对两个未知因变量使用 curve_fit，此方法是否正确?

Question

我是python的新手，我正在使用curve_fit在我的模型中进行不同的计算，我的一个方程包括两个未知变量，所以我不可能手动检查我的代码是否正确。

我的等式是

ln(S(b)/S(b50))= -b D + 1/6 b**2 D**2 K

我的未知数是D和K

ln(S(b(/S(b50(( 是我的 ydata b 是我的 xdata

所以我使用了以下内容：

xdata = np.array([50,300,600,1000])
ydata_beforelog = np.array([426.0938, 259.2896, 166.8042, 80.9248])
ydata = np.log(ydata_before/426.0938)
def func(x, D, K):
return (-x * D) + (1/6 * (x **2)* (D **2) * K)
popt, pcov = curve_fit(func, xdata, ydata)
popt[0] = popt[0] * 1000 # I need that for unit scaling
popt = ([ 1.48687053, -0.46540487])
'''
I would assume that those are my D and K?

Answer 1

请允许我建议使用lmfit(https://lmfit.github.io/lmfit-py( - 免责声明：我是主要作者。 这会改变你的适合看起来像这样：

import numpy as np
from lmfit import Model
xdata = np.array([50,300,600,1000])
ydata_before = np.array([426.0938, 259.2896, 166.8042, 80.9248])
ydata = np.log(ydata_before/426.0938)
def func(x, D, K):
return (-x * D) + (1/6 * (x **2)* (D **2) * K)
# create model from this function
mymodel = Model(func)
# create a dictionary of named parameters using the argument names of 
# your model function, so 'D' and 'K'.  Give initial values:
params = mymodel.make_params(D=1.5, K=-1)
# do the fit: fitting ydata with the parameters and
# independent variable "x" as defined by your model function:
result = mymodel.fit(ydata, params, x=xdata)
# print the results and fit statistics or just get the best-fit parameters:
print(result.fit_report())
for key, param in result.params.items():
print(param)

也就是说，您可以按名称引用参数。 此填充将打印出来：

[[Model]]
Model(func)
[[Fit Statistics]]
# fitting method   = leastsq
# function evals   = 12
# data points      = 4
# variables        = 2
chi-square         = 0.00709560
reduced chi-square = 0.00354780
Akaike info crit   = -21.3382982
Bayesian info crit = -22.5657095
[[Variables]]
D:  0.00148687 +/- 1.9408e-04 (13.05%) (init = 1.5)
K: -0.46540487 +/- 0.71331763 (153.27%) (init = -1)
[[Correlations]] (unreported correlations are < 0.100)
C(D, K) =  0.977
<Parameter 'D', value=0.0014868705336113577 +/- 0.000194, bounds=[-inf:inf]>
<Parameter 'K', value=-0.4654048673207782 +/- 0.713, bounds=[-inf:inf]>