Question

的Javascript＆GT;

如果您从事数据科学行业，如果您没有正常的分发表，您会感到困扰。我在Stackoverflow中看到了将z-score转换为JavaScript概率的文章。我真正想知道的是这个函数的反向计算。

＆＃13;

/**
 * @param {number} z - Number of standard deviations from the mean.
 */
function GetZPercent(z) {
   // If z is greater than 6.5 standard deviations from the mean
   // the number of significant digits will be outside of a reasonable 
   // range.
   if (z < -6.5)
     return 0.0;

   if (z > 6.5)
     return 1.0;

   var factK    = 1;
   var sum      = 0;
   var term     = 1;
   var k        = 0;
   var loopStop = Math.exp(-23);
   
   while (Math.abs(term) > loopStop) {
     term = 0.3989422804 * Math.pow(-1, k) * Math.pow(z, k) / (2 * k + 1) /
            Math.pow(2, k) * Math.pow(z, k + 1) / factK;
     sum += term;
     k++;
     factK *= k;
   }

   sum += 0.5;

   return sum;
 }

＆＃13;

我对如何将z-score转换为概率感到了解。但是，我不知道如何从javascript中的相应概率计算z得分（标准偏差）。例如，如果我输入0.95（或95％），我可以期望获得2.25标准差。如果输入2.25，上面的代码给出了95％。

Answer 1

我发现这段代码也有效。使用critz（p）将概率转换为z得分。例如，我们可以预期1.65来自critz（0.95），因为95％对应于z分数中的1.65标准差。

/*  The following JavaScript functions for calculating normal and
    chi-square probabilities and critical values were adapted by
    John Walker from C implementations
    written by Gary Perlman of Wang Institute, Tyngsboro, MA
    01879.  Both the original C code and this JavaScript edition
    are in the public domain.  */

/*  POZ  --  probability of normal z value

    Adapted from a polynomial approximation in:
            Ibbetson D, Algorithm 209
            Collected Algorithms of the CACM 1963 p. 616
    Note:
            This routine has six digit accuracy, so it is only useful for absolute
            z values <= 6.  For z values > to 6.0, poz() returns 0.0.
*/
    var Z_MAX = 6;
function poz(z) {

    var y, x, w;

    if (z == 0.0) {
        x = 0.0;
    } else {
        y = 0.5 * Math.abs(z);
        if (y > (Z_MAX * 0.5)) {
            x = 1.0;
        } else if (y < 1.0) {
            w = y * y;
            x = ((((((((0.000124818987 * w
                     - 0.001075204047) * w + 0.005198775019) * w
                     - 0.019198292004) * w + 0.059054035642) * w
                     - 0.151968751364) * w + 0.319152932694) * w
                     - 0.531923007300) * w + 0.797884560593) * y * 2.0;
        } else {
            y -= 2.0;
            x = (((((((((((((-0.000045255659 * y
                           + 0.000152529290) * y - 0.000019538132) * y
                           - 0.000676904986) * y + 0.001390604284) * y
                           - 0.000794620820) * y - 0.002034254874) * y
                           + 0.006549791214) * y - 0.010557625006) * y
                           + 0.011630447319) * y - 0.009279453341) * y
                           + 0.005353579108) * y - 0.002141268741) * y
                           + 0.000535310849) * y + 0.999936657524;
        }
    }
    return z > 0.0 ? ((x + 1.0) * 0.5) : ((1.0 - x) * 0.5);
}


/*  CRITZ  --  Compute critical normal z value to
               produce given p.  We just do a bisection
               search for a value within CHI_EPSILON,
               relying on the monotonicity of pochisq().  */

function critz(p) {
    var Z_EPSILON = 0.000001;     /* Accuracy of z approximation */
    var minz = -Z_MAX;
    var maxz = Z_MAX;
    var zval = 0.0;
    var pval;
    if( p < 0.0 ) p = 0.0;
    if( p > 1.0 ) p = 1.0;

    while ((maxz - minz) > Z_EPSILON) {
        pval = poz(zval);
        if (pval > p) {
            maxz = zval;
        } else {
            minz = zval;
        }
        zval = (maxz + minz) * 0.5;
    }
    return(zval);
}

Answer 2

这是一个执行相反计算的函数（z得分概率）：

function percentile_z(p) {
    var a0= 2.5066282,  a1=-18.6150006,  a2= 41.3911977,   a3=-25.4410605,
        b1=-8.4735109,  b2= 23.0833674,  b3=-21.0622410,   b4=  3.1308291,
        c0=-2.7871893,  c1= -2.2979648,  c2=  4.8501413,   c3=  2.3212128,
        d1= 3.5438892,  d2=  1.6370678, r, z;

    if (p>0.42) {
        r=Math.sqrt(-Math.log(0.5-p));
        z=(((c3*r+c2)*r+c1)*r+c0)/((d2*r+d1)*r+1);
    } else {
        r=p*p;
        z=p*(((a3*r+a2)*r+a1)*r+a0)/((((b4*r+b3)*r+b2)*r+b1)*r+1);
    }
    return z;
}

从easycalculation.com

抓取

如何将概率转换为z分数

2 个答案: