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Corrélation de deux tableaux en C #

Ayant deux tableaux de valeurs doubles, je veux calculer le coefficient de corrélation (valeur double simple, tout comme la fonction CORREL dans MS Excel). Existe-t-il une solution simple en ligne en C #?

J'ai déjà découvert une bibliothèque de mathématiques appelée Meta Numerics. Selon cette SO question , il devrait faire le travail. Ici est des documents pour la méthode de corrélation Meta Numerics, que je ne reçois pas .

Quelqu'un pourrait-il me fournir un extrait de code simple ou un exemple comment utiliser la bibliothèque?

Remarque: à la fin, j'ai été obligé d'utiliser l'une des implémentations personnalisées. Mais si quelqu'un qui lit cette question connaît bien la bibliothèque/framework C # math bien documenté pour le faire, n'hésitez pas et postez un lien en réponse.

c#.netarrayscorrelation
19
teejay

Vous pouvez avoir les valeurs dans des listes séparées au même index et utiliser un simple Zip .

var fitResult = new FitResult();
var values1 = new List<int>();
var values2 = new List<int>();

var correls = values1.Zip(values2, (v1, v2) =>
                                       fitResult.CorrelationCoefficient(v1, v2));

Une deuxième façon consiste à écrire votre propre implémentation personnalisée (la mienne n'est pas optimisée pour la vitesse):

public double ComputeCoeff(double[] values1, double[] values2)
{
    if(values1.Length != values2.Length)
        throw new ArgumentException("values must be the same length");

    var avg1 = values1.Average();
    var avg2 = values2.Average();

    var sum1 = values1.Zip(values2, (x1, y1) => (x1 - avg1) * (y1 - avg2)).Sum();

    var sumSqr1 = values1.Sum(x => Math.Pow((x - avg1), 2.0));
    var sumSqr2 = values2.Sum(y => Math.Pow((y - avg2), 2.0));

    var result = sum1 / Math.Sqrt(sumSqr1 * sumSqr2);

    return result;
}

Usage:

var values1 = new List<double> { 3, 2, 4, 5 ,6 };
var values2 = new List<double> { 9, 7, 12 ,15, 17 };

var result = ComputeCoeff(values1.ToArray(), values2.ToArray());
// 0.997054485501581

Debug.Assert(result.ToString("F6") == "0.997054");

Une autre façon consiste à utiliser directement la fonction Excel:

var values1 = new List<double> { 3, 2, 4, 5 ,6 };
var values2 = new List<double> { 9, 7, 12 ,15, 17 };

// Make sure to add a reference to Microsoft.Office.Interop.Excel.dll
// and use the namespace

var application = new Application();

var worksheetFunction = application.WorksheetFunction;

var result = worksheetFunction.Correl(values1.ToArray(), values2.ToArray());

Console.Write(result); // 0.997054485501581
28
Dustin Kingen

Math.NET Numerics est une bibliothèque mathématique bien documentée qui contient une classe de corrélation. Il calcule les corrélations classées par Pearson et Spearman: http://numerics.mathdotnet.com/api/MathNet.Numerics.Statistics/Correlation.htm

La bibliothèque est disponible sous la licence très libérale MIT/X11. Son utilisation pour calculer un coefficient de corrélation est aussi simple que suit:

using MathNet.Numerics.Statistics;

...

correlation = Correlation.Pearson(arrayOfValues1, arrayOfValues2);

Bonne chance!

20
Ruben Ramirez Padron

Afin de calculer le coefficient de corrélation produit-moment de Pearson

http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient

Vous pouvez utiliser ce code simple:

  public static Double Correlation(Double[] Xs, Double[] Ys) {
    Double sumX = 0;
    Double sumX2 = 0;
    Double sumY = 0;
    Double sumY2 = 0;
    Double sumXY = 0;

    int n = Xs.Length < Ys.Length ? Xs.Length : Ys.Length;

    for (int i = 0; i < n; ++i) {
      Double x = Xs[i];
      Double y = Ys[i];

      sumX += x;
      sumX2 += x * x;
      sumY += y;
      sumY2 += y * y;
      sumXY += x * y;
    }

    Double stdX = Math.Sqrt(sumX2 / n - sumX * sumX / n / n);
    Double stdY = Math.Sqrt(sumY2 / n - sumY * sumY / n / n);
    Double covariance = (sumXY / n - sumX * sumY / n / n);

    return covariance / stdX / stdY; 
  }
8
Dmitry Bychenko

Si vous ne souhaitez pas utiliser une bibliothèque tierce, vous pouvez utiliser la méthode de cet article (code de publication ici pour la sauvegarde).

