I want to run logistic regression 100 times with random splitting into test and training. I want to then save the performance metrics for individual runs and then later use them for gaining insight into the performance.
for (index <- 1 to 100) {
val splits = training_data.randomSplit(Array(0.90, 0.10), seed = index)
val training = splits(0).cache()
val test = splits(1)
logrmodel = train_LogisticRegression_model(training)
performLogisticRegressionRuns(logrmodel, test, index)
}
spark.stop()
}
def performLogisticRegressionRuns(model: LogisticRegressionModel, test: RDD[LabeledPoint], iterationcount: Int) {
private val sb = StringBuilder.newBuilder
// Compute raw scores on the test set. Once I cle
model.clearThreshold()
val predictionAndLabels = test.map { case LabeledPoint(label, features) =>
val prediction = model.predict(features)
(prediction, label)
}
val bcmetrics = new BinaryClassificationMetrics(predictionAndLabels)
// I am showing two sample metrics, but I am collecting more including recall, area under roc, f1 score etc....
val precision = bcmetrics.precisionByThreshold()
precision.foreach { case (t, p) =>
// If threshold is 0.5 as what we want, then get the precision and append it to the string. Idea is if score is <0.5 class 0, else class 1.
if (t == 0.5) {
println(s"Threshold is: $t, Precision is: $p")
sb ++= p.toString() + "\t"
}
}
val auROC = bcmetrics.areaUnderROC
sb ++= iteration + auPRC.toString() + "\t"
I want to save the performance results of each iteration in separate file. I tried this, but it does not work, any help with this will be great
val data = spark.parallelize(sb)
val filename = "logreg-metrics" + iterationcount.toString() + ".txt"
data.saveAsTextFile(filename)
}
I was able to resolve this, I did the following. I converted the String to a list.
val data = spark.parallelize(List(sb))
val filename = "logreg-metrics" + iterationcount.toString() + ".txt"
data.saveAsTextFile(filename)
Related
I would like to ask you for help to identify which part of my code is not efficient. I am comparing the QuickSort algorithm with the CountingSort algorithm, assuming that the number of elements in an Array[Byte] is less than 16.
However, the CountingSort time is much higher than the QuickSort time, in all the tests I had performed sequentially. Then, I wanted to test this code in Spark to compute the Median Filter, but the results of the distributed execution times are consistent with the sequential execution times. What I mean is that QuickSort is always faster than CountingSort, even for smaller arrays.
Evidently something in my code is hanging the final processing.
This is the code:
def Histogram(Input: Array[Byte]) : Array[Int] = {
val result = Array.ofDim[Int](256)
val range = Input.distinct.map(x => x & 0xFF)
val mx = Input.map(x => x & 0xFF).max
for (h <- range)
result(h) = Input.count(x => (x & 0xFF) == h)
result.slice(0, mx + 1)
}
def CummulativeSum(Input: Array[Int]): Array[Long] = Input.map(x => x.toLong).scanLeft(0.toLong)(_ + _).drop(1)
def CountingSort(Input: Array[Byte]): Array[Byte] = {
val hist = Histogram(Input)
val cum = CummulativeSum(hist)
val Output = Array.fill[Byte](Input.length)(0)
for (i <- Input.indices) {
Output(cum(Input(i) & 0xFF).toInt - 1) = Input(i)
cum(Input(i) & 0xFF) -= 1
}
Output
}
You can build your histogram without traversing the input quite so many times.
def histogram(input :Array[Byte]) :Array[Int] = {
val inputMap :Map[Int,Array[Byte]] = input.groupBy(_ & 0xFF)
.withDefaultValue(Array())
Array.tabulate(inputMap.keys.max+1)(inputMap(_).length)
}
I'm not sure if this is much faster, but it is certainly more concise.
def countingSort(input :Array[Byte]) :Array[Byte] =
histogram(input).zipWithIndex.flatMap{case (v,x) => Seq.fill(v)(x.toByte)}
My tests show it produces the same results but there could be edge conditions that I've missed.
I'm programming a K-means algorithm in Spark-Scala.
My model predicts in which cluster is each point.
Data
-6.59 -44.68
-35.73 39.93
47.54 -52.04
23.78 46.82
....
