最小化给出奇怪的结果(多参数拟合)
我正在努力拟合实验数据。为了适应它,我使用残留功能的最小化。一切都很微不足道,但是这次我找不到出了什么问题,为什么配合的结果如此奇怪。与原始问题相比,简化了示例。但是无论如何,即使我将使用的参数值设置为初始猜测,它也会给出错误的参数。
import matplotlib.pyplot as plt
import numpy as np
import csv
from scipy.optimize import curve_fit, minimize
x=np.arange(0,10,0.5)
a=0.5
b=3
ini_pars=[a, b]
def func(x, a, b):
return a*x+b
plt.plot(x, func(x,a,b))
plt.show()
def fit(pars):
A,B = pars
res = (func(x,a, b)-func(x, *pars))**2
s=sum(res)
return s
bnds=[(0.1,0.5),(1,5)]
x0=[0.1,4]
opt = minimize(fit, x0, bounds=bnds)
new_pars=[opt.x[0], opt.x[0]]
example = fit(ini_pars)
print(example)
example = fit(new_pars)
print(example)
print(new_pars)
plt.plot(x, func(x, *ini_pars))
plt.plot(x, func(x, *new_pars))
plt.show()
```[enter image description here][1]
[1]: https://i.sstatic.net/qc1Nu.png
I'm working on fitting of the experimental data. In order to fit it I use the minimization of the function of residual. Everything is quite trivial, but but this time I can't find what's wrong and why the result of fitting is so weird. The example is simplified in comparison with original problem. But anyway it gives wrong parameters even when I set used values of parameters as initial guess.
import matplotlib.pyplot as plt
import numpy as np
import csv
from scipy.optimize import curve_fit, minimize
x=np.arange(0,10,0.5)
a=0.5
b=3
ini_pars=[a, b]
def func(x, a, b):
return a*x+b
plt.plot(x, func(x,a,b))
plt.show()
def fit(pars):
A,B = pars
res = (func(x,a, b)-func(x, *pars))**2
s=sum(res)
return s
bnds=[(0.1,0.5),(1,5)]
x0=[0.1,4]
opt = minimize(fit, x0, bounds=bnds)
new_pars=[opt.x[0], opt.x[0]]
example = fit(ini_pars)
print(example)
example = fit(new_pars)
print(example)
print(new_pars)
plt.plot(x, func(x, *ini_pars))
plt.plot(x, func(x, *new_pars))
plt.show()
```[enter image description here][1]
[1]: https://i.sstatic.net/qc1Nu.png
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它应该是
new_pars = [opt.x [0],opt.x [1]]
而不是new_pars = [opt.x [0],opt.x [0]] < /代码>。还要注意,您可以通过
new_pars = opt.x
直接提取值。It should be
new_pars=[opt.x[0], opt.x[1]]
instead ofnew_pars=[opt.x[0], opt.x[0]]
. Note also that you can directly extract the values bynew_pars = opt.x
.