具有双值的两个阵列，我想计算相关系数（单，双值，就像在MS Excel中CORREL功能）。 C＃中有一些简单的单行解决方案吗？相关的两个阵列的
我已经发现了称为元Numerics的数学库。根据this SO question，它应该完成这项工作。 Here是Meta Numerics相关方法的文档，我不明白。
能请别人为我提供了简单的代码片段或例子，如何使用图书馆？
注意：最后，我不得不使用自定义实现之一。 但如果有人读这个问题知道好，有据可查的C＃ 数学库/框架，要做到这一点，请不要犹豫，在 答案张贴链接。
您可以在同一指数在单独的列表中的值，并使用一个简单的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));
第二种方式是写自己的定制实现（我是不是速度优化）：
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库，您可以使用this post中的方法（在此处发布代码进行备份）。
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));
}
为了计算皮尔逊积矩相关系数
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;
}
Math.NET Numerics的是一个包含相关类证据充分的数学库。它计算皮尔森和斯皮尔曼排名的相关性：http://numerics.mathdotnet.com/api/MathNet.Numerics.Statistics/Correlation.htm
该库可下的非常宽松的MIT/X11许可。使用它来计算相关系数非常简单，如下所示：
using MathNet.Numerics.Statistics;
...
correlation = Correlation.Pearson(arrayOfValues1, arrayOfValues2);
祝你好运！
感谢您的链接！这可能实际上是迄今为止最好的库，方法的使用真的不会更容易:-) – teejay
作为一个更新，Math.NET Numerics 3.5版添加了一种方法来相关类来计算加权皮尔逊相关性。 –
在我的测试中，@Dmitry Bychenko和@ keyboardP的上述代码发布通常与Microsoft Excel通过几次手动测试产生相同的相关性，并且不需要任何外部库。
例如运行此一次（数据为这个运行在底部列出）：
@Dmitry Bychenko：-0。00418479432051121
@keyboardP：______- 0.00418479432051131
MS Excel中：_________- 0.004184794
下面是测试线束：
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);
}
}
}
数据为上述的实施例号：
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
这也许可以帮助你也http://www.codeproject.com/Articles/8750/A-computational-statistics这是用于相关系数的代码http://www.functionx.com/vcsharp/applications/lcc.htm – terrybozzio
有一个来自http://ta-lib.org/的库，它有“CORREL”函数。它非常易于使用，并且可以为您提供与excel相同的结果。它像Excel一样返回结果数组而不是单个值。 –