#region License Information
/* HeuristicLab
* Copyright (C) 2002-2012 Heuristic and Evolutionary Algorithms Laboratory (HEAL)
*
* This file is part of HeuristicLab.
*
* HeuristicLab is free software: you can redistribute it and/or modify
* it under the terms of the GNU General Public License as published by
* the Free Software Foundation, either version 3 of the License, or
* (at your option) any later version.
*
* HeuristicLab is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU General Public License for more details.
*
* You should have received a copy of the GNU General Public License
* along with HeuristicLab. If not, see .
*/
#endregion
using System;
using System.Collections.Generic;
using System.Linq;
namespace HeuristicLab.Problems.DataAnalysis {
public class HoeffdingsDependenceCalculator : IDependencyCalculator {
public double Maximum { get { return 1.0; } }
public double Minimum { get { return -0.5; } }
public string Name { get { return "Hoeffdings Dependence"; } }
public double Calculate(IEnumerable originalValues, IEnumerable estimatedValues, out OnlineCalculatorError errorState) {
return HoeffdingsDependenceCalculator.CalculateHoeffdings(originalValues, estimatedValues, out errorState);
}
public static double CalculateHoeffdings(IEnumerable originalValues, IEnumerable estimatedValues, out OnlineCalculatorError errorState) {
double d = HoeffD(originalValues, estimatedValues, out errorState);
if (errorState != OnlineCalculatorError.None) return double.NaN;
return d;
}
///
/// computes Hoeffding's dependence coefficient.
/// Source: hoeffd.r from R package hmisc http://cran.r-project.org/web/packages/Hmisc/index.html
///
private static double HoeffD(IEnumerable xs, IEnumerable ys, out OnlineCalculatorError errorState) {
double[] rx = TiedRank(xs);
double[] ry = TiedRank(ys);
if (rx.Length != ry.Length) throw new ArgumentException("The number of elements in xs and ys does not match");
double[] rxy = TiedRank(xs, ys);
int n = rx.Length;
double q = 0, r = 0, s = 0;
double scaling = 1.0 / (n * (n - 1));
for (int i = 0; i < n; i++) {
q += (rx[i] - 1) * (rx[i] - 2) * (ry[i] - 1) * (ry[i] - 2) * scaling;
r += (rx[i] - 2) * (ry[i] - 2) * rxy[i] * scaling;
s += rxy[i] * (rxy[i] - 1) * scaling;
}
errorState = OnlineCalculatorError.None;
// return 30.0 * (q - 2 * (n - 2) * r + (n - 2) * (n - 3) * s) / n / (n - 1) / (n - 2) / (n - 3) / (n - 4);
double t0 = q / (n - 2) / (n - 3) / (n - 4);
double t1 = 2 * r / (n - 3) / (n - 4);
double t2 = s / (n - 4);
return 30.0 * (t0 - t1 + t2);
}
private static double[] TiedRank(IEnumerable xs) {
var xsArr = xs.ToArray();
var idx = Enumerable.Range(1, xsArr.Length).ToArray();
Array.Sort(xsArr, idx);
CRank(xsArr);
Array.Sort(idx, xsArr);
return xsArr;
}
///
/// Calculates the joint rank with midranks for ties. Source: hoeffd.r from R package hmisc http://cran.r-project.org/web/packages/Hmisc/index.html
///
///
///
///
private static double[] TiedRank(IEnumerable xs, IEnumerable ys) {
var xsArr = xs.ToArray();
var ysArr = ys.ToArray();
var r = new double[xsArr.Length];
int n = r.Length;
for (int i = 0; i < n; i++) {
var xi = xsArr[i];
var yi = ysArr[i];
double ri = 0.0;
for (int j = 0; j < n; j++) {
if (i != j) {
double cx;
if (xsArr[j] < xi) cx = 1.0;
else if (xsArr[j] > xi) cx = 0.0;
else cx = 0.5; // eq
double cy;
if (ysArr[j] < yi) cy = 1.0;
else if (ysArr[j] > yi) cy = 0.0;
else cy = 0.5; // eq
ri = ri + cx * cy;
}
}
r[i] = ri;
}
return r;
}
///
/// Calculates midranks. Source: Numerical Recipes in C. p 642
///
/// Sorted array of elements, replaces the elements by their rank, including midranking of ties
///
private static void CRank(double[] w) {
int i = 0;
int n = w.Length;
while (i < n - 1) {
if (w[i + 1] > w[i]) { // w[i+1] must be larger or equal w[i] as w must be sorted
// not a tie
w[i] = i + 1;
i++;
} else {
int j;
for (j = i + 1; j < n && w[j] <= w[i]; j++) ; // how far does it go (<= effectively means == as w must be sorted, side-step equality for double values)
double rank = 1 + 0.5 * (i + j - 1);
int k;
for (k = i; k < j; k++) w[k] = rank; // set the rank for all tied entries
i = j;
}
}
if (i == n - 1) w[n - 1] = n; // if the last element was not tied, this is its rank
}
}
}