Attemping to parallelize a nested loop in Scala - scala

I am comparing 2 dataframes in scala/spark using a nested loop and an external jar.
for (nrow <- dfm.rdd.collect) {
var mid = nrow.mkString(",").split(",")(0)
var mfname = nrow.mkString(",").split(",")(1)
var mlname = nrow.mkString(",").split(",")(2)
var mlssn = nrow.mkString(",").split(",")(3)
for (drow <- dfn.rdd.collect) {
var nid = drow.mkString(",").split(",")(0)
var nfname = drow.mkString(",").split(",")(1)
var nlname = drow.mkString(",").split(",")(2)
var nlssn = drow.mkString(",").split(",")(3)
val fNameArray = Array(mfname,nfname)
val lNameArray = Array (mlname,nlname)
val ssnArray = Array (mlssn,nlssn)
val fnamescore = Main.resultSet(fNameArray)
val lnamescore = Main.resultSet(lNameArray)
val ssnscore = Main.resultSet(ssnArray)
val overallscore = (fnamescore +lnamescore +ssnscore) /3
if(overallscore >= .95) {
println("MeditechID:".concat(mid)
.concat(" MeditechFname:").concat(mfname)
.concat(" MeditechLname:").concat(mlname)
.concat(" MeditechSSN:").concat(mlssn)
.concat(" NextGenID:").concat(nid)
.concat(" NextGenFname:").concat(nfname)
.concat(" NextGenLname:").concat(nlname)
.concat(" NextGenSSN:").concat(nlssn)
.concat(" FnameScore:").concat(fnamescore.toString)
.concat(" LNameScore:").concat(lnamescore.toString)
.concat(" SSNScore:").concat(ssnscore.toString)
.concat(" OverallScore:").concat(overallscore.toString))
}
}
}
What I'm hoping to do is add some parallelism to the outer loop so that I can create a threadpool of 5 and pull 5 records from the collection of the outerloop and compare them to the collection of the inner loop, rather than doing this serially. So the outcome would be I can specify the number of threads, have 5 records from the outerloop's collection processing at any given time against the collection in the inner loop. How would I go about doing this?

Let's start by analyzing what you are doing. You collect the data of dfm to the driver. Then, for each element you collect the data from dfn, transform it and compute a score for each pair of elements.
That's problematic in many ways. First even without considering parallel computing, the transformations on the elements of dfn are made as many times as dfm as elements. Also, you collect the data of dfn for every row of dfm. That's a lot of network communications (between the driver and the executors).
If you want to use spark to parallelize you computations, you need to use the API (RDD , SQL or Datasets). You seem to want to use RDDs to perform a cartesian product (this is O(N*M) so be careful, it may take a while).
Let's start by transforming the data before the Cartesian product to avoid performing them more than once per element. Also, for clarity, let's define a case class to contain your data and a function that transform your dataframes into RDDs of that case class.
case class X(id : String, fname : String, lname : String, lssn : String)
def toRDDofX(df : DataFrame) = {
df.rdd.map(row => {
// using pattern matching to convert the array to the case class X
row.mkString(",").split(",") match {
case Array(a, b, c, d) => X(a, b, c, d)
}
})
}
Then, I use filter to keep only the tuples whose score is more than .95 but you could use map, foreach... depending on what you intend to do.
val rddn = toRDDofX(dfn)
val rddm = toRDDofX(dfm)
rddn.cartesian(rddm).filter{ case (xn, xm) => {
val fNameArray = Array(xm.fname,xn.fname)
val lNameArray = Array(xm.lname,xn.lname)
val ssnArray = Array(xm.lssn,xn.lssn)
val fnamescore = Main.resultSet(fNameArray)
val lnamescore = Main.resultSet(lNameArray)
val ssnscore = Main.resultSet(ssnArray)
val overallscore = (fnamescore +lnamescore +ssnscore) /3
// and then, let's say we filter by score
overallscore > .95
}}

This is not a right way of iterating over spark dataframe. The major concern is the dfm.rdd.collect. If the dataframe is arbitrarily large, you would end up exception. This due to the fact that the collect function essentially brings all the data into the master node.
Alternate way would be use the foreach or map construct of the rdd.
dfm.rdd.foreach(x => {
// your logic
}
Now you are trying to iterate the second dataframe here. I am afraid that won't be possible. The elegant way is to join the dfm and dfn and iterate over the resulting dataset to compute your function.

