在 true_divide 错误中遇到零除而我的数据中没有零
divide by zero encountered in true_divide error without having zeros in my data
这是我的代码，这是我的数据，这是代码的输出。我试过在 x 轴上添加一个值，认为可能值太少可以解释为零。我不知道 true_divide 可能是什么，我无法解释这个被零除的错误，因为我的数据中没有一个零，检查了我所有的 2500 个数据点。希望你们中的一些人可以提供一些澄清。提前致谢。
import pandas as pd
import matplotlib.pyplot as plt
from iminuit import cost, Minuit
import numpy as np
frame = pd.read_excel('/Users/lorenzotecchia/Desktop/Analisi Laboratorio/Analisi dati/Quinta Esperienza/500Hz/F0000CH2.xlsx', 'F0000CH2')
data = pd.read_excel('/Users/lorenzotecchia/Desktop/Analisi Laboratorio/Analisi dati/Quinta Esperienza/500Hz/F0000CH1.xlsx', 'F0000CH1')
# tempi_500Hz = pd.DataFrame(frame,columns=['x'])
# Vout_500Hz = pd.DataFrame(frame,columns=['y'])
tempi_500Hz = pd.DataFrame(frame,columns=['x1'])
Vout_500Hz = pd.DataFrame(frame,columns=['y1'])
# Vin_500Hz = pd.DataFrame(data,columns=['y'])
def fit_esponenziale(x, α, β):
return α * (1 - np.exp(-x / β))
plt.xlabel('ω(Hz)')
plt.ylabel('Attenuazioni')
plt.title('Fit Parabolico')
plt.scatter(tempi_500Hz, Vout_500Hz)
least_squares = cost.LeastSquares(tempi_500Hz, Vout_500Hz, np.sqrt(Vout_500Hz), fit_esponenziale)
m = Minuit(least_squares, α=0, β=0)
m.migrad()
m.hesse()
plt.errorbar(tempi_500Hz, Vout_500Hz, fmt="o", label="data")
plt.plot(tempi_500Hz, fit_esponenziale(tempi_500Hz, *m.values), label="fit")
fit_info = [
f"$\chi^2$ / $n_\mathrm{{dof}}$ = {m.fval:.1f} / {len(tempi_500Hz) - m.nfit}",]
for p, v, e in zip(m.parameters, m.values, m.errors):
fit_info.append(f"{p} = ${v:.3f} \pm {e:.3f}$")
plt.legend()
plt.show()
input
output and example of data
这是一个有效的 Minuit 对比 curve_fit 示例。我对函数进行了缩放，使得指数衰减的数量级为 1（通常对于非线性拟合来说是个好主意）。最终，这两种方法都给出了非常相似的结果。
注意：无论错误是否有意义，我都将其打开。起始值等于零绝对是个坏主意。
import pandas as pd
import matplotlib.pyplot as plt
from iminuit import cost, Minuit
from scipy.optimize import curve_fit
import numpy as np
def fit_exp(x, a, b, c):
return a * (1 - np.exp(- 1000 * b * x) ) + c
nn = 170
xl = np.linspace( 0, 0.001, nn )
yl = fit_exp( xl, 15, 5.3, -8.1 ) + np.random.normal( size=nn, scale=0.05 )
#######################
### Minuit
#######################
least_squares = cost.LeastSquares(xl, yl, np.sqrt( np.abs( yl ) ), fit_exp )
print(least_squares)
m = Minuit(least_squares, a=1, b=5, c=-7)
print( "grad: ")
print( m.migrad() ) ### needs to be called to get fit values
print( m.values )### gives slightly different output
print("Hesse:")
print( m.hesse() )
#######################
### curve_fit
#######################
opt, cov = curve_fit(
fit_exp, xl, yl, sigma=np.sqrt( np.abs( yl ) ),
absolute_sigma=True
)
print( " curve_fit: ")
print( opt )
print( " covariance ")
print( cov )
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1 )
ax.plot(xl, yl, marker ='+', ls='' )
ax.plot(xl, fit_exp(xl, *m.values), ls="--")
ax.plot(xl, fit_exp(xl, *opt), ls=":")
plt.show()
实际上有一种方法可以完全线性拟合。
参见 here
import matplotlib.pyplot as plt
import numpy as np
from scipy.integrate import cumtrapz
def fit_exp(x, a, b, c):
return a * (1 - np.exp( -b * x) ) + c
nn = 170
xl = np.linspace( 0, 0.001, nn )
yl = fit_exp( xl, 15, 5300, -8.1 ) + np.random.normal( size=nn, scale=0.05 )
"""
with y = a( 1- exp(-bx) ) + c
we have Y = int y = -1/b y + d x + h ....try it out or see below
so we get a linear equation for b (actually 1/b ) to optimize
this goes as:
"""
Yl = cumtrapz( yl, xl, initial=0 )
ST = [xl, yl, np.ones( nn ) ]
S = np.transpose( ST )
eta = np.dot( ST, Yl )
A = np.dot( ST, S )
sol = np.linalg.solve( A, eta )
bFit = -1/sol[1]
print( bFit )
"""
now we can do a normal linear fit
"""
ST = [ fit_exp(xl, 1, bFit, 0), np.ones( nn ) ]
S = np.transpose( ST )
A = np.dot( ST, S )
eta = np.dot( ST, yl )
aFit, cFit = np.linalg.solve( A, eta )
print( aFit, cFit )
print(aFit + cFit, sol[0] ) ### consistency check
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1 )
ax.plot(xl, yl, marker ='+', ls='' )
## at best a sufficient fit, at worst a good start for a non-linear fit
ax.plot(xl, fit_exp( xl, aFit, bFit, cFit ) )
plt.show()
"""
a ( 1 - exp(-b x)) + c = a + c - a exp(-b x) = d - a exp( -b x )
int y = d x + a/b exp( -b x ) + g
= d x +a/b exp( -b x ) + a/b - a/b + c/b - c/b + g
= d x - 1/b ( a - a exp( -b x ) + c ) + c/b + a/b + g
= d x + k y + h
with k = -1/b and h = g + c/b + a/b.
d and h are fitted but not used, but as a+c = d we can check
for consistency
"""
这是我的代码，这是我的数据，这是代码的输出。我试过在 x 轴上添加一个值，认为可能值太少可以解释为零。我不知道 true_divide 可能是什么，我无法解释这个被零除的错误，因为我的数据中没有一个零，检查了我所有的 2500 个数据点。希望你们中的一些人可以提供一些澄清。提前致谢。
import pandas as pd
import matplotlib.pyplot as plt
from iminuit import cost, Minuit
import numpy as np
frame = pd.read_excel('/Users/lorenzotecchia/Desktop/Analisi Laboratorio/Analisi dati/Quinta Esperienza/500Hz/F0000CH2.xlsx', 'F0000CH2')
data = pd.read_excel('/Users/lorenzotecchia/Desktop/Analisi Laboratorio/Analisi dati/Quinta Esperienza/500Hz/F0000CH1.xlsx', 'F0000CH1')
# tempi_500Hz = pd.DataFrame(frame,columns=['x'])
# Vout_500Hz = pd.DataFrame(frame,columns=['y'])
tempi_500Hz = pd.DataFrame(frame,columns=['x1'])
Vout_500Hz = pd.DataFrame(frame,columns=['y1'])
# Vin_500Hz = pd.DataFrame(data,columns=['y'])
def fit_esponenziale(x, α, β):
return α * (1 - np.exp(-x / β))
plt.xlabel('ω(Hz)')
plt.ylabel('Attenuazioni')
plt.title('Fit Parabolico')
plt.scatter(tempi_500Hz, Vout_500Hz)
least_squares = cost.LeastSquares(tempi_500Hz, Vout_500Hz, np.sqrt(Vout_500Hz), fit_esponenziale)
m = Minuit(least_squares, α=0, β=0)
m.migrad()
m.hesse()
plt.errorbar(tempi_500Hz, Vout_500Hz, fmt="o", label="data")
plt.plot(tempi_500Hz, fit_esponenziale(tempi_500Hz, *m.values), label="fit")
fit_info = [
f"$\chi^2$ / $n_\mathrm{{dof}}$ = {m.fval:.1f} / {len(tempi_500Hz) - m.nfit}",]
for p, v, e in zip(m.parameters, m.values, m.errors):
