我想将此代码转换为Python，但在值的后半部分有很大的不同。
results是xdata中值的上下2σ置信度。也就是说，xdata应该始终落在两个相应的results值之间。
navs = {10, 10, 18, 30, 52, 87, 147, 245, 410, 684, 1141, 1903, 3173, 5290, 8816};
freqs = {0.00002, 0.00004, 0.0000666667, 0.000111111, 0.000185185, 0.000308642, 0.000514403, 0.000857339, 0.00142893, 0.00238166, 0.00396944, 0.00661594, 0.0165426, 0.0220568, 0.027571}
xdata = {0.578064980346793, 0.030812200935204, 0.316777979844816,
0.353718150091612, 0.287659600326548, 0.269254388840293,
0.16545714457921, 0.138759871084825, 0.0602382519940077,
0.10120771961, 0.065311134782518, 0.105235790998594,
0.124642033979457, 0.0271909963701794, 0.0686653810421847};
data = MapThread[{#1, #2, #3} &, {navs, freqs, xdata}]
post[x_, n_, y_] =
(n - 1) (1 - x)^n (1 - y)^(n - 2) Hypergeometric2F1[n, n, 1, x*y]
integral = Map[(values = #; mesh = Subdivide[0, 1, 1000];
Interpolation[
DeleteDuplicates[{Map[
SetPrecision[post[#, values[[1]], values[[3]]^2], 100] &,
mesh] // (Accumulate[#] - #/2 - #[[1]]/
2) & // #/#[[-1]] &,
mesh}\[Transpose], (#1[[1]] == #2[[1]] &)],
InterpolationOrder -> 1]) &, data];
results =
MapThread[{Sqrt[#1[.025]], Sqrt[#1[0.975]]} &, {integral, data}]
{{0.207919, 0.776508}, {0.0481485, 0.535278}, {0.0834002, 0.574447},
{0.137742, 0.551035}, {0.121376, 0.455097}, {0.136889, 0.403306},
{0.0674029, 0.279408}, {0.0612534, 0.228762}, {0.0158357, 0.134521},
{0.0525374, 0.156055}, {0.0270589, 0.108861}, {0.0740978, 0.137691},
{0.100498, 0.149646}, {0.00741129, 0.0525161}, {0.0507748, 0.0850961}}
这是我的翻译：{{1}}与以前一样，被截断为第7位以提高可读性。
我开始从第7对值开始偏离results值，而results的最后四个点不落在两个相应的{{1 }}值。
xdata由于对这个问题的评论，我发现：
results给出了import numpy as np
from scipy.integrate import cumtrapz
from scipy.interpolate import interp1d
from mpmath import *
mesh = list(np.linspace(0,1,1000));
navs = [10, 10, 18, 30, 52, 87, 147, 245, 410, 684, 1141, 1903, 3173, 5290, 8816]
freqs = [0.00002, 0.00004, 0.0000666667, 0.000111111, 0.000185185, 0.000308642, 0.000514403, 0.000857339, 0.00142893, 0.00238166, 0.00396944, 0.00661594, 0.0165426, 0.0220568, 0.027571]
xdata = [0.578064980346793, 0.030812200935204, 0.316777979844816,
0.353718150091612,0.287659600326548, 0.269254388840293,
0.16545714457921, 0.138759871084825, 0.0602382519940077,
0.10120771961, 0.065311134782518, 0.105235790998594,
0.124642033979457, 0.0271909963701794, 0.0686653810421847]
def post(x,n,y):
post = (n-1)*((1-x)**n)*((1-y)**(n-2))*hyp2f1(n,n,1,x*y)
return post
# setting the numeric precision to 100 as in Mathematica
# trying to get the most precise hypergeometric function values
mp.dps = 100
mp.pretty = True
results = []
for i in range(len(navs)):
postprob = [];
for j in range(len(mesh)):
posterior = post(mesh[j], navs[i], xdata[i]**2)
postprob.append(posterior)
# calculate the norm of the pdf for integration
norm = np.trapz(np.array(postprob),mesh);
# integrate pdf/norm to obtain cdf
integrate = list(np.unique(cumtrapz(np.array(postprob)/norm, mesh, initial=0)));
mesh2 = list(np.linspace(0,1,len(integrate)));
# interpolate inverse cdf to obtain the 2sigma quantiles
icdf = interp1d(integrate, mesh2, bounds_error=False, fill_value='extrapolate');
results.append(list(np.sqrt(icdf([0.025, 0.975]))))
results
[[0.2079198, 0.7765088], [0.0481485, 0.5352773], [0.0834, 0.5744489],
[0.1377413, 0.5510352], [0.1218029, 0.4566994], [0.1399324, 0.4122767],
[0.0733743, 0.3041607], [0.0739691, 0.2762597], [0.0230135, 0.1954886],
[0.0871462, 0.2588804], [0.05637, 0.2268962], [0.1731199, 0.3217401],
[0.2665897, 0.3969059], [0.0315915, 0.2238736], [0.2224567, 0.3728803]]
的结果，而在Python中，Hypergeometric2F1给出了1.0588267的结果。这是网格的第二个点，其差值在小数点后第五位。对于某个我找不到的特殊功能有更好的定义吗？
（我是Python的新手，也许代码有点天真）