Load the data
val data = sc.textFile("/home/borja/flink/kmeans/points")
val parsedData = data.map(s => Vectors.dense(s.split(' ').map(_.toDouble))).cache()
Cluster the data into two classes using KMeans
val numClusters = 10
val numIterations = 100
val clusters = KMeans.train(parsedData, numClusters, numIterations)
Predict
val prediction = clusters.predict(parsedData)
However, I need to put the result and the points in the same file, in the next format:
[no title, numberOfCluster (1,2,3,..10), pointX, pointY]:
6 -6.59 -44.68
8 -35.73 39.93
10 47.54 -52.04
7 23.78 46.82
This is the entry of this executable in Python to print really nice the result.
But my best effort has got just this:
(you can check the first numbers are wrong: 68, 384, ...)
var i = 0
val c = sc.parallelize(data.collect().map(x => {
val tuple = (i, x)
i += 1
tuple
}))
i = 0
val c2 = sc.parallelize(prediction.collect().map(x => {
val tuple = (i, x)
i += 1
tuple
}))
val result = c.join(c2)
result.take(5)
Result:
res94: Array[(Int, (String, Int))] = Array((68,(17.79 13.69,0)), (384,(-33.47 -4.87,8)), (440,(-4.75 -42.21,1)), (4,(-33.31 -13.11,6)), (324,(-39.04 -16.68,6)))
Thanks for your help! :)
I don't have a spark cluster handy to test, but something like this should work:
val result = parsedData.map { v =>
val cluster = clusters.predict(v)
s"$cluster ${v(0)} ${v(1)}"
}
result.saveAsTextFile("/some/output/path")
Given 2 huge list of values, I am trying to compute jaccard similarity between them in Spark using Scala.
Assume colHashed1 contains the first list of values and colHashed2 contains the second list.
Approach 1(trivial approach):
val jSimilarity = colHashed1.intersection(colHashed2).distinct.count/(colHashed1.union(colHashed2).distinct.count.toDouble)
Approach 2(using minHashing):
I have used the approach explained here.
import java.util.zip.CRC32
def getCRC32 (s : String) : Int =
{
val crc=new CRC32
crc.update(s.getBytes)
return crc.getValue.toInt & 0xffffffff
}
val maxShingleID = Math.pow(2,32)-1
def pickRandomCoeffs(kIn : Int) : Array[Int] =
{
var k = kIn
val randList = Array.fill(k){0}
while(k > 0)
{
// Get a random shingle ID.
var randIndex = (Math.random()*maxShingleID).toInt
// Ensure that each random number is unique.
while(randList.contains(randIndex))
{
randIndex = (Math.random()*maxShingleID).toInt
}
// Add the random number to the list.
k = k - 1
randList(k) = randIndex
}
return randList
}
val colHashed1 = list1Values.map(a => getCRC32(a))
val colHashed2 = list2Values.map(a => getCRC32(a))
val nextPrime = 4294967311L
val numHashes = 10
val coeffA = pickRandomCoeffs(numHashes)
val coeffB = pickRandomCoeffs(numHashes)
var signature1 = Array.fill(numHashes){0}
for (i <- 0 to numHashes-1)
{
// Evaluate the hash function.
val hashCodeRDD = colHashed1.map(ele => ((coeffA(i) * ele + coeffB(i)) % nextPrime))
// Track the lowest hash code seen.
signature1(i) = hashCodeRDD.min.toInt
}
var signature2 = Array.fill(numHashes){0}
for (i <- 0 to numHashes-1)
{
// Evaluate the hash function.
val hashCodeRDD = colHashed2.map(ele => ((coeffA(i) * ele + coeffB(i)) % nextPrime))
// Track the lowest hash code seen.
signature2(i) = hashCodeRDD.min.toInt
}
var count = 0
// Count the number of positions in the minhash signature which are equal.
for(k <- 0 to numHashes-1)
{
if(signature1(k) == signature2(k))
count = count + 1
}
val jSimilarity = count/numHashes.toDouble
Approach 1 seems to outperform Approach 2 in terms of time always. When I analyzed the code, min() function call on the RDD in Approach 2 takes significant time and that function is called many times depending upon how many hash functions are used.
The intersection and union operations used in Approach 1 seems to work faster compared to the repeated min() function calls.
I don't understand why minHashing does not help here. I expected minHashing to work faster compared to trivial approach. Is there anything I am doing wrong here?
