Is there an easily available implementation of erf() for Python?

Question

I can implement the error function, erf, myself, but I'd prefer not to. Is there a python package with no external dependencies that contains an implementation of this function? I have found this but this seems to be part of some much larger package (and it's not even clear which one!).

Answer 1

From Python's math.erf function documentation, it uses up to 50 terms in the approximation:

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.

Answer 2

SciPy has an implementation of the erf function, see scipy.special.erf.

Answer 3

One note for those aiming for higher performance: vectorize, if possible.

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)

gives results

# vectorized
10 loops, best of 3: 108 ms per loop

# loops
1 loops, best of 3: 2.34 s per loop

Answer 4

Since v.2.7. the standard math module contains erf function. This should be the easiest way.

http://docs.python.org/2/library/math.html#math.erf

Answer 5

I would recommend you download numpy (to have efficiant matrix in python) and scipy (a Matlab toolbox substitute, which uses numpy). The erf function lies in scipy.

>>>from scipy.special import erf
>>>help(erf)

You can also use the erf function defined in pylab, but this is more intended at plotting the results of the things you compute with numpy and scipy. If you want an all-in-one installation of these software you can use directly the Python Enthought distribution.

Answer 6

I recommend SciPy for numerical functions in Python, but if you want something with no dependencies, here is a function with an error error is less than 1.5 * 10^-7 for all inputs.

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)

The algorithm comes from Handbook of Mathematical Functions, formula 7.1.26.

Answer 7

I have a function which does 10^5 erf calls. On my machine...

scipy.special.erf makes it time at 6.1s

erf Handbook of Mathematical Functions takes 8.3s

erf Numerical Recipes 6.2 takes 9.5s

(three-run averages, code taken from above posters).

Answer 8

A pure python implementation can be found in the mpmath module (http://code.google.com/p/mpmath/)

From the doc string:

>>> 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

For large real x, \mathrm{erf}(x) approaches 1 very rapidly::

>>> print erf(3)
0.999977909503001
>>> print erf(5)
0.999999999998463

The error function is an odd function::

>>> nprint(chop(taylor(erf, 0, 5)))
[0.0, 1.12838, 0.0, -0.376126, 0.0, 0.112838]

:func:erf implements arbitrary-precision evaluation and supports complex numbers::

>>> mp.dps = 50
>>> print erf(0.5)
0.52049987781304653768274665389196452873645157575796
>>> mp.dps = 25
>>> print erf(1+j)
(1.316151281697947644880271 + 0.1904534692378346862841089j)

Related functions

See also :func:erfc, which is more accurate for large x, and :func:erfi which gives the antiderivative of \exp(t^2).

The Fresnel integrals :func:fresnels and :func:fresnelc are also related to the error function.

Answer 9

To answer my own question, I have ended up using the following code, adapted from a Java version I found elsewhere on the web:

# 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