting numeric col


    val colCount = kvPairs.map((_,1)).reduceByKey(_+_)
    val e1 = colCount.map{case ((x,y),z) => (x,(y,z))}
    var numPairs = e1.filter{case (x,(y,z)) => str_col.contains(x) }

    //running for loops which needs to be parallelized/optimized as it sequentially operates on each column. Idea is to find the top10, bottom10 and number of distinct elements column wise
    for(i <- str_col){
       var total = numPairs.filter{case (x,(y,z)) => x==i}.sortBy(_._2._2)
       var leastOnes = total.take(10)
       println("leastOnes for Col" + i)
       leastOnes.foreach(println)
       var maxOnes = total.sortBy(-_._2._2).take(10)
       println("maxOnes for Col" + i)
       maxOnes.foreach(println)
       println("distinct for Col" + i + " is " + total.count)
    }

Answer 1

Let me simplify your question a bit. (A lot actually.) We have an RDD[(Int, String)] and we want to find the top 10 most common Strings for each Int (which are all in the 0–100 range).

Instead of sorting, as in your example, it is more efficient to use the Spark built-in RDD.top(n) method. Its run-time is linear in the size of the data, and requires moving much less data around than a sort.

Consider the implementation of top in RDD.scala. You want to do the same, but with one priority queue (heap) per Int key. The code becomes fairly complex:

import org.apache.spark.util.BoundedPriorityQueue // Pretend it's not private.

def top(n: Int, rdd: RDD[(Int, String)]): Map[Int, Iterable[String]] = {
  // A heap that only keeps the top N values, so it has bounded size.
  type Heap = BoundedPriorityQueue[(Long, String)]
  // Get the word counts.
  val counts: RDD[[(Int, String), Long)] =
    rdd.map(_ -> 1L).reduceByKey(_ + _)
  // In each partition create a column -> heap map.
  val perPartition: RDD[Map[Int, Heap]] =
    counts.mapPartitions { items =>
      val heaps =
        collection.mutable.Map[Int, Heap].withDefault(i => new Heap(n))
      for (((k, v), count) <- items) {
        heaps(k) += count -> v
      }
      Iterator.single(heaps)
    }
  // Merge the per-partition heap maps into one.
  val merged: Map[Int, Heap] =
    perPartition.reduce { (heaps1, heaps2) =>
      val heaps =
        collection.mutable.Map[Int, Heap].withDefault(i => new Heap(n))
      for ((k, heap) <- heaps1.toSeq ++ heaps2.toSeq) {
        for (cv <- heap) {
          heaps(k) += cv
        }
      }
      heaps
    }
  // Discard counts, return just the top strings.
  merged.mapValues(_.map { case(count, value) => value })
}

This is efficient, but made painful because we need to work with multiple columns at the same time. It would be way easier to have one RDD per column and just call rdd.top(10) on each.

Unfortunately the naive way to split up the RDD into N smaller RDDs does N passes:

def split(together: RDD[(Int, String)], columns: Int): Seq[RDD[String]] = {
  together.cache // We will make N passes over this RDD.
  (0 until columns).map {
    i => together.filter { case (key, value) => key == i }.values
  }
}

A more efficient solution could be to write out the data into separate files by key, then load it back into separate RDDs. This is discussed in Write to multiple outputs by key Spark - one Spark job.

Answer 2

Thanks for @Daniel Darabos's answer. But there are some mistakes.

1. mixed use of Map and collection.mutable.Map

2. withDefault((i: Int) => new Heap(n)) do not create a new Heap when you set heaps(k) += count -> v

3. mix uasage of parentheses

Here is the modified code:

//import org.apache.spark.util.BoundedPriorityQueue // Pretend it's not private. copy to your own folder and import it
import org.apache.log4j.{Level, Logger}
import org.apache.spark.rdd.RDD
import org.apache.spark.{SparkConf, SparkContext}


object BoundedPriorityQueueTest {

  //  https://stackoverflow.com/questions/28166190/spark-column-wise-word-count
  def top(n: Int, rdd: RDD[(Int, String)]): Map[Int, Iterable[String]] = {
    // A heap that only keeps the top N values, so it has bounded size.
    type Heap = BoundedPriorityQueue[(Long, String)]
    // Get the word counts.
    val counts: RDD[((Int, String), Long)] =
    rdd.map(_ -> 1L).reduceByKey(_ + _)
    // In each partition create a column -> heap map.
    val perPartition: RDD[collection.mutable.Map[Int, Heap]] =
    counts.mapPartitions { items =>
      val heaps =
        collection.mutable.Map[Int, Heap]() // .withDefault((i: Int) => new Heap(n))
      for (((k, v), count) <- items) {
        println("\n---")
        println("before add " + ((k, v), count) + ", the map is: ")
        println(heaps)
        if (!heaps.contains(k)) {
          println("not contains key " + k)
          heaps(k) = new Heap(n)
          println(heaps)
        }
        heaps(k) += count -> v
        println("after add " + ((k, v), count) + ", the map is: ")
        println(heaps)

      }
      println(heaps)
      Iterator.single(heaps)
    }
    // Merge the per-partition heap maps into one.
    val merged: collection.mutable.Map[Int, Heap] =
    perPartition.reduce { (heaps1, heaps2) =>
      val heaps =
        collection.mutable.Map[Int, Heap]() //.withDefault((i: Int) => new Heap(n))
      println(heaps)
      for ((k, heap) <- heaps1.toSeq ++ heaps2.toSeq) {
        for (cv <- heap) {
          heaps(k) += cv
        }
      }
      heaps
    }
    // Discard counts, return just the top strings.
    merged.mapValues(_.map { case (count, value) => value }).toMap
  }

  def main(args: Array[String]): Unit = {
    Logger.getRootLogger().setLevel(Level.FATAL) //http://stackoverflow.com/questions/27781187/how-to-stop-messages-displaying-on-spark-console
    val conf = new SparkConf().setAppName("word count").setMaster("local[1]")
    val sc = new SparkContext(conf)
    sc.setLogLevel("WARN") //http://stackoverflow.com/questions/27781187/how-to-stop-messages-displaying-on-spark-console


    val words = sc.parallelize(List((1, "s11"), (1, "s11"), (1, "s12"), (1, "s13"), (2, "s21"), (2, "s22"), (2, "s22"), (2, "s23")))
    println("# words：" + words.count())

    val result = top(1, words)

    println("\n--result:")
    println(result)
    sc.stop()

    print("DONE")
  }

}