Related

Selecting every 3rd element from a huge RDD [duplicate]

I'm looking for a way to split an RDD into two or more RDDs. The closest I've seen is Scala Spark: Split collection into several RDD? which is still a single RDD.
If you're familiar with SAS, something like this:
data work.split1, work.split2;
set work.preSplit;
if (condition1)
output work.split1
else if (condition2)
output work.split2
run;
which resulted in two distinct data sets. It would have to be immediately persisted to get the results I intend...
It is not possible to yield multiple RDDs from a single transformation*. If you want to split a RDD you have to apply a filter for each split condition. For example:
def even(x): return x % 2 == 0
def odd(x): return not even(x)
rdd = sc.parallelize(range(20))
rdd_odd, rdd_even = (rdd.filter(f) for f in (odd, even))
If you have only a binary condition and computation is expensive you may prefer something like this:
kv_rdd = rdd.map(lambda x: (x, odd(x)))
kv_rdd.cache()
rdd_odd = kv_rdd.filter(lambda kv: kv[1]).keys()
rdd_even = kv_rdd.filter(lambda kv: not kv[1]).keys()
It means only a single predicate computation but requires additional pass over all data.
It is important to note that as long as an input RDD is properly cached and there no additional assumptions regarding data distribution there is no significant difference when it comes to time complexity between repeated filter and for-loop with nested if-else.
With N elements and M conditions number of operations you have to perform is clearly proportional to N times M. In case of for-loop it should be closer to (N + MN) / 2 and repeated filter is exactly NM but at the end of the day it is nothing else than O(NM). You can see my discussion** with Jason Lenderman to read about some pros-and-cons.
At the very high level you should consider two things:
Spark transformations are lazy, until you execute an action your RDD is not materialized
Why does it matter? Going back to my example:
rdd_odd, rdd_even = (rdd.filter(f) for f in (odd, even))
If later I decide that I need only rdd_odd then there is no reason to materialize rdd_even.
If you take a look at your SAS example to compute work.split2 you need to materialize both input data and work.split1.
RDDs provide a declarative API. When you use filter or map it is completely up to Spark engine how this operation is performed. As long as the functions passed to transformations are side effects free it creates multiple possibilities to optimize a whole pipeline.
At the end of the day this case is not special enough to justify its own transformation.
This map with filter pattern is actually used in a core Spark. See my answer to How does Sparks RDD.randomSplit actually split the RDD and a relevant part of the randomSplit method.
If the only goal is to achieve a split on input it is possible to use partitionBy clause for DataFrameWriter which text output format:
def makePairs(row: T): (String, String) = ???
data
.map(makePairs).toDF("key", "value")
.write.partitionBy($"key").format("text").save(...)
* There are only 3 basic types of transformations in Spark:
RDD[T] => RDD[T]
RDD[T] => RDD[U]
(RDD[T], RDD[U]) => RDD[W]
where T, U, W can be either atomic types or products / tuples (K, V). Any other operation has to be expressed using some combination of the above. You can check the original RDD paper for more details.
** https://chat.stackoverflow.com/rooms/91928/discussion-between-zero323-and-jason-lenderman
*** See also Scala Spark: Split collection into several RDD?
As other posters mentioned above, there is no single, native RDD transform that splits RDDs, but here are some "multiplex" operations that can efficiently emulate a wide variety of "splitting" on RDDs, without reading multiple times:
http://silex.freevariable.com/latest/api/#com.redhat.et.silex.rdd.multiplex.MuxRDDFunctions
Some methods specific to random splitting:
http://silex.freevariable.com/latest/api/#com.redhat.et.silex.sample.split.SplitSampleRDDFunctions
Methods are available from open source silex project:
https://github.com/willb/silex
A blog post explaining how they work:
http://erikerlandson.github.io/blog/2016/02/08/efficient-multiplexing-for-spark-rdds/
def muxPartitions[U :ClassTag](n: Int, f: (Int, Iterator[T]) => Seq[U],
persist: StorageLevel): Seq[RDD[U]] = {
val mux = self.mapPartitionsWithIndex { case (id, itr) =>
Iterator.single(f(id, itr))
}.persist(persist)
Vector.tabulate(n) { j => mux.mapPartitions { itr => Iterator.single(itr.next()(j)) } }
}
def flatMuxPartitions[U :ClassTag](n: Int, f: (Int, Iterator[T]) => Seq[TraversableOnce[U]],
persist: StorageLevel): Seq[RDD[U]] = {
val mux = self.mapPartitionsWithIndex { case (id, itr) =>
Iterator.single(f(id, itr))
}.persist(persist)
Vector.tabulate(n) { j => mux.mapPartitions { itr => itr.next()(j).toIterator } }
}
As mentioned elsewhere, these methods do involve a trade-off of memory for speed, because they operate by computing entire partition results "eagerly" instead of "lazily." Therefore, it is possible for these methods to run into memory problems on large partitions, where more traditional lazy transforms will not.
One way is to use a custom partitioner to partition the data depending upon your filter condition. This can be achieved by extending Partitioner and implementing something similar to the RangePartitioner.
A map partitions can then be used to construct multiple RDDs from the partitioned RDD without reading all the data.
val filtered = partitioned.mapPartitions { iter => {
new Iterator[Int](){
override def hasNext: Boolean = {
if(rangeOfPartitionsToKeep.contains(TaskContext.get().partitionId)) {
false
} else {
iter.hasNext
}
}
override def next():Int = iter.next()
}
Just be aware that the number of partitions in the filtered RDDs will be the same as the number in the partitioned RDD so a coalesce should be used to reduce this down and remove the empty partitions.
If you split an RDD using the randomSplit API call, you get back an array of RDDs.
If you want 5 RDDs returned, pass in 5 weight values.
e.g.
val sourceRDD = val sourceRDD = sc.parallelize(1 to 100, 4)
val seedValue = 5
val splitRDD = sourceRDD.randomSplit(Array(1.0,1.0,1.0,1.0,1.0), seedValue)
splitRDD(1).collect()
res7: Array[Int] = Array(1, 6, 11, 12, 20, 29, 40, 62, 64, 75, 77, 83, 94, 96, 100)

