拟合曲线的显示问题：无法解决粗糙度

Question

尽管有一个使用lmfit库进行曲线拟合的工作脚本，但我无法解决显示问题。实际上，由于只有5个相关值，因此生成的图相当粗糙。你知道吗

在切换到lmfit之前，我使用的是curve_fit，只需使用np.linspace并绘制拟合过程产生的优化值，就可以解决显示问题。然后，我通过plt.errorbar显示“真实”值。使用lmfit，上述解决方案会产生不匹配错误，因为它会识别“伪”自变量并启动不匹配类型错误。你知道吗

我的完整脚本如下：

import lmfit as lf
from lmfit import Model, Parameters
import numpy as np
import matplotlib.pyplot as plt
from math import atan


def on_res(omega_eff, thetas, R2avg=5, k_ex=0.1, phi_ex=500):
    return R2avg*(np.sin(thetas))**2 + ((np.sin(thetas))**2)*(phi_ex*k_ex/(k_ex**2 + omega_eff**2))

model = Model(on_res,independent_vars=['omega_eff','thetas'])
params = model.make_params(R2avg=5, k_ex=0.01, phi_ex=1500)


carrier = 6146.53
O_1 = 5846
spin_locks = (1000, 2000, 3000, 4000, 5000) 
delta_omega = (O_1 - carrier)

omega_eff1 = ((delta_omega**2) + (spin_locks[0]**2))**0.5
omega_eff2 = ((delta_omega**2) + (spin_locks[1]**2))**0.5
omega_eff3 = ((delta_omega**2) + (spin_locks[2]**2))**0.5
omega_eff4 = ((delta_omega**2) + (spin_locks[3]**2))**0.5
omega_eff5 = ((delta_omega**2) + (spin_locks[4]**2))**0.5

theta_rad1 = atan(spin_locks[0]/delta_omega)
theta_rad2 = atan(spin_locks[1]/delta_omega)
theta_rad3 = atan(spin_locks[2]/delta_omega)
theta_rad4 = atan(spin_locks[3]/delta_omega)
theta_rad5 = atan(spin_locks[4]/delta_omega)

x = (omega_eff1/1000, omega_eff2/1000, omega_eff3/1000, omega_eff4/1000, omega_eff5/1000)# , omega_eff6/1000)# , omega_eff7/1000)
theta = (theta_rad1, theta_rad2, theta_rad3, theta_rad4, theta_rad5)
R1rho_vals = (7.9328, 6.2642, 6.0005, 5.9972, 5.988)
e = (0.2, 0.2, 0.2, 0.2, 0.2)

new_x = np.linspace(0, 6, 1000)
omega_eff = np.array(x, dtype=float)
thetas = np.array(theta, dtype=float)
R1rho_vals = np.array(R1rho_vals, dtype=float)
error = np.array(e, dtype=float)
R2avg = []
k_ex = []
phi_ex = []

result = model.fit(R1rho_vals, params, weights=1/error, thetas=thetas, omega_eff=omega_eff, method = "emcee", steps = 1000)

print(result.fit_report())

plt.errorbar(x, R1rho_vals, yerr = error, fmt = ".k", markersize = 8, capsize = 3)
plt.plot(new_x, result.best_fit)
plt.show()

如您所见，运行它时，它会启动“形状不匹配”错误消息。将plt.plot线更改为plt.plot(x, result.best_fit)可以正确生成图形，但显示的轮廓非常粗糙（正如人们所期望的，在x轴上只有5个点）。你知道吗

你知道有什么办法可以解决这个问题吗？查看文档时，我注意到提供的示例都通过实际自变量值绘制结果，因为它们有足够的实验值。你知道吗

Answer 1

您需要使用自变量的新值重新评估ModelResult：

plt.plot(new_x, result.eval(omega_eff=new_x/1000., thetas=thetas))