C#中两个数组的相关性

本文关键字：两个 数组 相关性 | 更新日期: 2023-09-27 18:21:01

有两个双值数组，我想计算相关系数（单双值，就像MS Excel中的CORREL函数一样）。C#中有一些简单的单行解决方案吗？

我已经发现了名为Meta Numerics的数学库。根据这个SO问题，它应该做这项工作。这是元数据相关方法的文档，我不知道。

请有人给我一个简单的代码片段或如何使用这个库的例子好吗？

注意：最后，我不得不使用一个自定义实现。但是，如果读过这个问题的人知道好的、有充分记录的C#要做到这一点，请不要犹豫，并在中发布链接答复

C#中两个数组的相关性

您可以在同一索引的单独列表中拥有值，并使用简单的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));

第二种方法是编写自己的自定义实现（我的没有针对速度进行优化）：

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;
}

用法：

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");

另一种方法是直接使用Excel函数：

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

Math.NET Numerics是一个文档丰富的数学库，包含Correlation类。它计算了Pearson和Spearman排名的相关性：http://numerics.mathdotnet.com/api/MathNet.Numerics.Statistics/Correlation.htm

该库可以在非常自由的MIT/X11许可证下使用。使用它来计算相关系数如下所示：

using MathNet.Numerics.Statistics;
...
correlation = Correlation.Pearson(arrayOfValues1, arrayOfValues2);

祝你好运！

为了计算Pearson乘积矩相关系数

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

你可以使用这个简单的代码：

  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; 
  }

如果您不想使用第三方库，您可以使用本文中的方法（此处发布代码用于备份）。

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));
}

在我的测试中，@Dmitry Bychenko和@keyboardP在上面发布的代码在我所做的一些手动测试中产生了与Microsoft Excel大致相同的相关性，并且不需要任何外部库。

例如，运行一次（底部列出了此运行的数据）：

@德米特里·拜琴科：-0.00418479432051121

@键盘p:_____-0.00418479432051131

MS Excel：__________________-0.004184794

这是一个测试线束：

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);
        }
    }
}

以上示例数字的数据：

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

Public Function Correlation(ByRef array1() As Double, ByRef array2() As Double) As Double
    'siehe https://stackoverflow.com/questions/17447817/correlation-of-two-arrays-in-c-sharp
    'der hier errechnete "Pearson correlation coefficient" muss noch quadriert werden, um R-Squared zu erhalten, siehe
    'https://en.wikipedia.org/wiki/Coefficient_of_determination

    Dim array_xy(array1.Length - 1) As Double
    Dim array_xp2(array1.Length - 1) As Double
    Dim array_yp2(array1.Length - 1) As Double
    Dim i As Integer
    For i = 0 To array1.Length - 1
        array_xy(i) = array1(i) * array2(i)
    Next i
    For i = 0 To array1.Length - 1
        array_xp2(i) = Math.Pow(array1(i), 2.0)
    Next i
    For i = 0 To array1.Length - 1
        array_yp2(i) = Math.Pow(array2(i), 2.0)
    Next i

    Dim sum_x As Double = 0
    Dim sum_y As Double = 0
    Dim EinDouble As Double
    For Each EinDouble In array1
        sum_x += EinDouble
    Next
    For Each EinDouble In array2
        sum_y += EinDouble
    Next
    Dim sum_xy As Double = 0
    For Each EinDouble In array_xy
        sum_xy += EinDouble
    Next
    Dim sum_xpow2 As Double = 0
    For Each EinDouble In array_xp2
        sum_xpow2 += EinDouble
    Next
    Dim sum_ypow2 As Double = 0
    For Each EinDouble In array_yp2
        sum_ypow2 += EinDouble
    Next
    Dim Ex2 As Double = Math.Pow(sum_x, 2.0)
    Dim Ey2 As Double = Math.Pow(sum_y, 2.0)
    Dim ReturnWert As Double
    ReturnWert = (array1.Length * sum_xy - sum_x * sum_y) / Math.Sqrt((array1.Length * sum_xpow2 - Ex2) * (array1.Length * sum_ypow2 - Ey2))
    Correlation = ReturnWert
End Function
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