public double Correlation(double[] array1, double[] array2)
{
    double[] array_xy = new double[array1.Length];
    double[] array_xp2 = new double[array1.Length];
    double[] array_yp2 = new double[array1.Length];
    for (int i = 0; i < array1.Length; i++)
    array_xy[i] = array1[i] * array2[i];
    for (int i = 0; i < array1.Length; i++)
    array_xp2[i] = Math.Pow(array1[i], 2.0);
    for (int i = 0; i < array1.Length; i++)
    array_yp2[i] = Math.Pow(array2[i], 2.0);
    double sum_x = 0;
    double sum_y = 0;
    foreach (double n in array1)
        sum_x += n;
    foreach (double n in array2)
        sum_y += n;
    double sum_xy = 0;
    foreach (double n in array_xy)
        sum_xy += n;
    double sum_xpow2 = 0;
    foreach (double n in array_xp2)
        sum_xpow2 += n;
    double sum_ypow2 = 0;
    foreach (double n in array_yp2)
        sum_ypow2 += n;
    double Ex2 = Math.Pow(sum_x, 2.00);
    double Ey2 = Math.Pow(sum_y, 2.00);

    return (array1.Length * sum_xy - sum_x * sum_y) /
           Math.Sqrt((array1.Length * sum_xpow2 - Ex2) * (array1.Length * sum_ypow2 - Ey2));
}
6
keyboardP

Lors de mes tests, les publications de code @Dmitry Bychenko et @ keyboardP ci-dessus ont généralement donné les mêmes corrélations que Microsoft Excel sur une poignée de tests manuels que j'ai faits, et n'ont pas eu besoin de bibliothèques externes.

par exemple. Exécuter cette fois (les données de cette analyse sont répertoriées en bas):

@Dmitry Bychenko: -0.00418479432051121

@keyboardP: ______- 0,00418479432051131

MS Excel: _________- 0,004184794

Voici un harnais de test:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace TestCorrel {
    class Program {

        static void Main(string[] args) {

            Random Rand = new Random(DateTime.Now.Millisecond);

            List<double> x = new List<double>();
            List<double> y = new List<double>();

            for (int i = 0; i < 100; i++) {

                x.Add(Rand.Next(1000) * Rand.NextDouble());
                y.Add(Rand.Next(1000) * Rand.NextDouble());

                Console.WriteLine(x[i] + "," + y[i]);
            }

            Console.WriteLine("Correl1: " + Correl1(x, y));
            Console.WriteLine("Correl2: " + Correl2(x, y));
        }

        public static double Correl1(List<double> x, List<double> y) {

            //https://stackoverflow.com/questions/17447817/correlation-of-two-arrays-in-c-sharp
            if (x.Count != y.Count)
                return (double.NaN); //throw new ArgumentException("values must be the same length");

            double sumX = 0;
            double sumX2 = 0;
            double sumY = 0;
            double sumY2 = 0;
            double sumXY = 0;

            int n = x.Count < y.Count ? x.Count : y.Count;

            for (int i = 0; i < n; ++i) {

                Double xval = x[i];
                Double yval = y[i];

                sumX += xval;
                sumX2 += xval * xval;
                sumY += yval;
                sumY2 += yval * yval;
                sumXY += xval * yval;
            }

            Double stdX = Math.Sqrt(sumX2 / n - sumX * sumX / n / n);
            Double stdY = Math.Sqrt(sumY2 / n - sumY * sumY / n / n);
            Double covariance = (sumXY / n - sumX * sumY / n / n);

            return covariance / stdX / stdY;
        }

        public static double Correl2(List<double> x, List<double> y) {

            double[] array_xy = new double[x.Count];
            double[] array_xp2 = new double[x.Count];
            double[] array_yp2 = new double[x.Count];

            for (int i = 0; i < x.Count; i++)
                array_xy[i] = x[i] * y[i];
            for (int i = 0; i < x.Count; i++)
                array_xp2[i] = Math.Pow(x[i], 2.0);
            for (int i = 0; i < x.Count; i++)
                array_yp2[i] = Math.Pow(y[i], 2.0);
            double sum_x = 0;
            double sum_y = 0;
            foreach (double n in x)
                sum_x += n;
            foreach (double n in y)
                sum_y += n;
            double sum_xy = 0;
            foreach (double n in array_xy)
                sum_xy += n;
            double sum_xpow2 = 0;
            foreach (double n in array_xp2)
                sum_xpow2 += n;
            double sum_ypow2 = 0;
            foreach (double n in array_yp2)
                sum_ypow2 += n;
            double Ex2 = Math.Pow(sum_x, 2.00);
            double Ey2 = Math.Pow(sum_y, 2.00);