Sample data can be viewed here
JaccardSimilarity with MinHash is not giving consistent results:
import java.util.zip.CRC32
object Jaccard {
def getCRC32(s: String): Int = {
val crc = new CRC32
crc.update(s.getBytes)
return crc.getValue.toInt & 0xffffffff
}
def pickRandomCoeffs(kIn: Int, maxShingleID: Double): Array[Int] = {
var k = kIn
val randList = Array.ofDim[Int](k)
while (k > 0) {
// Get a random shingle ID.
var randIndex = (Math.random() * maxShingleID).toInt
// Ensure that each random number is unique.
while (randList.contains(randIndex)) {
randIndex = (Math.random() * maxShingleID).toInt
}
// Add the random number to the list.
k = k - 1
randList(k) = randIndex
}
return randList
}
def approach2(list1Values: List[String], list2Values: List[String]) = {
val maxShingleID = Math.pow(2, 32) - 1
val colHashed1 = list1Values.map(a => getCRC32(a))
val colHashed2 = list2Values.map(a => getCRC32(a))
val nextPrime = 4294967311L
val numHashes = 10
val coeffA = pickRandomCoeffs(numHashes, maxShingleID)
val coeffB = pickRandomCoeffs(numHashes, maxShingleID)
val signature1 = for (i <- 0 until numHashes) yield {
val hashCodeRDD = colHashed1.map(ele => (coeffA(i) * ele + coeffB(i)) % nextPrime)
hashCodeRDD.min.toInt // Track the lowest hash code seen.
}
val signature2 = for (i <- 0 until numHashes) yield {
val hashCodeRDD = colHashed2.map(ele => (coeffA(i) * ele + coeffB(i)) % nextPrime)
hashCodeRDD.min.toInt // Track the lowest hash code seen
}
val count = (0 until numHashes)
.map(k => if (signature1(k) == signature2(k)) 1 else 0)
.fold(0)(_ + _)
val jSimilarity = count / numHashes.toDouble
jSimilarity
}
// def approach1(list1Values: List[String], list2Values: List[String]) = {
// val colHashed1 = list1Values.toSet
// val colHashed2 = list2Values.toSet
//
// val jSimilarity = colHashed1.intersection(colHashed2).distinct.count / (colHashed1.union(colHashed2).distinct.count.toDouble)
// jSimilarity
// }
def approach1(list1Values: List[String], list2Values: List[String]) = {
val colHashed1 = list1Values.toSet
val colHashed2 = list2Values.toSet
val jSimilarity = (colHashed1 & colHashed2).size / (colHashed1 ++ colHashed2).size.toDouble
jSimilarity
}
def main(args: Array[String]) {
val list1Values = List("a", "b", "c")
val list2Values = List("a", "b", "d")
for (i <- 0 until 5) {
println(s"Iteration ${i}")
println(s" - Approach 1: ${approach1(list1Values, list2Values)}")
println(s" - Approach 2: ${approach2(list1Values, list2Values)}")
}
}
}
OUTPUT:
Iteration 0
- Approach 1: 0.5
- Approach 2: 0.5
Iteration 1
- Approach 1: 0.5
- Approach 2: 0.5
Iteration 2
- Approach 1: 0.5
- Approach 2: 0.8
Iteration 3
- Approach 1: 0.5
- Approach 2: 0.8
Iteration 4
- Approach 1: 0.5
- Approach 2: 0.4
Why are you using it?
It seems to me that the overhead cost for minHashing approach just outweighs its functionality in Spark. Especially as numHashes increases.
Here are some observations I've found in your code:
First, while (randList.contains(randIndex)) this part will surely slow down your process as numHashes (which is by the way equal to the size of randList) increases.
Second, You can save some time if you rewrite this code:
var signature1 = Array.fill(numHashes){0}
for (i <- 0 to numHashes-1)
{
// Evaluate the hash function.
val hashCodeRDD = colHashed1.map(ele => ((coeffA(i) * ele + coeffB(i)) % nextPrime))
// Track the lowest hash code seen.
signature1(i) = hashCodeRDD.min.toInt
}
var signature2 = Array.fill(numHashes){0}
for (i <- 0 to numHashes-1)
{
// Evaluate the hash function.
val hashCodeRDD = colHashed2.map(ele => ((coeffA(i) * ele + coeffB(i)) % nextPrime))
// Track the lowest hash code seen.
signature2(i) = hashCodeRDD.min.toInt
}
var count = 0
// Count the number of positions in the minhash signature which are equal.
for(k <- 0 to numHashes-1)
{
if(signature1(k) == signature2(k))
count = count + 1
}
into
var count = 0
for (i <- 0 to numHashes - 1)
{
val hashCodeRDD1 = colHashed1.map(ele => ((coeffA(i) * ele + coeffB(i)) % nextPrime))
val hashCodeRDD2 = colHashed2.map(ele => ((coeffA(i) * ele + coeffB(i)) % nextPrime))
val sig1 = hashCodeRDD1.min.toInt
val sig2 = hashCodeRDD2.min.toInt
if (sig1 == sig2) { count = count + 1 }
}
This method simplifies the three loops into one. However, I am not sure if that would give a huge boost in computational time.
