CODEWITHSUNDEEP
pythonpandasscalanumpypercentile
Georg Heiler
Georg Heiler

Reputation: 17676

percentiles pandas vs. scala where is the bug?

For a list of numbers

val numbers = Seq(0.0817381355303346, 0.08907955219917718, 0.10581384008994665, 0.10970915785902469, 0.1530743353025532, 0.16728932033107657, 0.181932212814931, 0.23200826752868853, 0.2339654613723784, 0.2581657775305527, 0.3481071101229365, 0.5010850992326521, 0.6153244818101578, 0.6233250409474894, 0.6797744231690304, 0.6923891392381571, 0.7440316016776881, 0.7593186414698002, 0.8028091068764153, 0.8780699052482807, 0.8966649331194205)

python / pandas computes the following percentiles:

25%     0.167289
50%     0.348107
75%     0.692389

However, scala returns:

calcPercentiles(Seq(.25, .5, .75), sortedNumber.toArray)

25% 0.1601818278168149
50% 0.3481071101229365
75% 0.7182103704579226

The numbers are almost matching - but different. How can I get rid of the difference (and most likely fix a bug in my scala code?

val sortedNumber = numbers.sorted

import scala.collection.mutable
case class PercentileResult(percentile:Double, value:Double)

// https://github.com/scalanlp/breeze/blob/master/math/src/main/scala/breeze/stats/DescriptiveStats.scala#L537
def calculatePercentile(arr: Array[Double], p: Double)={
    // +1 so that the .5 == mean for even number of elements.
    val f = (arr.length + 1) * p
    val i = f.toInt
    if (i == 0) arr.head
    else if (i >= arr.length) arr.last
    else {
      arr(i - 1) + (f - i) * (arr(i) - arr(i - 1))
    }
  }

 def calcPercentiles(percentiles:Seq[Double], arr: Array[Double]):Array[PercentileResult] = {
    val results = new mutable.ListBuffer[PercentileResult]
    percentiles.foreach(p => {
      val r = PercentileResult(percentile = p, value = calculatePercentile(arr, p))
      results.append(r)
    })
    results.toArray
  }

python:

 import pandas as pd

df = pd.DataFrame({'foo':[0.0817381355303346, 0.08907955219917718, 0.10581384008994665, 0.10970915785902469, 0.1530743353025532, 0.16728932033107657, 0.181932212814931, 0.23200826752868853, 0.2339654613723784, 0.2581657775305527, 0.3481071101229365, 0.5010850992326521, 0.6153244818101578, 0.6233250409474894, 0.6797744231690304, 0.6923891392381571, 0.7440316016776881, 0.7593186414698002, 0.8028091068764153, 0.8780699052482807, 0.8966649331194205]})
display(df.head())
df.describe()

Upvotes: 1

Views: 167

Answers (1)

Mateusz Kubuszok
Mateusz Kubuszok

Reputation: 27535

After a bit trial and error I write this code that returns the same results as Panda (using linear interpolation as this is pandas default):

def calculatePercentile(numbers: Seq[Double], p: Double): Double = {
  // interpolate only - no special handling of the case when rank is integer
  val rank = (numbers.size - 1) * p
  val i = numbers(math.floor(rank).toInt)
  val j = numbers(math.ceil(rank).toInt)
  val fraction = rank - math.floor(rank)
  i + (j - i) * fraction
}

From that I would say that the errors was here:

(arr.length + 1) * p

Percentile of 0 should be 0, and percentile at 100% should be a maximal index.

So for numbers (.size == 21) that would be indices 0 and 20. However, for 100% you would get index value of 22 - bigger than the size of array! If not for these guard clauses:

else if (i >= arr.length) arr.last

you would get error and you could suspect that something is wrong. Perhaps authors of the code:

https://github.com/scalanlp/breeze/blob/master/math/src/main/scala/breeze/stats/DescriptiveStats.scala#L537

used a different definition of percentile... (?) or they might simply have a bug. I cannot tell.

BTW: This:

def calcPercentiles(percentiles:Seq[Double], arr: Array[Double]): Array[PercentileResult]

could be much easier to write like this:

def calcPercentiles(percentiles:Seq[Double], numbers: Seq[Double]): Seq[PercentileResult] =
  percentiles.map { p =>
    PercentileResult(p, calculatePercentile(numbers, p))
  }

Upvotes: 1

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