Collection.partition in spark [duplicate]

I'm looking for a way to split an RDD into two or more RDDs. The closest I've seen is Scala Spark: Split collection into several RDD? which is still a single RDD.
If you're familiar with SAS, something like this:
data work.split1, work.split2;
set work.preSplit;
if (condition1)
output work.split1
else if (condition2)
output work.split2
run;
which resulted in two distinct data sets. It would have to be immediately persisted to get the results I intend...
It is not possible to yield multiple RDDs from a single transformation*. If you want to split a RDD you have to apply a filter for each split condition. For example:
def even(x): return x % 2 == 0
def odd(x): return not even(x)
rdd = sc.parallelize(range(20))
rdd_odd, rdd_even = (rdd.filter(f) for f in (odd, even))
If you have only a binary condition and computation is expensive you may prefer something like this:
kv_rdd = rdd.map(lambda x: (x, odd(x)))
kv_rdd.cache()
rdd_odd = kv_rdd.filter(lambda kv: kv[1]).keys()
rdd_even = kv_rdd.filter(lambda kv: not kv[1]).keys()
It means only a single predicate computation but requires additional pass over all data.
It is important to note that as long as an input RDD is properly cached and there no additional assumptions regarding data distribution there is no significant difference when it comes to time complexity between repeated filter and for-loop with nested if-else.
With N elements and M conditions number of operations you have to perform is clearly proportional to N times M. In case of for-loop it should be closer to (N + MN) / 2 and repeated filter is exactly NM but at the end of the day it is nothing else than O(NM). You can see my discussion** with Jason Lenderman to read about some pros-and-cons.
At the very high level you should consider two things:
Spark transformations are lazy, until you execute an action your RDD is not materialized
Why does it matter? Going back to my example:
rdd_odd, rdd_even = (rdd.filter(f) for f in (odd, even))
If later I decide that I need only rdd_odd then there is no reason to materialize rdd_even.
If you take a look at your SAS example to compute work.split2 you need to materialize both input data and work.split1.
RDDs provide a declarative API. When you use filter or map it is completely up to Spark engine how this operation is performed. As long as the functions passed to transformations are side effects free it creates multiple possibilities to optimize a whole pipeline.
At the end of the day this case is not special enough to justify its own transformation.
This map with filter pattern is actually used in a core Spark. See my answer to How does Sparks RDD.randomSplit actually split the RDD and a relevant part of the randomSplit method.
If the only goal is to achieve a split on input it is possible to use partitionBy clause for DataFrameWriter which text output format:
def makePairs(row: T): (String, String) = ???
data
.map(makePairs).toDF("key", "value")
.write.partitionBy($"key").format("text").save(...)
* There are only 3 basic types of transformations in Spark:
RDD[T] => RDD[T]
RDD[T] => RDD[U]
(RDD[T], RDD[U]) => RDD[W]
