使用python将erf函数拟合到数据

Question

我已阅读此article中的erf函数。

我也尝试过实现此thread中提供的代码。

但考虑到erf函数的定义，我不了解如何实现对以下数据的拟合：

X   Y
 33     21.09
 35     21.14
 37     21.21
 39     21.32
 41     21.46
 43     23
 45     27
 47     31
 49     36
 51     40
 53     45
 55     49
 57     53
 59     57
 61     61
 63     65
 65     69
 67     73
 69     77
 71     78
 73     79
 75     79

编辑：

我更改了数据，使其不再是步骤。数据图看起来像：

此图应该适合sigmoid，tanh等功能。但是使用下面评论中的代码我并不合适。

Answer 1

你很容易适应erf。上面评论中的解决方案不能像修改后的问题一样工作，数据已经改变，提供的起始值不起作用。然而，从斜坡到水平时的曲率非常大，即过渡非常尖锐，使得erf，tanh和sigmoid都不起作用。下面我将展示拟合以及一些更适合数据的替代/修改函数。

import matplotlib.pyplot as plt
import numpy as np
from scipy.special import erf
from scipy.optimize import curve_fit

data = np.array( [
    33, 21.09,
    35, 21.14,
    37, 21.21,
    39, 21.32,
    41, 21.46,
    43, 23,
    45, 27,
    47, 31,
    49, 36,
    51, 40,
    53, 45,
    55, 49,
    57, 53,
    59, 57,
    61, 61,
    63, 65,
    65, 69,
    67, 73,
    69, 77,
    71, 78,
    73, 79,
    75, 79 ] )

xData = data[ ::2 ]
yData = data[ 1::2 ]


def sign_pow( x, p):
    return np.copysign( abs( x )**p, x )


guessErf = [ 32., 55., 10., 20. ]
solErf , pcov = curve_fit( lambda x, a, b, c, d:  a * (1 + erf( ( x- b ) / c ) ) + d , xData, yData, p0=guessErf )
print solErf

guessTh = [ 32., 55., 10., 20. ]
solTh , pcov = curve_fit( lambda x, a, b, c, d:  a * (1 + np.tanh( ( x- b ) / c ) ) + d , xData, yData, p0=guessTh )
print solTh

guessThP = [ 32., 55., 10., 20., 1 ]
solThP , pcov = curve_fit( lambda x, a, b, c, d, p:  a * (1 + sign_pow( np.tanh( sign_pow( ( x- b ) / c , p ) ) , (1./p) ) ) + d , xData, yData, p0=guessThP )
print solThP


guessLz = [ 32., 55., 10., 20.]
solLz , pcov = curve_fit( lambda x, a, b, c, d:  a * ( 1 + (x - b) / np.sqrt( 1 + ( x - b )**2 / c ) )+ d , xData, yData, p0=guessLz )
print solLz

guessLzP = [ 2., 55., 200., 47., 2]
solLzP , pcov = curve_fit( lambda x, a, b, c, d, p:  a * ( 1 + (x - b) / ( 1 + abs( ( x - b ) / c )**p )**( 1. / p ) ) + d , xData, yData, p0=guessLzP, maxfev=5000 )
print solLzP
#~ solLzP=guessLzP


testX = np.linspace( 30, 80, 150 )

erfList = [ solErf[0] * ( 1 + erf( ( x - solErf[1] ) / solErf[2] ) ) + solErf[3] for x in testX ]
thList = [ solTh[0] * ( 1 + np.tanh( ( x - solTh[1] ) / solTh[2] ) ) + solTh[3] for x in testX ]
thPList = [ solThP[0] * ( 1 + sign_pow( np.tanh( sign_pow( ( x - solThP[1] ) / solThP[2] , solThP[4] ) ) , 1. / solThP[4] ) ) + solThP[3] for x in testX ]
lzList = [ solLz[0] * ( 1 + (x - solLz[1] ) / np.sqrt( 1 + ( x - solLz[1] )**2 / solLz[2] ) ) + solLz[3] for x in testX ]
lzPList = [ solLzP[0] * ( 1 + (x - solLzP[1] ) / ( 1 + abs( ( x - solLzP[1] ) / solLzP[2] )**solLzP[4] )**( 1. / solLzP[4] ) ) + solLzP[3] for x in testX ]

fig = plt.figure()
ax = fig.add_subplot( 1, 1, 1 )
ax.plot( xData, yData , ls='', marker='d', label='data')
ax.plot( testX, erfList, label='erf')
ax.plot( testX, thList, label='tanh' )
ax.plot( testX, thPList, label= 'tanh power' )
ax.plot( testX, lzList, label= 'x/1+x^2' )
ax.plot( testX, lzPList, label= 'x/1+x^P' , ls='--')
ax.legend( loc=0 )
plt.show()

提供：

>> [31.57905297 55.87319052 14.64143581 18.66393531]
>> [33.34190267 55.90472589 13.54119241 16.97367616]
>> [28.77092734 55.75011551 13.6999564  21.16422056  7.22849049]
>> [  2.49108137  55.93839349 232.504463    47.90421787]
>> [  2.09892962  55.75526439 -13.72758215  47.8461409   17.7915103 ]

可以很容易地看到曲率问题。修改后的tanh和x/(1+x**p)**(1/p)同样适用。