One other suggestion I have, assuming that the first approach still turns out to be much faster is to use the property of sets to modify the first approach:
val colHashed1_dist = colHashed1.distinct
val colHashed2_dist = colHashed2.distinct
val intersect_cnt = colHashed1_dist.intersection(colHashed2_dist).distinct.count
val jSimilarity = intersect_cnt/(colHashed1_dist.count + colHashed2_dist.count - intersect_cnt).toDouble
with that, instead of getting the union, you can just reuse the value of the intersection.
Actually, in LSH apporach you would calculate minHash only once for each of your documents and then compare two minHases for each possible pair of documents. And in case of trivial approach you would perform full comparison of documents for each possible pair of documents. Which is roughly N^2/2 number of comparisons. Hence extra cost of calculating minHashes is negligible for large enough number of documents.
You should actually compare the performance of the trivial approach:
val jSimilarity = colHashed1.intersection(colHashed2).distinct.count/(colHashed1.union(colHashed2).distinct.count.toDouble)
and performance of the Jaccard distance calculation (last lines in your code):
var count = 0
// Count the number of positions in the minhash signature which are equal.
for(k <- 0 to numHashes-1)
{
if(signature1(k) == signature2(k))
count = count + 1
}
val jSimilarity = count/numHashes.toDouble
I'm running a Bernoulli Naive Bayes using code:
val splits = MyData.randomSplit(Array(0.75, 0.25), seed = 2L)
val training = splits(0).cache()
val test = splits(1)
val model = NaiveBayes.train(training, lambda = 3.0, modelType = "bernoulli")
My question is how can I get the probability of membership to class 0 (or 1) and count AUC. I want to get similar result to LogisticRegressionWithSGD or SVMWithSGD where I was using this code:
val numIterations = 100
val model = SVMWithSGD.train(training, numIterations)
model.clearThreshold()
// Compute raw scores on the test set.
val labelAndPreds = test.map { point =>
val prediction = model.predict(point.features)
(prediction, point.label)
}
// Get evaluation metrics.
val metrics = new BinaryClassificationMetrics(labelAndPreds)
val auROC = metrics.areaUnderROC()
Unfortunately this code isn't working for NaiveBayes.
Concerning the probabilities for Bernouilli Naive Bayes, here is an example :
// Building dummy data
val data = sc.parallelize(List("0,1 0 0", "1,0 1 0", "1,0 0 1", "0,1 0 1","1,1 1 0"))
// Transforming dummy data into LabeledPoint
val parsedData = data.map { line =>
val parts = line.split(',')
LabeledPoint(parts(0).toDouble, Vectors.dense(parts(1).split(' ').map(_.toDouble)))
}
// Prepare data for training
val splits = parsedData.randomSplit(Array(0.75, 0.25), seed = 2L)
val training = splits(0).cache()
val test = splits(1)
val model = NaiveBayes.train(training, lambda = 3.0, modelType = "bernoulli")
// labels
val labels = model.labels
// Probabilities for all feature vectors
val features = parsedData.map(lp => lp.features)
model.predictProbabilities(features).take(10) foreach println
// For one specific vector, I'm taking the first vector in the parsedData
val testVector = parsedData.first.features
println(s"For vector ${testVector} => probability : ${model.predictProbabilities(testVector)}")
As for the AUC :
// Compute raw scores on the test set.
val labelAndPreds = test.map { point =>
val prediction = model.predict(point.features)
(prediction, point.label)
}
// Get evaluation metrics.