            double Correl = 
            (x.Count * sum_xy - sum_x * sum_y) /
            Math.Sqrt((x.Count * sum_xpow2 - Ex2) * (x.Count * sum_ypow2 - Ey2));

            return (Correl);
        }
    }
}

Données pour les numéros d'exemple ci-dessus:

287.688269702572,225.610842817282
618.9313498167,177.955550192835
25.7778882802361,27.6549569366756
140.847984766051,714.618547504125
438.618761728806,533.48764902702
481.347431274758,214.381256273194
21.6406916848573,393.559209519792
135.30397563209,158.419851317732
334.314685154853,814.275162949821
764.614904770914,50.1435267264692
42.8179292282173,47.8631582287434
237.216836650491,370.488416981179
388.849658539449,134.961087643151
305.903013161804,441.926902444068
10.6625048679591,369.567569480076
36.9316453891488,24.8947204607049
2.10067253471383,491.941975629861
7.94887068492774,573.037801189831
341.738006353722,653.497146697015
98.8424873439793,475.215988045193
272.248712629196,36.1088809138671
122.336823399801,169.158256422336
9.32281673202422,631.076001565473
201.118425176068,803.724831627554
415.514343714115,64.248651454341
227.791637123,230.512133914284
25.3438658925443,396.854282886188
596.238994411304,72.543763144195
230.239735877253,933.983901697669
796.060099040186,689.952468971234
9.30882684202344,269.22063744125
16.5005430148451,8.96549091859045
536.324005148524,358.829873788557
519.694526420764,17.3212184707267
552.628357889423,12.5541588051962
210.516099897454,388.57537739937
141.341571405689,268.082028986924
503.880356335491,753.447006912645
515.494990213539,444.451280259737
973.8670776076,168.922799013985
85.7111146094795,36.3784999169309
37.2147129193017,108.040356312432
504.590177939548,50.3934166889607
482.821039277511,888.984586256083
5.52549206350255,156.717087003271
405.833169031345,394.099059180868
459.249365587835,11.68776424494
429.421127440604,314.216759666901
126.908422469584,331.907062556551
62.1416232716952,3.19765723645578
4.16058817699579,604.04046284223
484.262182311277,220.177370167886
58.6774453314382,339.09660232677
463.482149892246,199.181594849183
344.128297473829,268.531428258182
0.883430369609702,209.346384477963
77.9462970131758,255.221325168955
583.629439312792,235.557751925922
358.409186083083,376.046612200349
81.2148325150902,10.7696774717279
53.7315618049966,274.171515094196
111.284646992239,130.174321939319
317.280491961763,338.077288461885
177.454564264722,7.53587801919127
69.2239431670047,233.693477620228
823.419546454875,0.111916855029723
23.7174749401014,200.989081544331
44.9598299125022,102.633862571155
74.1602278468945,292.485449988155
130.11182449251,23.4682153367755
243.088760058903,335.807090202722
13.3974915991526,436.983231269281
73.3900805168739,252.352352472186
592.144630201228,92.3395205570103
57.7306153447044,47.1416798900541
522.649018382024,584.427794722108
15.3662010204821,60.1693953262499
16.8335716728277,851.401980430541
33.9869734449251,0.930781653584345
116.66608504982,146.126050951949
92.8896130355492,711.765618208687
317.91980889529,322.186540377413
44.8574470732629,209.275617858058
751.201537871362,37.935519233316
161.817758424588,2.83156183493862
531.64078452142,79.1750782491523
114.803219681048,283.106988439852
123.472725123853,154.125248027558
89.9276725453919,63.4626924192825
105.623296753328,111.234188702067
435.72981759707,23.7058234576629
259.324810619152,69.3535200857341
719.885234421531,381.086239833891
24.2674900099018,198.408173349876
57.7761600361095,146.52277489124
77.4594609157459,710.746080866431
636.671781979814,538.894185951396
56.6035279932448,58.2563265684323
485.16099039333,427.849954283261
91.9552873247095,576.92944263617
2
Mike