SQL Server中的BetaInv函数
类似于问题： MySQL中的反转beta我需要在SQL Server存储过程中使用BetaInv函数。
函数在这里描述： Excel的BETAINV
是否有人知道TSQL中的任何类似内容，或者您将它包装在CLR .NET托管SQL用户定义函数中？
我确实需要在存储过程中使用它，而不是在使用存储过程检索数据后在C＃端执行代码，因为我应该将所有逻辑保留在数据库服务器上以便更好地重用。
我可以假设在SQL Server中运行的.NET托管udf的执行速度与普通的本机TSQL函数一样快吗？
谢谢！
我最终自己实现了整个function，这里有源代码以防有人需要它：
public static class UDFs { private const int MAXIT = 100; private const double EPS = 0.0000003; private const double FPMIN = 1.0E-30; [SqlFunction(Name = "BetaInv", DataAccess = DataAccessKind.Read)] public static SqlDouble BetaInv(SqlDouble p, SqlDouble alpha, SqlDouble beta, SqlDouble A, SqlDouble B) { return InverseBeta(p.Value, alpha.Value, beta.Value, A.Value, B.Value); } private static double InverseBeta(double p, double alpha, double beta, double A, double B) { double x = 0; double a = 0; double b = 1; double precision = Math.Pow(10, -6); // converge until there is 6 decimal places precision while ((b - a) > precision) { x = (a + b) / 2; if (IncompleteBetaFunction(x, alpha, beta) > p) { b = x; } else { a = x; } } if ((B > 0) && (A > 0)) { x = x * (B - A) + A; } return x; } private static double IncompleteBetaFunction(double x, double a, double b) { double bt = 0; if (x <= 0.0) { return 0; } if (x >= 1) { return 1; } bt = System.Math.Exp(Gammln(a + b) - Gammln(a) - Gammln(b) + a * System.Math.Log(x) + b * System.Math.Log(1.0 - x)); if (x < ((a + 1.0) / (a + b + 2.0))) { // Use continued fraction directly. return (bt * betacf(a, b, x) / a); } else { // Use continued fraction after making the symmetry transformation. return (1.0 - bt * betacf(b, a, 1.0 - x) / b); } } private static double betacf(double a, double b, double x) { int m, m2; double aa, c, d, del, h, qab, qam, qap; qab = a + b; // These q's will be used in factors that occur in the coe.cients (6.4.6). qap = a + 1.0; qam = a - 1.0; c = 1.0; // First step of Lentz's method. d = 1.0 - qab * x / qap; if (System.Math.Abs(d) < FPMIN) { d = FPMIN; } d = 1.0 / d; h = d; for (m = 1; m <= MAXIT; ++m) { m2 = 2 * m; aa = m * (b - m) * x / ((qam + m2) * (a + m2)); d = 1.0 + aa * d; //One step (the even one) of the recurrence. if (System.Math.Abs(d) < FPMIN) { d = FPMIN; } c = 1.0 + aa / c; if (System.Math.Abs(c) < FPMIN) { c = FPMIN; } d = 1.0 / d; h *= d * c; aa = -(a + m) * (qab + m) * x / ((a + m2) * (qap + m2)); d = 1.0 + aa * d; // Next step of the recurrence (the odd one). if (System.Math.Abs(d) < FPMIN) { d = FPMIN; } c = 1.0 + aa / c; if (System.Math.Abs(c) < FPMIN) { c = FPMIN; } d = 1.0 / d; del = d * c; h *= del; if (System.Math.Abs(del - 1.0) < EPS) { // Are we done? break; } } if (m > MAXIT) { return 0; } else { return h; } } public static double Gammln(double xx) { double x, y, tmp, ser; double[] cof = new double[] { 76.180091729471457, -86.505320329416776, 24.014098240830911, -1.231739572450155, 0.001208650973866179, -0.000005395239384953 }; y = xx; x = xx; tmp = x + 5.5; tmp -= (x + 0.5) * System.Math.Log(tmp); ser = 1.0000000001900149; for (int j = 0; j <= 5; ++j) { y += 1; ser += cof[j] / y; } return -tmp + System.Math.Log(2.5066282746310007 * ser / x); } } } 正如您在代码中看到的那样，SqlFunction正在调用InverseBeta私有方法，该方法使用其他几种方法完成工作。
结果与Excel.BetaInv相同，逗号后最多5或6位数。