val metrics = new BinaryClassificationMetrics(labelAndPreds)
val auROC = metrics.areaUnderROC()
Concerning the inquiry from the chat :
val results = parsedData.map { lp =>
val probs: Vector = model.predictProbabilities(lp.features)
(for (i <- 0 to (probs.size - 1)) yield ((lp.label, labels(i), probs(i))))
}.flatMap(identity)
results.take(10).foreach(println)
// (0.0,0.0,0.59728640251696)
// (0.0,1.0,0.40271359748304003)
// (1.0,0.0,0.2546873180388961)
// (1.0,1.0,0.745312681961104)
// (1.0,0.0,0.47086939671877026)
// (1.0,1.0,0.5291306032812298)
// (0.0,0.0,0.6496075621805428)
// (0.0,1.0,0.3503924378194571)
// (1.0,0.0,0.4158585282373076)
// (1.0,1.0,0.5841414717626924)
and if you are only interested in the argmax classes :
val results = training.map { lp => val probs: Vector = model.predictProbabilities(lp.features)
val bestClass = probs.argmax
(labels(bestClass), probs(bestClass))
}
results.take(10) foreach println
// (0.0,0.59728640251696)
// (1.0,0.745312681961104)
// (1.0,0.5291306032812298)
// (0.0,0.6496075621805428)
// (1.0,0.5841414717626924)
Note: Works with Spark 1.5+
EDIT: (for Pyspark users)
It seems like some are having troubles getting probabilities using pyspark and mllib. Well that's normal, spark-mllib doesn't present that function for pyspark.
Thus you'll need to use the spark-ml DataFrame-based API :
from pyspark.sql import Row
from pyspark.ml.linalg import Vectors
from pyspark.ml.classification import NaiveBayes
df = spark.createDataFrame([
Row(label=0.0, features=Vectors.dense([0.0, 0.0])),
Row(label=0.0, features=Vectors.dense([0.0, 1.0])),
Row(label=1.0, features=Vectors.dense([1.0, 0.0]))])
nb = NaiveBayes(smoothing=1.0, modelType="bernoulli")
model = nb.fit(df)
model.transform(df).show(truncate=False)
# +---------+-----+-----------------------------------------+----------------------------------------+----------+
# |features |label|rawPrediction |probability |prediction|
# +---------+-----+-----------------------------------------+----------------------------------------+----------+
# |[0.0,0.0]|0.0 |[-1.4916548767777167,-2.420368128650429] |[0.7168141592920354,0.28318584070796465]|0.0 |
# |[0.0,1.0]|0.0 |[-1.4916548767777167,-3.1135153092103742]|[0.8350515463917526,0.16494845360824742]|0.0 |
# |[1.0,0.0]|1.0 |[-2.5902671654458262,-1.7272209480904837]|[0.29670329670329676,0.7032967032967034]|1.0 |
# +---------+-----+-----------------------------------------+----------------------------------------+----------+
You'll just need to select your prediction column and compute your AUC.
For more information about Naive Bayes in spark-ml, please refer to the official documentation here.
I have an rdd of (String,Int) which is sorted by key
val data = Array(("c1",6), ("c2",3),("c3",4))
val rdd = sc.parallelize(data).sortByKey
Now I want to start the value for the first key with zero and the subsequent keys as sum of the previous keys.
Eg: c1 = 0 , c2 = c1's value , c3 = (c1 value +c2 value) , c4 = (c1+..+c3 value)
expected output:
(c1,0), (c2,6), (c3,9)...
Is it possible to achieve this ?
I tried it with map but the sum is not preserved inside the map.
var sum = 0 ;
val t = keycount.map{ x => { val temp = sum; sum = sum + x._2 ; (x._1,temp); }}
Compute partial results for each partition:
val partials = rdd.mapPartitionsWithIndex((i, iter) => {
val (keys, values) = iter.toSeq.unzip
val sums = values.scanLeft(0)(_ + _)
Iterator((keys.zip(sums.tail), sums.last))
})
Collect partials sums
val partialSums = partials.values.collect
Compute cumulative sum over partitions and broadcast it:
val sumMap = sc.broadcast(
(0 until rdd.partitions.size)
.zip(partialSums.scanLeft(0)(_ + _))
.toMap
)
Compute final results:
val result = partials.keys.mapPartitionsWithIndex((i, iter) => {
val offset = sumMap.value(i)
if (iter.isEmpty) Iterator()
else iter.next.map{case (k, v) => (k, v + offset)}.toIterator
})
Spark has buit-in supports for hive ANALYTICS/WINDOWING functions and the cumulative sum could be achieved easily using ANALYTICS functions.
Hive wiki ANALYTICS/WINDOWING functions.
