CODEWITHSUNDEEP
scalaapache-sparkmachine-learningk-meansapache-spark-mllib
Igor Masternoy
Igor Masternoy

Reputation: 436

KMeans|| for sentiment analysis on Spark

I'm trying to write sentiment analysis program based on Spark. To do this I'm using word2vec and KMeans clustering. From word2Vec I've got 20k word/vectors collection in 100 dimension space and now I'm trying to clusterize this vectors space. When I run KMeans with default parallel implementation the algorithm worked 3 hours! But with random initialization strategy it was like 8 minutes. What am I doing wrong? I have mac book pro machine with 4 kernels processor and 16 GB of RAM.

K ~= 4000 maxInteration was 20

var vectors: Iterable[org.apache.spark.mllib.linalg.Vector] =
      model.getVectors.map(entry => new VectorWithLabel(entry._1, entry._2.map(_.toDouble)))
    val data = sc.parallelize(vectors.toIndexedSeq).persist(StorageLevel.MEMORY_ONLY_2)
    log.info("Clustering data size {}",data.count())
    log.info("==================Train process started==================");
    val clusterSize = modelSize/5

    val kmeans = new KMeans()
    kmeans.setInitializationMode(KMeans.K_MEANS_PARALLEL)
    kmeans.setK(clusterSize)
    kmeans.setRuns(1)
    kmeans.setMaxIterations(50)
    kmeans.setEpsilon(1e-4)

    time = System.currentTimeMillis()
    val clusterModel: KMeansModel = kmeans.run(data)

And spark context initialization is here:

val conf = new SparkConf()
      .setAppName("SparkPreProcessor")
      .setMaster("local[4]")
      .set("spark.default.parallelism", "8")
      .set("spark.executor.memory", "1g")
    val sc = SparkContext.getOrCreate(conf)

Also few updates about running this program. I'm running it inside Intelij IDEA. I don't have real Spark cluster. But I thought that your personal machine can be Spark cluster

I saw that the program hangs inside this loop from Spark code LocalKMeans.scala:

// Initialize centers by sampling using the k-means++ procedure.
    centers(0) = pickWeighted(rand, points, weights).toDense
    for (i <- 1 until k) {
      // Pick the next center with a probability proportional to cost under current centers
      val curCenters = centers.view.take(i)
      val sum = points.view.zip(weights).map { case (p, w) =>
        w * KMeans.pointCost(curCenters, p)
      }.sum
      val r = rand.nextDouble() * sum
      var cumulativeScore = 0.0
      var j = 0
      while (j < points.length && cumulativeScore < r) {
        cumulativeScore += weights(j) * KMeans.pointCost(curCenters, points(j))
        j += 1
      }
      if (j == 0) {
        logWarning("kMeansPlusPlus initialization ran out of distinct points for centers." +
          s" Using duplicate point for center k = $i.")
        centers(i) = points(0).toDense
      } else {
        centers(i) = points(j - 1).toDense
      }
    }

Upvotes: 2

Views: 1211

Answers (2)

Igor Masternoy
Igor Masternoy

Reputation: 436

I've run spark on AWS with 3 slaves (c3.xlarge) and the result is the same - problem is that parallel KMeans initialize algo in N parallel runs, but it's still extremely slow for small amount of data, my solution is to contionue using Random initialization. Data size approximately: 4k clusters for 21k 100-dim vectors.

Upvotes: 1

CAFEBABE
CAFEBABE

Reputation: 4101

Initialisation using KMeans.K_MEANS_PARALLEL is more complicated then random. However, it shouldn't make such a big difference. I would recommend to investigate, whether it is the parallel algorithm which takes to much time (it should actually be more efficient then KMeans itself).

For information on profiling see: http://spark.apache.org/docs/latest/monitoring.html

If it is not the initialisation which takes up the time there is something seriously wrong. However, using random initialisation shouldn't be any worse for the final result (just less efficient!).

Actually when you use KMeans.K_MEANS_PARALLEL to initialise you should get reasonable results with 0 iterations. If this is not the case there might be some regularities in the distribution of the data which send KMeans offtrack. Hence, if you haven't distributed your data randomly you could also change this. However, such an impact would surprise me give a fixed number of iterations.

Upvotes: 1

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