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）。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 question，它应该可以完成这项工作。这是 Meta Numerics 相关方法的文档，我不明白。
有人可以为我提供简单的代码片段或如何使用该库的示例吗？
注意：最后，我被迫使用其中一种自定义实现。但是，如果阅读此问题的人知道良好的、有据可查的 C# 数学库/框架来执行此操作，请不要犹豫，并在答案中发布链接。
您可以将值放在同一索引处的不同列表中，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 是一个文档齐全的数学库，其中包含一个 Correlation 类。它计算 Pearson 和 Spearman 排名的相关性：http: //numerics.mathdotnet.com/api/MathNet.Numerics.Statistics/Correlation.htm
该库在非常宽松的 MIT/X11 许可下可用。使用它来计算相关系数很简单，如下所示：
using MathNet.Numerics.Statistics;
...
correlation = Correlation.Pearson(arrayOfValues1, arrayOfValues2);
祝你好运！
为了计算皮尔逊积矩相关系数
http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient
您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，可以使用这篇文章中的方法（在此处发布代码以供备份）。
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 的上述代码发布在我所做的少数手动测试中与 Microsoft Excel 的相关性通常相同，并且不需要任何外部库。
例如运行一次（此运行的数据列在底部）：
@Dmitry Bychenko：-0.00418479432051121
@keyboardP:______-0.00418479432051131
MS Excel:_________-0.004184794
这是一个测试工具：
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);
}
}
}
上述示例数字的数据：
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