Example:
Assuming you have sqlContext object-
val datardd = sqlContext.sparkContext.parallelize(Seq(("a",1),("b",2), ("c",3),("d",4),("d",5),("d",6)))
import sqlContext.implicits._
//Register as test table
datardd.toDF("id","val").createOrReplaceTempView("test")
//Calculate Cumulative sum
sqlContext.sql("select id,val, " +
"SUM(val) over ( order by id rows between unbounded preceding and current row ) cumulative_Sum " +
"from test").show()
This approach cause to below warning. In case executor runs outOfMemory, tune job’s memory parameters accordingly to work with huge dataset.
WARN WindowExec: No Partition Defined for Window operation! Moving
all data to a single partition, this can cause serious performance
degradation
I hope this helps.
Here is a solution in PySpark. Internally it's essentially the same as #zero323's Scala solution, but it provides a general-purpose function with a Spark-like API.
import numpy as np
def cumsum(rdd, get_summand):
"""Given an ordered rdd of items, computes cumulative sum of
get_summand(row), where row is an item in the RDD.
"""
def cumsum_in_partition(iter_rows):
total = 0
for row in iter_rows:
total += get_summand(row)
yield (total, row)
rdd = rdd.mapPartitions(cumsum_in_partition)
def last_partition_value(iter_rows):
final = None
for cumsum, row in iter_rows:
final = cumsum
return (final,)
partition_sums = rdd.mapPartitions(last_partition_value).collect()
partition_cumsums = list(np.cumsum(partition_sums))
partition_cumsums = [0] + partition_cumsums
partition_cumsums = sc.broadcast(partition_cumsums)
def add_sums_of_previous_partitions(idx, iter_rows):
return ((cumsum + partition_cumsums.value[idx], row)
for cumsum, row in iter_rows)
rdd = rdd.mapPartitionsWithIndex(add_sums_of_previous_partitions)
return rdd
# test for correctness by summing numbers, with and without Spark
rdd = sc.range(10000,numSlices=10).sortBy(lambda x: x)
cumsums, values = zip(*cumsum(rdd,lambda x: x).collect())
assert all(cumsums == np.cumsum(values))
I came across a similar problem and implemented #Paul 's solution. I wanted to do cumsum on a integer frequency table sorted by key(the integer), and there was a minor problem with np.cumsum(partition_sums), error being unsupported operand type(s) for +=: 'int' and 'NoneType'.
Because if the range is big enough, the probability of each partition having something is thus big enough(no None values). However, if the range is much smaller than count, and number of partitions remains the same, some of the partitions would be empty. Here comes the modified solution:
def cumsum(rdd, get_summand):
"""Given an ordered rdd of items, computes cumulative sum of
get_summand(row), where row is an item in the RDD.
"""
def cumsum_in_partition(iter_rows):
total = 0
for row in iter_rows:
total += get_summand(row)
yield (total, row)
rdd = rdd.mapPartitions(cumsum_in_partition)
def last_partition_value(iter_rows):
final = None
for cumsum, row in iter_rows:
final = cumsum
return (final,)
partition_sums = rdd.mapPartitions(last_partition_value).collect()
# partition_cumsums = list(np.cumsum(partition_sums))
#----from here are the changed lines
partition_sums = [x if x is not None else 0 for x in partition_sums]
temp = np.cumsum(partition_sums)
partition_cumsums = list(temp)
#----
partition_cumsums = [0] + partition_cumsums
partition_cumsums = sc.broadcast(partition_cumsums)
def add_sums_of_previous_partitions(idx, iter_rows):
return ((cumsum + partition_cumsums.value[idx], row)
for cumsum, row in iter_rows)
rdd = rdd.mapPartitionsWithIndex(add_sums_of_previous_partitions)
return rdd
#test on random integer frequency
x = np.random.randint(10, size=1000)
D = sqlCtx.createDataFrame(pd.DataFrame(x.tolist(),columns=['D']))
c = D.groupBy('D').count().orderBy('D')
c_rdd = c.rdd.map(lambda x:x['count'])
cumsums, values = zip(*cumsum(c_rdd,lambda x: x).collect())
you can want to try out with windows over using rowsBetween. hope still helpful.
import org.apache.spark.sql.functions._
import org.apache.spark.sql.expressions.Window
val data = Array(("c1",6), ("c2",3),("c3",4))
val df = sc.parallelize(data).sortByKey().toDF("c", "v")
val w = Window.orderBy("c")
val r = df.select( $"c", sum($"v").over(w.rowsBetween(-2, -1)).alias("cs"))
display(r)