我能够自己实现误差函数erf,但我不希望这样做。有没有一个Python包含有这个函数的实现且不需要外部依赖?我找到了这个,但它似乎是某个较大包的一部分(而且不清楚是哪个!)。
我建议使用SciPy进行Python中的数值函数,但如果您想要没有依赖项的内容,这里有一个函数,其误差小于所有输入的1.5 * 10-7。
def erf(x):
# save the sign of x
sign = 1 if x >= 0 else -1
x = abs(x)
# constants
a1 = 0.254829592
a2 = -0.284496736
a3 = 1.421413741
a4 = -1.453152027
a5 = 1.061405429
p = 0.3275911
# A&S formula 7.1.26
t = 1.0/(1.0 + p*x)
y = 1.0 - (((((a5*t + a4)*t) + a3)*t + a2)*t + a1)*t*math.exp(-x*x)
return sign*y # erf(-x) = -erf(x)
该算法来自于数学函数手册,公式7.1.26。
我建议您下载NumPy(以在Python中拥有高效的矩阵)和SciPy(Matlab工具箱替代品,使用NumPy)。 erf函数位于SciPy中。
>>>from scipy.special import erf
>>>help(erf)
你也可以使用在pylab中定义的erf函数,但这更适用于绘制使用numpy和scipy计算出来的结果。如果你想要一个包含这些软件的完整安装包,你可以直接使用Python Enthought distribution。
可以在 mpmath 模块中找到纯 Python 实现(http://code.google.com/p/mpmath/)
从文档字符串中:
>>> from mpmath import *
>>> mp.dps = 15
>>> print erf(0)
0.0
>>> print erf(1)
0.842700792949715
>>> print erf(-1)
-0.842700792949715
>>> print erf(inf)
1.0
>>> print erf(-inf)
-1.0
对于较大的实数x
,\mathrm{erf}(x)
会非常迅速地接近1:
>>> print erf(3)
0.999977909503001
>>> print erf(5)
0.999999999998463
>>> nprint(chop(taylor(erf, 0, 5)))
[0.0, 1.12838, 0.0, -0.376126, 0.0, 0.112838]
:func:erf
实现任意精度计算并支持复数:
>>> mp.dps = 50
>>> print erf(0.5)
0.52049987781304653768274665389196452873645157575796
>>> mp.dps = 25
>>> print erf(1+j)
(1.316151281697947644880271 + 0.1904534692378346862841089j)
相关函数
另请参阅 :func:erfc
,它对于大的x
更加精确,
以及 :func:erfi
,它给出了\exp(t^2)
的反导数。
Fresnel积分 :func:fresnels
和 :func:fresnelc
也与误差函数相关。
# from: http://www.cs.princeton.edu/introcs/21function/ErrorFunction.java.html
# Implements the Gauss error function.
# erf(z) = 2 / sqrt(pi) * integral(exp(-t*t), t = 0..z)
#
# fractional error in math formula less than 1.2 * 10 ^ -7.
# although subject to catastrophic cancellation when z in very close to 0
# from Chebyshev fitting formula for erf(z) from Numerical Recipes, 6.2
def erf(z):
t = 1.0 / (1.0 + 0.5 * abs(z))
# use Horner's method
ans = 1 - 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 z >= 0.0:
return ans
else:
return -ans
from math import erf
(为了可移植性、准确性、速度等)。 - smcierf
函数的?通过以下设置:from scipy.special import erf; import numpy as np; data = np.random.randn(10e5)
,我从result = erf(data)
中获得非常快的运行时间。特别是在这种情况下,每个循环需要32毫秒。如果我天真地循环遍历numpy
数组中的所有元素,唯一会导致运行时间> 1秒的方法就是这样。 - 8one6对于那些追求更高性能的人,有一个提示:如果可能的话,进行向量化。
import numpy as np
from scipy.special import erf
def vectorized(n):
x = np.random.randn(n)
return erf(x)
def loopstyle(n):
x = np.random.randn(n)
return [erf(v) for v in x]
%timeit vectorized(10e5)
%timeit loopstyle(10e5)
提供结果
# vectorized
10 loops, best of 3: 108 ms per loop
# loops
1 loops, best of 3: 2.34 s per loop
从Python的math.erf函数文档中可以看出,它在近似计算中使用了高达50个术语:
Implementations of the error function erf(x) and the complementary error
function erfc(x).
Method: we use a series approximation for erf for small x, and a continued
fraction approximation for erfc(x) for larger x;
combined with the relations erf(-x) = -erf(x) and erfc(x) = 1.0 - erf(x),
this gives us erf(x) and erfc(x) for all x.
The series expansion used is:
erf(x) = x*exp(-x*x)/sqrt(pi) * [
2/1 + 4/3 x**2 + 8/15 x**4 + 16/105 x**6 + ...]
The coefficient of x**(2k-2) here is 4**k*factorial(k)/factorial(2*k).
This series converges well for smallish x, but slowly for larger x.
The continued fraction expansion used is:
erfc(x) = x*exp(-x*x)/sqrt(pi) * [1/(0.5 + x**2 -) 0.5/(2.5 + x**2 - )
3.0/(4.5 + x**2 - ) 7.5/(6.5 + x**2 - ) ...]
after the first term, the general term has the form:
k*(k-0.5)/(2*k+0.5 + x**2 - ...).
This expansion converges fast for larger x, but convergence becomes
infinitely slow as x approaches 0.0. The (somewhat naive) continued
fraction evaluation algorithm used below also risks overflow for large x;
but for large x, erfc(x) == 0.0 to within machine precision. (For
example, erfc(30.0) is approximately 2.56e-393).
Parameters: use series expansion for abs(x) < ERF_SERIES_CUTOFF and
continued fraction expansion for ERF_SERIES_CUTOFF <= abs(x) <
ERFC_CONTFRAC_CUTOFF. ERFC_SERIES_TERMS and ERFC_CONTFRAC_TERMS are the
numbers of terms to use for the relevant expansions.
#define ERF_SERIES_CUTOFF 1.5
#define ERF_SERIES_TERMS 25
#define ERFC_CONTFRAC_CUTOFF 30.0
#define ERFC_CONTFRAC_TERMS 50
Error function, via power series.
Given a finite float x, return an approximation to erf(x).
Converges reasonably fast for small x.