最小二乘法进行曲线拟合 Python

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本文不手动实现最小二乘，调用scipy库中实现好的相关优化函数。

考虑如下的含有4个参数的函数式：

f(x)=A−D1+(x/C)B+D

构造数据

import numpy as npfrom scipy import optimizeimport matplotlib.pyplot as pltdef logistic4(x, A, B, C, D):    return (A-D)/(1+(x/C)**B)+Ddef residuals(p, y, x):     A, B, C, D = p    return y - logisctic4(x, A, B, C, D)def peval(x, p):    A, B, C, D = p    return logistic4(x, A, B, C, D) A, B, C, D = .5, 2.5, 8, 7.3x = np.linspace(0, 20, 20)y_true = logistic4(x, A, B, C, D)y_meas = y_true + 0.2 * np.random.randn(len(y_true))
   
   

    
    
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调用工具箱函数，进行优化

p0 = [1/2]*4plesq = optimize.leastsq(residuals, p0, args=(y_meas, x))                        # leastsq函数的功能其实是根据误差（y_meas-y_true）                        # 估计模型（也即函数）的参数
   
   

    
    
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绘图

plt.figure(figsize=(6, 4.5))plt.plot(x, peval(x, plesq[0]), x, y_meas, 'o', x, y_true)plt.legend(['Fit', 'Noisy', 'True'], loc='upper left')plt.title('least square for the noisy data (measurements)')for i, (param, true, est) in enumerate(zip('ABCD', [A, B, C, D], plesq[0])):    plt.text(11, 2-i*.5, '{} = {:.2f}, est({:.2f}) = {:.2f}'.format(param, true, param, est))plt.savefig('./logisitic.png')plt.show()
   
   

    
    
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