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Scalable Data Science

prepared by Raazesh Sainudiin and Sivanand Sivaram

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sds/uji/week2/02_SparkEssentials/005_RDDsTransformationsActionsHOMEWORK

HOMEWORK notebook - RDDs Transformations and Actions

Just go through the notebook and familiarize yourself with these transformations and actions.

1. Perform the takeOrdered action on the RDD

To illustrate take and takeOrdered actions, let’s create a bigger RDD named rdd0_1000000 that is made up of a million integers from 0 to 1000000.
We will sc.parallelize the Seq Scala collection by using its .range(startInteger,stopInteger) method.


val rdd0_1000000 = sc.parallelize(Seq.range(0, 1000000)) // <Shift+Enter> to create an RDD of million integers: 0,1,2,...,10^6


rdd0_1000000.take(5) // <Ctrl+Enter> gives the first 5 elements of the RDD, (0, 1, 2, 3, 4)

takeordered(n) returns n elements ordered in ascending order (by default) or as specified by the optional key function, as shown below.


rdd0_1000000.takeOrdered(5) // <Shift+Enter> is same as rdd0_1000000.take(5) 


rdd0_1000000.takeOrdered(5)(Ordering[Int].reverse) // <Ctrl+Enter> to get the last 5 elements of the RDD 999999, 999998, ..., 999995


// HOMEWORK: edit the numbers below to get the last 20 elements of an RDD made of a sequence of integers from 669966 to 969696
sc.parallelize(Seq.range(0, 10)).takeOrdered(5)(Ordering[Int].reverse) // <Ctrl+Enter> evaluate this cell after editing it for the right answer

2. More examples of map

val rdd = sc.parallelize(Seq(1, 2, 3, 4))    // <Shift+Enter> to evaluate this cell (using default number of partitions)


rdd.map( x => x*2) // <Ctrl+Enter> to transform rdd by map that doubles each element

To see what’s in the transformed RDD, let’s perform the actions of count and collect on the rdd.map( x => x*2), the transformation of rdd by the map given by the closure x => x*2.


rdd.map( x => x*2).count()    // <Shift+Enter> to perform count (action) the element of the RDD = 4


rdd.map( x => x*2).collect()    // <Shift+Enter> to perform collect (action) to show 2, 4, 6, 8


// HOMEWORK: modify the '???' in the code below to collect and display the square (x*x) of each element of the RDD
// the answer should be Array[Int] = Array(1, 4, 9, 16) Press <Cntrl+Enter> to evaluate the cell after modifying '???'
sc.parallelize(Seq(1, 2, 3, 4)).map( x => ???).collect()

3. More examples of filter

Let’s declare another val RDD named rddFiltered by transforming our first RDD named rdd via the filter transformation x%2==0 (of being even).

This filter transformation based on the closure x => x%2==0 will return true if the element, modulo two, equals zero. The closure is automatically passed on to the workers for evaluation (when an action is called later). So this will take our RDD of (1,2,3,4) and return RDD of (2, 4).


val rddFiltered = rdd.filter( x => x%2==0 )    // <Ctrl+Enter> to declare rddFiltered from transforming rdd


rddFiltered.collect()    // <Ctrl+Enter> to collect (action) elements of rddFiltered; should be (2, 4)

4. More examples of reduce

val rdd = sc.parallelize(Array(1,2,3,4,5))


rdd.reduce( (x,y)=>x+y ) // <Shift+Enter> to do reduce (action) to sum and return Int = 15


rdd.reduce( _ + _ )    // <Shift+Enter> to do same sum as above and return Int = 15 (undescore syntax)


rdd.reduce( (x,y)=>x*y ) // <Shift+Enter> to do reduce (action) to multiply and return Int = 120


val rdd0_1000000 = sc.parallelize(Seq.range(0, 1000000)) // <Shift+Enter> to create an RDD of million integers: 0,1,2,...,10^6


rdd0_1000000.reduce( (x,y)=>x+y ) // <Ctrl+Enter> to do reduce (action) to sum and return Int 1783293664


// the following correctly returns Int = 0 although for wrong reason 
// we have flowed out of Int's numeric limits!!! (but got lucky with 0*x=0 for any Int x)
// <Shift+Enter> to do reduce (action) to multiply and return Int = 0
rdd0_1000000.reduce( (x,y)=>x*y ) 


// <Ctrl+Enter> to do reduce (action) to multiply 1*2*...*9*10 and return correct answer Int = 3628800
sc.parallelize(Seq.range(1, 11)).reduce( (x,y)=>x*y ) 

CAUTION: Know the limits of your numeric types!