where T, U, W can be either atomic types or products / tuples (K, V). Any other operation has to be expressed using some combination of the above. You can check the original RDD paper for more details.
** https://chat.stackoverflow.com/rooms/91928/discussion-between-zero323-and-jason-lenderman
*** See also Scala Spark: Split collection into several RDD?
As other posters mentioned above, there is no single, native RDD transform that splits RDDs, but here are some "multiplex" operations that can efficiently emulate a wide variety of "splitting" on RDDs, without reading multiple times:
http://silex.freevariable.com/latest/api/#com.redhat.et.silex.rdd.multiplex.MuxRDDFunctions
Some methods specific to random splitting:
http://silex.freevariable.com/latest/api/#com.redhat.et.silex.sample.split.SplitSampleRDDFunctions
Methods are available from open source silex project:
https://github.com/willb/silex
A blog post explaining how they work:
http://erikerlandson.github.io/blog/2016/02/08/efficient-multiplexing-for-spark-rdds/
def muxPartitions[U :ClassTag](n: Int, f: (Int, Iterator[T]) => Seq[U],
persist: StorageLevel): Seq[RDD[U]] = {
val mux = self.mapPartitionsWithIndex { case (id, itr) =>
Iterator.single(f(id, itr))
}.persist(persist)
Vector.tabulate(n) { j => mux.mapPartitions { itr => Iterator.single(itr.next()(j)) } }
}
def flatMuxPartitions[U :ClassTag](n: Int, f: (Int, Iterator[T]) => Seq[TraversableOnce[U]],
persist: StorageLevel): Seq[RDD[U]] = {
val mux = self.mapPartitionsWithIndex { case (id, itr) =>
Iterator.single(f(id, itr))
}.persist(persist)
Vector.tabulate(n) { j => mux.mapPartitions { itr => itr.next()(j).toIterator } }
}
As mentioned elsewhere, these methods do involve a trade-off of memory for speed, because they operate by computing entire partition results "eagerly" instead of "lazily." Therefore, it is possible for these methods to run into memory problems on large partitions, where more traditional lazy transforms will not.
One way is to use a custom partitioner to partition the data depending upon your filter condition. This can be achieved by extending Partitioner and implementing something similar to the RangePartitioner.
A map partitions can then be used to construct multiple RDDs from the partitioned RDD without reading all the data.
val filtered = partitioned.mapPartitions { iter => {
new Iterator[Int](){
override def hasNext: Boolean = {
if(rangeOfPartitionsToKeep.contains(TaskContext.get().partitionId)) {
false
} else {
iter.hasNext
}
}
override def next():Int = iter.next()
}
Just be aware that the number of partitions in the filtered RDDs will be the same as the number in the partitioned RDD so a coalesce should be used to reduce this down and remove the empty partitions.
If you split an RDD using the randomSplit API call, you get back an array of RDDs.
If you want 5 RDDs returned, pass in 5 weight values.
e.g.
val sourceRDD = val sourceRDD = sc.parallelize(1 to 100, 4)
val seedValue = 5
val splitRDD = sourceRDD.randomSplit(Array(1.0,1.0,1.0,1.0,1.0), seedValue)
splitRDD(1).collect()
res7: Array[Int] = Array(1, 6, 11, 12, 20, 29, 40, 62, 64, 75, 77, 83, 94, 96, 100)

Creating multiple RDDs out of one RDD [duplicate]

