Question

我正在搜索一个函数或一些代码，它们返回c中给定值的INVERSE累积正态分布。所以，如果我输入0.5，我得到0，0.157给我-1 aso。

有没有办法在c中实现它？

Answer 1

这应该可以解决问题。它的Objective-c代码但应该可以轻松转换为c。我用它进行统计计算，效果很好。

- (double)getInverseCDFValue:(double)p {

double a1 = -39.69683028665376;
double a2 = 220.9460984245205;
double a3 = -275.9285104469687;
double a4 = 138.3577518672690;
double a5 =-30.66479806614716;
double a6 = 2.506628277459239;

double b1 = -54.47609879822406;
double b2 = 161.5858368580409;
double b3 = -155.6989798598866;
double b4 = 66.80131188771972;
double b5 = -13.28068155288572;

double c1 = -0.007784894002430293;
double c2 = -0.3223964580411365;
double c3 = -2.400758277161838;
double c4 = -2.549732539343734;
double c5 = 4.374664141464968;
double c6 = 2.938163982698783;

double d1 = 0.007784695709041462;
double d2 = 0.3224671290700398;
double d3 = 2.445134137142996;
double d4 = 3.754408661907416;

//Define break-points.

double p_low =  0.02425;
double p_high = 1 - p_low;
long double  q, r, e, u;
long double x = 0.0;


//Rational approximation for lower region.

if (0 < p && p < p_low) {
    q = sqrt(-2*log(p));
    x = (((((c1*q+c2)*q+c3)*q+c4)*q+c5)*q+c6) / ((((d1*q+d2)*q+d3)*q+d4)*q+1);
}

//Rational approximation for central region.

if (p_low <= p && p <= p_high) {
    q = p - 0.5;
    r = q*q;
    x = (((((a1*r+a2)*r+a3)*r+a4)*r+a5)*r+a6)*q / (((((b1*r+b2)*r+b3)*r+b4)*r+b5)*r+1);
}

//Rational approximation for upper region.

if (p_high < p && p < 1) {
    q = sqrt(-2*log(1-p));
    x = -(((((c1*q+c2)*q+c3)*q+c4)*q+c5)*q+c6) / ((((d1*q+d2)*q+d3)*q+d4)*q+1);
}


//Pseudo-code algorithm for refinement

if(( 0 < p)&&(p < 1)){
    e = 0.5 * erfc(-x/sqrt(2)) - p;
    u = e * sqrt(2*M_PI) * exp(x*x/2);
    x = x - u/(1 + x*u/2);
}


iCFDValue = x;
return iCFDValue;
}

c中的逆累积分布函数？

1 个答案: