为什么我尝试将 tanh(x) 函数拟合到数据效果不佳?

Why is my attempt to fit a tanh(x) function to data not working well?

我有不同阳极电压下的阳极电流数据。我正在尝试使用 curve_fit 将 tanh(x) 曲线拟合到所得的 I-V 曲线,但我一直得到一条线。

因为我试图根据 log10(x) 将曲线拟合到 y,所以我做了 curve_fit 2方式:

  1. 我先取数据的log10,其次拟合曲线
  2. 我先拟合曲线,再取数据的log10。

方法一代码及输出:

import matplotlib.pyplot as plt
from scipy.optimize import curve_fit

def fitfunction(v, a, b, c, d):
    return a * np.tanh(b * v + c) + d

# x data = V90
# y data = np.log10(I90)

pars, cov = curve_fit(fitfunction, V90, np.log10(I90))

plt.plot(V90, fitfunction(V90, *pars), 'r-', linewidth='3', label='Line of Best Fit')
plt.scatter(V90, np.log10(I90), marker='.', label='Data')

plt.title('Graph of Line of Best Fit of Anode Current against Anode Potential')
plt.grid(True)
plt.xlabel('Voltage (V)')
plt.ylabel('Current (A)')
plt.legend()
plt.show()

使用上述代码生成的图表是:

方法二代码及输出:

pars, cov = curve_fit(fitfunction, V90, I90)
print(pars)

y = fitfunction(V90, *pars)

if any(i < 0 for i in y) == True:
    y = y + abs(min(y))

y = y[377:]
x = V90[377:]

plt.plot(x, np.log10(y), 'r-', linewidth='3', label='Line of Best Fit')
plt.scatter(V90, np.log10(I90), marker='.', label='Data')

plt.title('Graph of Line of Best Fit of Anode Current against Anode Potential')
plt.grid(True)
plt.xlabel('Voltage (V)')
plt.ylabel('Current (A)')
plt.legend()
plt.show()

使用此代码生成的图表是:

我不太清楚为什么在第二种方法中,即使我切断了 0 值,当 V90 < 0.

时仍然存在如此大的偏差

导致问题的数据部分似乎在 2.5V-5.0V 的某处,大约 500 行数据:

#Voltages(V) Currents(A) Af=89.9mA Vf=24.3V
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4.81000,0.0003933869315912
4.81500,0.00039350763159120005
4.82000,0.00039353273159120004
4.82500,0.0003935321315912
4.83000,0.0003935798315912
4.83500,0.0003935836315912
4.84000,0.0003936641315912
4.84500,0.00039368763159120006
4.85000,0.00039365193159120003
4.85500,0.00039369873159120005
4.86000,0.0003937326315912
4.86500,0.00039382033159120005
4.87000,0.0003938480315912
4.87500,0.00039383903159120003
4.88000,0.0003938897315912
4.88500,0.0003940000315912
4.89000,0.0003939517315912
4.89500,0.0003940104315912
4.90000,0.0003940086315912
4.90500,0.0003939877315912
4.91000,0.00039403513159120006
4.91500,0.00039411413159120006
4.92000,0.0003941618315912
4.92500,0.00039414903159120003
4.93000,0.00039421843159120006
4.93500,0.00039421603159120006
4.94000,0.00039430313159120005
4.94500,0.0003943028315912
4.95000,0.0003943370315912
4.95500,0.0003943808315912
4.96000,0.0003943910315912
4.96500,0.00039442853159120004
4.97000,0.0003944494315912
4.97500,0.00039449833159120004
4.98000,0.00039456143159120004
4.98500,0.0003945686315912
4.99000,0.0003945677315912
4.99500,0.0003946011315912

这是一些基于您的代码的代码。在这里,有一个选项可以从文件中读取,或者曲线拟合动态创建的完美数据。如果您有“不错”的值,例如 (a,b,c,d)=1,2,3,4,并记录数据,则曲线拟合效果很好。但是,如果您 fiddle 使用参数,即使是“完美”的数据,您也可能会得出不合适的结果。我确实注意到,当拒绝数据 V<-3.0,并且 not 采用 log10(I90) 时,您得到的拟合看起来不错,但并不完美。当 V>-2.5 时尝试数据,并取 I90 数据的日志,你会得到这个,这还不错:

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""


test code to check tanh fitting

"""

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


def fitfunction(v,a,b,c,d):
    return (a * np.tanh((b * v) + c) + d)


# x data = V90
# y data = np.log10(I90)

# read data from file
readdata=False

if(readdata==True):
    # arr = np.loadtxt("anode_90ma_test1.csv", delimiter=",")
    arr = np.loadtxt("test_data4.csv", delimiter=",")
    V90=arr[:,0]
    # option of using raw data or log of raw data
    I90=arr[:,1]
    # I90=np.log10(arr[:,1])
    
else:
    a_coeff=100
    b_coeff=1
    c_coeff=0.2
    d_coeff=-99
    V90=np.arange(-4,4,0.1)
    createvalues = np.vectorize(fitfunction)
    I90=createvalues(V90,a_coeff,b_coeff,c_coeff,d_coeff)
    # optionally take log10
    # I90=np.log10(I90)
        

# pars, cov = curve_fit(fitfunction,V90,np.log10(I90))
pars, cov = curve_fit(fitfunction,V90,I90)
print("fit pars = ",pars)

plt.plot(V90,fitfunction(V90,*pars),'r-',linewidth='3',label='Line of Best Fit')
plt.scatter(V90,I90,marker='.',label='Data')


plt.title('Graph of Line of Best Fit of Anode Current against Anode Potential')
plt.grid(True)
plt.xlabel('Voltage (V)')
plt.ylabel('Current (A)')
plt.legend()
plt.show() 

我必须承认,我没有发现数据和拟合有任何特别的问题。我实际上会说,由于初始值的问题,可能会出现任何问题。异常值可以忽略不计,但可以尝试使用 robust fitting。可以像这样自动猜测起始值:

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

def f( x, a, b, c, d ):
    return a * np.tanh( b * x + c ) + d

def g( x, a, b, c, d, p ):
    """
    sharpen the transition to the flat part with parameter p
    """
    w = np.copysign( np.ones( len( x ) ), b * x + c )
    return a * w * np.tanh( np.abs( b * x + c )**p )**( 1 / p ) + d


data = np.loadtxt( 
    "anode_90ma_test1.txt",
    skiprows=1, delimiter=","
)

xl, yl = data[:,0], np.log10( data[:,1] )

### simple guesses for a, c and d
d0 = ( min( yl ) + max( yl ) ) / 2
a0 = ( max( yl ) - min( yl ) ) / 2
npa = np.argwhere( np.heaviside( yl - d0, 0 ) == 0 )
c0 = -xl[ npa[-1,0] ]
print ("a0, c0, d0 = ", a0, c0, d0 )

### best guess for b via differential equation
### and alternative guesses for a and d
### uses: dy/dx = u * y**2 + v * y + w
### with u = -b/a, v = 2 b / a d and w = -b/a d^2 + a b
dy = np.gradient( yl, xl )

VT = np.array([
    yl**2, yl, np.ones( len( yl ) )
])
V = np.transpose( VT )
eta = np.dot( VT, dy )
A = np.dot( VT, V )
sol = np.linalg.solve( A, eta )
print( sol )
u, v, w = sol

df = -v / 2 / u 
bf = np.sqrt( u**2 * df**2 - w * u )
af = -bf / u 
print( "d = ", df )
print( "b = ", bf )
print( "a = ", af )

### non-linear fit

sol, cov = curve_fit( f, xl, yl, p0=[ a0, bf, c0, d0 ] )
print( sol )
### with sharpened edges
sol2, cov2 = curve_fit( g, xl, yl, p0=np.append( sol, 1 ) )

### plotting
fig = plt.figure()
ax = fig.add_subplot( 1, 1, 1 )

ax.plot(
    data[:,0], np.log10( data[:,1] ),
    ls='', marker='+', label='data', alpha=0.5
)
ax.plot( data[:,0], f( data[:,0], *sol ), label="round edges" )
ax.plot( data[:,0], g( data[:,0], *sol2 ), label="sharp edges" )
ax.plot( data[:,0], f( data[:,0], af, bf, c0, df ), ls=':',label="guess" )
ax.axhline( y=d0, color='k', ls=':' )
ax.grid()
ax.legend( loc=0 )
plt.show()

提供

a0, c0, d0 =  3.9470363024506527 0.205 -7.350877976113524
[ -0.62643791  -8.42148481 -21.38742817]
d =  -6.721723517680841
b =  2.0814552335484855
a =  3.3226840415005183
[ 3.30445925  2.18844797  0.19235933 -6.7106928 ]

最终 non-linear 完美契合。