fit_info.append(f"{p} = ${v:.3f} \pm {e:.3f}$")
plt.legend()
plt.show()
input
output and example of data
这是一个有效的 Minuit 对比 curve_fit 示例。我对函数进行了缩放，使得指数衰减的数量级为 1（通常对于非线性拟合来说是个好主意）。最终，这两种方法都给出了非常相似的结果。
注意：无论错误是否有意义，我都将其打开。起始值等于零绝对是个坏主意。
import pandas as pd
import matplotlib.pyplot as plt
from iminuit import cost, Minuit
from scipy.optimize import curve_fit
import numpy as np
def fit_exp(x, a, b, c):
return a * (1 - np.exp(- 1000 * b * x) ) + c
nn = 170
xl = np.linspace( 0, 0.001, nn )
yl = fit_exp( xl, 15, 5.3, -8.1 ) + np.random.normal( size=nn, scale=0.05 )
#######################
### Minuit
#######################
least_squares = cost.LeastSquares(xl, yl, np.sqrt( np.abs( yl ) ), fit_exp )
print(least_squares)
m = Minuit(least_squares, a=1, b=5, c=-7)
print( "grad: ")
print( m.migrad() ) ### needs to be called to get fit values
print( m.values )### gives slightly different output
print("Hesse:")
print( m.hesse() )
#######################
### curve_fit
#######################
opt, cov = curve_fit(
fit_exp, xl, yl, sigma=np.sqrt( np.abs( yl ) ),
absolute_sigma=True
)
print( " curve_fit: ")
print( opt )
print( " covariance ")
print( cov )
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1 )
ax.plot(xl, yl, marker ='+', ls='' )
ax.plot(xl, fit_exp(xl, *m.values), ls="--")
ax.plot(xl, fit_exp(xl, *opt), ls=":")
plt.show()
实际上有一种方法可以完全线性拟合。 参见 here
import matplotlib.pyplot as plt
import numpy as np
from scipy.integrate import cumtrapz
def fit_exp(x, a, b, c):
return a * (1 - np.exp( -b * x) ) + c
nn = 170
xl = np.linspace( 0, 0.001, nn )
yl = fit_exp( xl, 15, 5300, -8.1 ) + np.random.normal( size=nn, scale=0.05 )
"""
with y = a( 1- exp(-bx) ) + c
we have Y = int y = -1/b y + d x + h ....try it out or see below
so we get a linear equation for b (actually 1/b ) to optimize
this goes as:
"""
Yl = cumtrapz( yl, xl, initial=0 )
ST = [xl, yl, np.ones( nn ) ]
S = np.transpose( ST )
eta = np.dot( ST, Yl )
A = np.dot( ST, S )
sol = np.linalg.solve( A, eta )
bFit = -1/sol[1]
print( bFit )
"""
now we can do a normal linear fit
"""
ST = [ fit_exp(xl, 1, bFit, 0), np.ones( nn ) ]
S = np.transpose( ST )
A = np.dot( ST, S )
eta = np.dot( ST, yl )
aFit, cFit = np.linalg.solve( A, eta )
print( aFit, cFit )
print(aFit + cFit, sol[0] ) ### consistency check
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1 )
ax.plot(xl, yl, marker ='+', ls='' )
## at best a sufficient fit, at worst a good start for a non-linear fit
ax.plot(xl, fit_exp( xl, aFit, bFit, cFit ) )
plt.show()
"""
a ( 1 - exp(-b x)) + c = a + c - a exp(-b x) = d - a exp( -b x )
int y = d x + a/b exp( -b x ) + g
= d x +a/b exp( -b x ) + a/b - a/b + c/b - c/b + g
= d x - 1/b ( a - a exp( -b x ) + c ) + c/b + a/b + g
= d x + k y + h
with k = -1/b and h = g + c/b + a/b.
d and h are fitted but not used, but as a+c = d we can check
for consistency
"""