I'm looking for a way to split an RDD into two or more RDDs. The closest I've seen is Scala Spark: Split collection into several RDD? which is still a single RDD.
If you're familiar with SAS, something like this:
data work.split1, work.split2;
set work.preSplit;
if (condition1)
output work.split1
else if (condition2)
output work.split2
run;
which resulted in two distinct data sets. It would have to be immediately persisted to get the results I intend...
It is not possible to yield multiple RDDs from a single transformation*. If you want to split a RDD you have to apply a filter for each split condition. For example:
def even(x): return x % 2 == 0
def odd(x): return not even(x)
rdd = sc.parallelize(range(20))
rdd_odd, rdd_even = (rdd.filter(f) for f in (odd, even))
If you have only a binary condition and computation is expensive you may prefer something like this:
kv_rdd = rdd.map(lambda x: (x, odd(x)))
kv_rdd.cache()
rdd_odd = kv_rdd.filter(lambda kv: kv[1]).keys()
rdd_even = kv_rdd.filter(lambda kv: not kv[1]).keys()
It means only a single predicate computation but requires additional pass over all data.
It is important to note that as long as an input RDD is properly cached and there no additional assumptions regarding data distribution there is no significant difference when it comes to time complexity between repeated filter and for-loop with nested if-else.
With N elements and M conditions number of operations you have to perform is clearly proportional to N times M. In case of for-loop it should be closer to (N + MN) / 2 and repeated filter is exactly NM but at the end of the day it is nothing else than O(NM). You can see my discussion** with Jason Lenderman to read about some pros-and-cons.
At the very high level you should consider two things:
Spark transformations are lazy, until you execute an action your RDD is not materialized
Why does it matter? Going back to my example:
rdd_odd, rdd_even = (rdd.filter(f) for f in (odd, even))
If later I decide that I need only rdd_odd then there is no reason to materialize rdd_even.
If you take a look at your SAS example to compute work.split2 you need to materialize both input data and work.split1.
RDDs provide a declarative API. When you use filter or map it is completely up to Spark engine how this operation is performed. As long as the functions passed to transformations are side effects free it creates multiple possibilities to optimize a whole pipeline.
At the end of the day this case is not special enough to justify its own transformation.
This map with filter pattern is actually used in a core Spark. See my answer to How does Sparks RDD.randomSplit actually split the RDD and a relevant part of the randomSplit method.
If the only goal is to achieve a split on input it is possible to use partitionBy clause for DataFrameWriter which text output format:
def makePairs(row: T): (String, String) = ???
data
.map(makePairs).toDF("key", "value")
.write.partitionBy($"key").format("text").save(...)
* There are only 3 basic types of transformations in Spark:
RDD[T] => RDD[T]
RDD[T] => RDD[U]
(RDD[T], RDD[U]) => RDD[W]
where T, U, W can be either atomic types or products / tuples (K, V). Any other operation has to be expressed using some combination of the above. You can check the original RDD paper for more details.
** https://chat.stackoverflow.com/rooms/91928/discussion-between-zero323-and-jason-lenderman
*** See also Scala Spark: Split collection into several RDD?
As other posters mentioned above, there is no single, native RDD transform that splits RDDs, but here are some "multiplex" operations that can efficiently emulate a wide variety of "splitting" on RDDs, without reading multiple times:
http://silex.freevariable.com/latest/api/#com.redhat.et.silex.rdd.multiplex.MuxRDDFunctions
Some methods specific to random splitting:
http://silex.freevariable.com/latest/api/#com.redhat.et.silex.sample.split.SplitSampleRDDFunctions
Methods are available from open source silex project:
https://github.com/willb/silex
A blog post explaining how they work:
http://erikerlandson.github.io/blog/2016/02/08/efficient-multiplexing-for-spark-rdds/
def muxPartitions[U :ClassTag](n: Int, f: (Int, Iterator[T]) => Seq[U],
persist: StorageLevel): Seq[RDD[U]] = {
val mux = self.mapPartitionsWithIndex { case (id, itr) =>
Iterator.single(f(id, itr))
}.persist(persist)
Vector.tabulate(n) { j => mux.mapPartitions { itr => Iterator.single(itr.next()(j)) } }
}
def flatMuxPartitions[U :ClassTag](n: Int, f: (Int, Iterator[T]) => Seq[TraversableOnce[U]],
persist: StorageLevel): Seq[RDD[U]] = {
val mux = self.mapPartitionsWithIndex { case (id, itr) =>
Iterator.single(f(id, itr))
}.persist(persist)
Vector.tabulate(n) { j => mux.mapPartitions { itr => itr.next()(j).toIterator } }
}
As mentioned elsewhere, these methods do involve a trade-off of memory for speed, because they operate by computing entire partition results "eagerly" instead of "lazily." Therefore, it is possible for these methods to run into memory problems on large partitions, where more traditional lazy transforms will not.
One way is to use a custom partitioner to partition the data depending upon your filter condition. This can be achieved by extending Partitioner and implementing something similar to the RangePartitioner.
A map partitions can then be used to construct multiple RDDs from the partitioned RDD without reading all the data.
val filtered = partitioned.mapPartitions { iter => {
new Iterator[Int](){
override def hasNext: Boolean = {
if(rangeOfPartitionsToKeep.contains(TaskContext.get().partitionId)) {
false
} else {
iter.hasNext
}
}
override def next():Int = iter.next()
}
Just be aware that the number of partitions in the filtered RDDs will be the same as the number in the partitioned RDD so a coalesce should be used to reduce this down and remove the empty partitions.
If you split an RDD using the randomSplit API call, you get back an array of RDDs.
If you want 5 RDDs returned, pass in 5 weight values.
e.g.
val sourceRDD = val sourceRDD = sc.parallelize(1 to 100, 4)
val seedValue = 5
val splitRDD = sourceRDD.randomSplit(Array(1.0,1.0,1.0,1.0,1.0), seedValue)
splitRDD(1).collect()
res7: Array[Int] = Array(1, 6, 11, 12, 20, 29, 40, 62, 64, 75, 77, 83, 94, 96, 100)

