具有雙值的兩個陣列，我想計算相關係數（單，雙值，就像在MS Excel中CORREL功能）。 C＃中有一些簡單的單行解決方案嗎？相關的兩個陣列的
我已經發現了稱爲元Numerics的數學庫。根據this SO question，它應該完成這項工作。 Here是Meta Numerics相關方法的文檔，我不明白。
能請別人爲我提供了簡單的代碼片段或例子，如何使用圖書館？
注意：最後，我不得不使用自定義實現之一。 但如果有人讀這個問題知道好，有據可查的C＃ 數學庫/框架，要做到這一點，請不要猶豫，在 答案張貼鏈接。
您可以在同一指數在單獨的列表中的值，並使用一個簡單的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));
第二種方式是寫自己的定製實現（我是不是速度優化）：
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");
另一種方法是使用Excel直接功能：
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
如果您不想使用第三方庫，您可以使用this post中的方法（在此處發佈代碼進行備份）。
double[] array1 = { 3, 2, 4, 5, 6 };
double[] array2 = { 9, 7, 12, 15, 17 };
double correl = Correlation(array1, array2);
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));
}
爲了計算皮爾遜積矩相關係數
http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient
你可以使用這個簡單的代碼：
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;
}
Math.NET Numerics的是一個包含相關類證據充分的數學庫。它計算皮爾森和斯皮爾曼排名的相關性：http://numerics.mathdotnet.com/api/MathNet.Numerics.Statistics/Correlation.htm
該庫可下的非常寬鬆的MIT/X11許可。使用它來計算相關係數非常簡單，如下所示：
using MathNet.Numerics.Statistics;
...
correlation = Correlation.Pearson(arrayOfValues1, arrayOfValues2);
祝你好運！
感謝您的鏈接！這可能實際上是迄今爲止最好的庫，方法的使用真的不會更容易:-) – teejay
作爲一個更新，Math.NET Numerics 3.5版添加了一種方法來相關類來計算加權皮爾遜相關性。 –
在我的測試中，@Dmitry Bychenko和@ keyboardP的上述代碼發佈通常與Microsoft Excel通過幾次手動測試產生相同的相關性，並且不需要任何外部庫。
例如運行此一次（數據爲這個運行在底部列出）：
@Dmitry Bychenko：-0。00418479432051121
@keyboardP：______- 0.00418479432051131
MS Excel中：_________- 0.004184794
下面是測試線束：
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);
}
}
}
數據爲上述的實施例號：
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
這也許可以幫助你也http://www.codeproject.com/Articles/8750/A-computational-statistics這是用於相關係數的代碼http://www.functionx.com/vcsharp/applications/lcc.htm – terrybozzio
有一個來自http://ta-lib.org/的庫，它有「CORREL」函數。它非常易於使用，並且可以爲您提供與excel相同的結果。它像Excel一樣返回結果數組而不是單個值。 –