python – `scipy.stat.distributions`的内置概率密度函数是否比用户提供的慢?

假设我有一个数组:adata = array([0.5,1.,2.,3.,6.,10.]),我想计算这个数组的Weibull分布的对数似然,给定参数[5., 1.5]和[5.1,1.6].我从未想过我需要为此任务编写自己的Weibull概率密度函数,因为它已在scipy.stat.distributions中提供.所以,这应该这样做:

from scipy import stats
from numpy import *
adata=array([0.5, 1.,2.,3.,6.,10.])
def wb2LL(p, x): #log-likelihood of 2 parameter Weibull distribution
    return sum(log(stats.weibull_min.pdf(x, p[1], 0., p[0])), axis=1)

和:

>>> wb2LL(array([[5.,1.5],[5.1,1.6]]).T[...,newaxis], adata)
array([-14.43743911, -14.68835298])

或者我重新发明轮子并编写一个新的Weibull pdf函数,例如:

wbp=lambda p, x: p[1]/p[0]*((x/p[0])**(p[1]-1))*exp(-((x/p[0])**p[1]))
def wb2LL1(p, x): #log-likelihood of 2 paramter Weibull
    return sum(log(wbp(p,x)), axis=1)

和:

>>> wb2LL1(array([[5.,1.5],[5.1,1.6]]).T[...,newaxis], adata)
array([-14.43743911, -14.68835298])

不可否认,我总是理所当然地认为,如果某些东西已经处于scipy状态,那么它应该得到很好的优化,并且重新发明轮子很少会让它变得更快.但令人惊讶的是:如果我计时,100万次调用wb2LL1(数组([[5.,1.5],[5.1,1.6]])[…,newaxis],adata)需要2.156s而100000次调用wb2LL (array([[5.,1.5],[5.1,1.6]])[…,newaxis],adata)需要5.219s,内置stats.weibull_min.pdf()的速度要慢两倍.

检查源代码python_path / sitepackage / scipy / stat / distributions.py并没有提供简单的答案,至少对我而言.如果有的话,我希望stats.weibull_min.pdf()几乎与wbp()一样快.

相关源代码:2999-3033行:

class frechet_r_gen(rv_continuous):
    """A Frechet right (or Weibull minimum) continuous random variable.

    %(before_notes)s

    See Also
    --------
    weibull_min : The same distribution as `frechet_r`.
    frechet_l, weibull_max

    Notes
    -----
    The probability density function for `frechet_r` is::

        frechet_r.pdf(x, c) = c * x**(c-1) * exp(-x**c)

    for ``x > 0``, ``c > 0``.

    %(example)s

    """
    def _pdf(self, x, c):
        return c*pow(x,c-1)*exp(-pow(x,c))
    def _logpdf(self, x, c):
        return log(c) + (c-1)*log(x) - pow(x,c)
    def _cdf(self, x, c):
        return -expm1(-pow(x,c))
    def _ppf(self, q, c):
        return pow(-log1p(-q),1.0/c)
    def _munp(self, n, c):
        return special.gamma(1.0+n*1.0/c)
    def _entropy(self, c):
        return -_EULER / c - log(c) + _EULER + 1
frechet_r = frechet_r_gen(a=0.0, name='frechet_r', shapes='c')
weibull_min = frechet_r_gen(a=0.0, name='weibull_min', shapes='c')

所以,问题是:stats.weibull_min.pdf()真的慢了吗?如果是这样,怎么样?

最佳答案 pdf()方法在rv_continuous类中定义,该类调用frechet_r_gen._pdf(). pdf()的代码是:

def pdf(self,x,*args,**kwds):
    loc,scale=map(kwds.get,['loc','scale'])
    args, loc, scale = self._fix_loc_scale(args, loc, scale)
    x,loc,scale = map(asarray,(x,loc,scale))
    args = tuple(map(asarray,args))
    x = asarray((x-loc)*1.0/scale)
    cond0 = self._argcheck(*args) & (scale > 0)
    cond1 = (scale > 0) & (x >= self.a) & (x <= self.b)
    cond = cond0 & cond1
    output = zeros(shape(cond),'d')
    putmask(output,(1-cond0)+np.isnan(x),self.badvalue)
    if any(cond):
        goodargs = argsreduce(cond, *((x,)+args+(scale,)))
        scale, goodargs = goodargs[-1], goodargs[:-1]
        place(output,cond,self._pdf(*goodargs) / scale)
    if output.ndim == 0:
        return output[()]
    return output

因此,它有许多参数处理代码,这使得它变慢.

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