Calculating pearson for Breeze vectors

pearson.scala

import breeze.linalg._
import breeze.stats._
import scala.math.sqrt


/**
 * Effecient for sparse vectors. Scales in O(activeSize)
 */
// Must take SparseVector, for implicits to be linked correctly
def pearson(a: SparseVector[Double], b: SparseVector[Double]): Double = {
  if (a.length != b.length)
    throw new IllegalArgumentException("Vectors not of the same length.")

  val n = a.length

  val dot = a.dot(b)
  val adot = a.dot(a)
  val bdot = b.dot(b)
  val amean = mean(a)
  val bmean = mean(b)

  // See Wikipedia http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient#For_a_sample
  (dot - n * amean * bmean ) / ( sqrt(adot - n * amean * amean)  * sqrt(bdot - n * bmean * bmean) )
}

/**
 * Works for all all vectors. Scales in O(length)
 */
def pearson(a: Vector[Double], b: Vector[Double]): Double = {
  // Delegate to efficient method if possible,
  if (a.isInstanceOf[SparseVector] && b.isInstanceOf[SparseVector]) {
    return pearson(a.asInstanceOf[SparseVector[Double]], b.asInstanceOf[SparseVector[Double]])
  }

  if (a.length != b.length)
    throw new IllegalArgumentException("Vectors not of the same length.")

  val n = a.length

  val dot = a.dot(b)
  val adot = a.dot(a)
  val bdot = b.dot(b)
  val amean = mean(a)
  val bmean = mean(b)

  // See Wikipedia http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient#For_a_sample
  (dot - n * amean * bmean ) / ( sqrt(adot - n * amean * amean)  * sqrt(bdot - n * bmean * bmean) )
}

The efficient method can be several thousand times faster. For two vectors with approximately 1000 non-zero elements, this is the difference:

Length	Effecient method	Inefficient method
1 048 576	<0.005 s	0.22 s
4 194 304	<0.005 s	1.05 s
16 777 216	<0.005 s	2.86 s
67 108 864	~0.01 s	Throws OutOfMemory