The minimum and maximum value of Int and Long types are as follows:


(Int.MinValue , Int.MaxValue)


(Long.MinValue, Long.MaxValue)


// <Ctrl+Enter> to do reduce (action) to multiply 1*2*...*20 and return wrong answer as Int = -2102132736
//  we have overflowed out of Int's in a circle back to negative Ints!!! (rigorous distributed numerics, anyone?)
sc.parallelize(Seq.range(1, 21)).reduce( (x,y)=>x*y ) 


//<Ctrl+Enter> we can accomplish the multiplication using Long Integer types 
// by adding 'L' ro integer values, Scala infers that it is type Long
sc.parallelize(Seq.range(1L, 21L)).reduce( (x,y)=>x*y ) 

As the following products over Long Integers indicate, they are limited too!


 // <Shift+Enter> for wrong answer Long = -8718968878589280256 (due to Long's numeric limits)
sc.parallelize(Seq.range(1L, 61L)).reduce( (x,y)=>x*y )


// <Cntrl+Enter> for wrong answer Long = 0 (due to Long's numeric limits)
sc.parallelize(Seq.range(1L, 100L)).reduce( (x,y)=>x*y ) 


5. Let us do a bunch of transformations to our RDD and perform an action
  • start from a Scala Seq,
  • sc.parallelize the list to create an RDD,
  • filter that RDD, creating a new filtered RDD,
  • do a map transformation that maps that RDD to a new mapped RDD,
  • and finally, perform a reduce action to sum the elements in the RDD.

This last reduce action causes the parallelize, the filter, and the map transformations to actually be executed, and return a result back to the driver machine.


sc.parallelize(Seq(1, 2, 3, 4))    // <Ctrl+Enter> will return Array(4, 8)
  .filter(x => x%2==0)             // (2, 4) is the filtered RDD
  .map(x => x*2)                   // (4, 8) is the mapped RDD
  .reduce(_+_)                     // 4+8=12 is the final result from reduce

6. Transform the RDD by distinct to make another RDD

Let’s declare another RDD named rdd2 that has some repeated elements to apply the distinct transformation to it. That would give us a new RDD that only contains the distinct elements of the input RDD.


val rdd2 = sc.parallelize(Seq(4, 1, 3, 2, 2, 2, 3, 4))    // <Ctrl+Enter> to declare rdd2

Let’s apply the distinct transformation to rdd2 and have it return a new RDD named rdd2Distinct that contains the distinct elements of the source RDD rdd2.


val rdd2Distinct = rdd2.distinct() // <Ctrl+Enter> transformation: distinct gives distinct elements of rdd2


rdd2Distinct.collect()    // <Ctrl+Enter> to collect (action) as Array(4, 2, 1, 3)

7. more flatMap

val rdd = sc. parallelize(Array(1,2,3)) // <Shift+Enter> to create an RDD of three Int elements 1,2,3

Let us pass the rdd above to a map with a closure that will take in each element x and return Array(x, x+5). So each element of the mapped RDD named rddOfArrays is an Array[Int], an array of integers.


// <Shift+Enter> to make RDD of Arrays, i.e., RDD[Array[int]]
val rddOfArrays = rdd.map( x => Array(x, x+5) ) 


rddOfArrays.collect() // <Ctrl+Enter> to see it is RDD[Array[int]] = (Array(1, 6), Array(2, 7), Array(3, 8))

Now let’s observer what happens when we use flatMap to transform the same rdd and create another RDD called rddfM.

Interestingly, flatMap flattens our rdd by taking each Array (or sequence in general) and truning it into individual elements.

Thus, we end up with the RDD rddfM consisting of the elements (1, 6, 2, 7, 3, 8) as shown from the output of rddfM.collect below.


val rddfM = rdd.flatMap(x => Array(x, x+5))    // <Shift+Enter> to flatMap the rdd using closure (x => Array(x, x+5))


rddfM.collect    // <Ctrl+Enter> to collect rddfM = (1, 6, 2, 7, 3, 8)

Scalable Data Science

prepared by Raazesh Sainudiin and Sivanand Sivaram

supported by and