在python中計算F分布p值？

Question

假設我有一個F值和相關的自由度，df1和df2。 如何使用python以編程方式計算與這些數字相關的p值？

注意：我不接受使用scipy或statsmodels的解決方案。

Answer 1

F分布的CDF（以及因此p值）可以用正則化（不完全）β函數I(x; a, b) ，參見例如MathWorld 。 使用此博客的 I(x; a, b)代碼，該代碼僅使用math ，p值為

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

這里是一些樣本值的結果，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

為了防止博客消失，這里的代碼如下：

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.'