这几天做一个数值的线性回归分析，用.net实现了几个有关于计算线性回归的几个函数
分位值函数PERCENTILE
算法说明：
//Excel中 percentile(array,p)算法是：
//将数组array从小到大排序，计算(n-1)*p的整数部分为i，小数部分为j，其中n为数组大小，
//则percentile的值是：(1-j)*array第i+1个数+j*array第i+2个数。
//例如：{1，3，4，5，6，7，8，9，19，29，39，49，59，69，79，80}计算30%的分位数：
//(16-1)*30%=4.5, i= 4, j =0.5
//percentile(a,30%)=(1-0.5)*6+0.5*7=6.5
//ref http://office.microsoft.com/zh-cn/excel-help/HP005209211.aspx
/// <summary>
/// Excel中PERCENTILE的算法，求分位值
/// </summary>
/// <param name="array">数据数组</param>
/// <param name="p">要求的分位值的值</param>
/// <returns></returns>
public static double PERCENTILE(ArrayList array,double p)
{
array.Sort(); //从小到大排序
double d = (array.Count - 1) * p;
string s = d.ToString("F");
int iLen = s.LastIndexOf(".");
int i=int.Parse(s.Substring(0, iLen));
string sValue = "0." + s.Substring(iLen + 1, s.Length-iLen-1);
double j=Double.Parse(sValue);
double fristNumber = Double.Parse(array[i].ToString());
double nextNumber = Double.Parse(array[i+1].ToString());
double valueNumber = (1 - j) * fristNumber + j * nextNumber;
return Math.Round(valueNumber,4);
}
/// <summary>
/// 求一个数的对数，保留6位小数
/// </summary>
/// <param name="d"></param>
/// <returns></returns>
public static double LN(double d)
{
return Math.Round(Math.Log(d),6);
}
/// <summary>
/// Excel中的SLOPE函数,求斜率，保留6位小数
/// </summary>
/// <param name="yArray">y点的数据数组</param>
/// <param name="xArray">x点的数据数组</param>
/// <returns></returns>
public static double SLOPE(double[] yArray, double[] xArray)
{
//算法说明
//先求xArray和yArray的分别平均值
//然后
if (xArray == null)
throw new ArgumentNullException("xArray");
if (yArray == null)
throw new ArgumentNullException("yArray");
if (xArray.Length != yArray.Length)
throw new ArgumentException("Array Length Mismatch");
if (xArray.Length < 2)
throw new ArgumentException("Arrays too short.");
double n = xArray.Length;
double sumxy = 0;
double sumx = 0;
double sumy = 0;

double sumx2 = 0;
for (int i = 0; i < xArray.Length; i++)
{
sumx += xArray[i];
sumy += yArray[i];
}
////求平均值
sumx = sumx / n;
sumy = sumy / n;
for (int i = 0; i < xArray.Length; i++)
{
sumxy += (xArray[i]-sumx) * (yArray[i]-sumy);
sumx2 += (xArray[i]-sumx) * (xArray[i]-sumx);
}
return Math.Round(sumxy / sumx2, 6);
}
/// <summary>
/// Excel中的INTERCEPT函数,求截距，保留6位小数
/// </summary>
/// <param name="yArray">y点的数据数组</param>
/// <param name="xArray">x点的数据数组</param>
/// <returns></returns>
public static double INTERCEPT(double[] yArray, double[] xArray)
{
if (xArray == null)
throw new ArgumentNullException("xArray");
if (yArray == null)
throw new ArgumentNullException("yArray");
if (xArray.Length != yArray.Length)
throw new ArgumentException("Array Length Mismatch");
if (xArray.Length < 2)
throw new ArgumentException("Arrays too short.");
double n = xArray.Length;
double sumxy = 0;
double sumx = 0;
double sumy = 0;
double sumx2 = 0;
for (int i = 0; i < xArray.Length; i++)
{
sumxy += xArray[i] * yArray[i];
sumx += xArray[i];
sumy += yArray[i];
sumx2 += xArray[i] * xArray[i];
}
double k = Math.Round(((sumxy - sumx * sumy / n) / (sumx2 - sumx * sumx / n)), 6); //斜率
//直线方程：y=kx+b b=y-kx
return Math.Round((sumy - k * sumx) / n,6);
}
/// <summary>
/// Excel中的EXP函数,求e的n次幂，保留6位小数
/// </summary>
/// <param name="value"></param>
/// <returns></returns>
public static double EXP(double value)
{
return Math.Round(Math.Exp(value), 6);
}