def lerfcc(x):
    z = abs(x)
    t = 1.0 / (1.0+0.5*z)
    ans = t * math.exp(-z*z-1.26551223 + t*(1.00002368+t*(0.37409196+t*(0.09678418+t*(-0.18628806+t*(0.27886807+t*(-1.13520398+t*(1.48851587+t*(-0.82215223+t*0.17087277)))))))))
    if x >= 0:
        return ans
    else:
        return 2.0 - ans
Returns the complementary error function erfc(x) with fractional
dataset/ETHPy150Open radlab/sparrow/src/main/python/third_party/stats.py/lerfcc