How can i avoid for loop for KNN search?

My goal is to have the k nearest neighbours of each data point. I would like to avoid the use of a for loop with lookup and use something else simultaneously on each rdd_distance point, but I can't figure out how to do this.
parsedData = RDD[Object]
//Object have an id and a vector as attribute
//sqdist1 output is a Double
var rdd_distance = parsedData.cartesian(parsedData)
.flatMap { case (x,y) =>
if(x.get_id != y.get_id)
Some((x.get_id,(y.get_id,sqdist1(x.get_vector,y.get_vector))))
else None
}
for(ind1 <- 1 to size) {
val ind2 = ind1.toString
val tab1 = rdd_distance.lookup(ind2)
val rdd_knn0 = sc.parallelize(tab1)
val tab_knn = rdd_knn0.takeOrdered(k)(Ordering[(Double)].on(x=>x._2))
}
Is that possible without use a for loop with lookup ?
This code solves your question (but inefficient when the number of parsedData is big).
rdd_distance.groupByKey().map {
case (x, iterable) =>
x -> iterable.toSeq.sortBy(_._2).take(k)
}
So this is more appropriate solution.
import org.apache.spark.mllib.rdd.MLPairRDDFunctions._
rdd_distance.topByKey(k)(Ordering.by(-_._2)) // because smaller is better.
Note that this code is included Spark 1.4.0. If you use the earlier version, use this code instead https://github.com/apache/spark/blob/master/mllib/src/main/scala/org/apache/spark/mllib/rdd/MLPairRDDFunctions.scala
The idea of topBykey is to use BoundedPriorityQueue with aggregateByKey which retains top k items.

My RDD change his values himself

I have a basic RDD[Object] on which i apply a map with a hashfunction on Object values using nextGaussian and nextDouble scala function. And when i print values there change at each print
def hashmin(x:Data_Object, w:Double) = {
val x1 = x.get_vector.toArray
var a1 = Array(0.0).tail
val b = Random.nextDouble * w
for( ind <- 0 to x1.size-1) {
val nG = Random.nextGaussian
a1 = a1 :+ nG
}
var sum = 0.0
for( ind <- 0 to x1.size-1) {
sum = sum + (x1(ind)*a1(ind))
}
val hash_val = (sum+b)/w
val hash_val1 = (x.get_id,hash_val)
hash_val1
}
val w = 8
val rddhash = parsedData.map(x => hashmin(x,w))
rddhash.foreach(println)
rddhash.foreach(println)
I don't understand why. Thank you in advance.
RDDs are merely a "pointer" to the data + operations to be applied to it. Actions materialize those operations by executing the RDD lineage.
So, RDDs are basically recomputed when an action is requested. In this case, the map function calling hashmin is being evaluated every time the foreach action is called.
There're few options:
Cache the RDD - this will cause the lineage to be broken and the results of the first transformation will be preserved:
val rddhash = parsedData.map(x => hashmin(x,w)).cache()
Use a seed for your random function, sothat the pseudo-random sequence generated is each time the same.
RDDs are lazy - they're computed when they're used. So the calls to Random.nextGaussian are made again each time you call foreach.
You can use persist() to store an RDD if you want to keep fixed values.