python - 在 python 中计算 F 分布 p 值？

Question

假设我有一个 F 值和相关联的自由度 df1 和 df2。我如何使用 python 以编程方式计算与这些数字关联的 p 值？

注意:我不会接受使用 scipy 或 statsmodels 的解决方案。

Answer 1

F 分布的 CDF(以及因此的 p 值)可以使用正则化(不完整)beta 函数 I(x; a, b) 计算，参见例如 MathWorld .使用来自此 blog 的 I(x; a, b) 的代码，它仅使用 math，p 值为

1 - incompbeta(.5*df1, .5*df2, float(df1)*F/(df1*F+df2))

这里是一些样本值的结果，匹配scipy.stats.f.sf:

In [57]: F, df1, df2 = 5, 20, 18

In [58]: 1 - incompbeta(.5*df1, .5*df2, float(df1)*F/(df1*F+df2))
Out[58]: 0.0005812207389501722

In [59]: st.f.sf(F, df1, df2)
Out[59]: 0.00058122073922042188

以防博客消失，这里是代码:

import math

def incompbeta(a, b, x):

    ''' incompbeta(a,b,x) evaluates incomplete beta function, here a, b > 0 and 0 <= x <= 1. This function requires contfractbeta(a,b,x, ITMAX = 200) 
    (Code translated from: Numerical Recipes in C.)'''

    if (x == 0):
        return 0;
    elif (x == 1):
        return 1;
    else:
        lbeta = math.lgamma(a+b) - math.lgamma(a) - math.lgamma(b) + a * math.log(x) + b * math.log(1-x)
        if (x < (a+1) / (a+b+2)):
            return math.exp(lbeta) * contfractbeta(a, b, x) / a;
        else:
            return 1 - math.exp(lbeta) * contfractbeta(b, a, 1-x) / b;

def contfractbeta(a,b,x, ITMAX = 200):

    """ contfractbeta() evaluates the continued fraction form of the incomplete Beta function; incompbeta().  
    (Code translated from: Numerical Recipes in C.)"""

    EPS = 3.0e-7
    bm = az = am = 1.0
    qab = a+b
    qap = a+1.0
    qam = a-1.0
    bz = 1.0-qab*x/qap

    for i in range(ITMAX+1):
        em = float(i+1)
        tem = em + em
        d = em*(b-em)*x/((qam+tem)*(a+tem))
        ap = az + d*am
        bp = bz+d*bm
        d = -(a+em)*(qab+em)*x/((qap+tem)*(a+tem))
        app = ap+d*az
        bpp = bp+d*bz
        aold = az
        am = ap/bpp
        bm = bp/bpp
        az = app/bpp
        bz = 1.0
        if (abs(az-aold)<(EPS*abs(az))):
            return az

    print 'a or b too large or given ITMAX too small for computing incomplete beta function.'