數據集的簡單線性迴歸

Question

我期待在C＃中爲一組數據創建一個趨勢函數，看起來像使用大型數學庫對我的需求有點矯枉過正。

給出一個值列表，如6,13,7,9,12,4,2,2,1。我想得到簡單線性迴歸的斜率（看它是遞減還是遞增）和下一個估計值。我知道有大量的圖書館可以做到這一點，但我想要一個更簡單的方法。

我對統計數據不太瞭解，所以如果有人能夠以我的方式來做到這一點，將不勝感激。

Answer 1

你不需要大量的庫。公式相對簡單。

給定一對爲你的X數組和y的數據，你會計算最小二乘法擬合係數是這樣的：

http://mathworld.wolfram.com/LeastSquaresFitting.html

公式（27）和（28）你希望兩個。編碼只涉及輸入數組值的平方和和。

Answer 2

（從第一天示例第15天），爲未來的預測我自己的代碼

static void Main(string[] args) 
    { 
     double[] xVal = new double[9] 
     { 

    ... 


      }; 
     double[] yVal = new double[9] { 
    ... 

     }; 
     double rsquared; 
     double yintercept; 
     double slope; 
     LinearRegression(xVal, yVal,0,9, out rsquared, out yintercept, out slope); 
     Console.WriteLine(yintercept + (slope*15));//15 is xvalue of future(no of day from 1) 

     Console.ReadKey(); 
    } 
    public static void LinearRegression(double[] xVals, double[] yVals, 
             int inclusiveStart, int exclusiveEnd, 
             out double rsquared, out double yintercept, 
             out double slope) 
    { 
     Debug.Assert(xVals.Length == yVals.Length); 
     double sumOfX = 0; 
     double sumOfY = 0; 
     double sumOfXSq = 0; 
     double sumOfYSq = 0; 
     double ssX = 0; 
     double ssY = 0; 
     double sumCodeviates = 0; 
     double sCo = 0; 
     double count = exclusiveEnd - inclusiveStart; 

     for (int ctr = inclusiveStart; ctr < exclusiveEnd; ctr++) 
     { 
      double x = xVals[ctr]; 
      double y = yVals[ctr]; 
      sumCodeviates += x * y; 
      sumOfX += x; 
      sumOfY += y; 
      sumOfXSq += x * x; 
      sumOfYSq += y * y; 
     } 
     ssX = sumOfXSq - ((sumOfX * sumOfX)/count); 
     ssY = sumOfYSq - ((sumOfY * sumOfY)/count); 
     double RNumerator = (count * sumCodeviates) - (sumOfX * sumOfY); 
     double RDenom = (count * sumOfXSq - (sumOfX * sumOfX)) 
     * (count * sumOfYSq - (sumOfY * sumOfY)); 
     sCo = sumCodeviates - ((sumOfX * sumOfY)/count); 

     double meanX = sumOfX/count; 
     double meanY = sumOfY/count; 
     double dblR = RNumerator/Math.Sqrt(RDenom); 
     rsquared = dblR * dblR; 
     yintercept = meanY - ((sCo/ssX) * meanX); 
     slope = sCo/ssX; 
    }