gra*_*ger 3 python optimization scipy
我正在尝试用高斯曲线拟合我的数据。这是我的代码:
\n\nimport numpy as np\nfrom scipy import optimize\n\n# The independent variable where the data is measured\nx_coord = np.array([-0.1216 , -0.11692308, -0.11224615, -0.10756923, -0.10289231,\n -0.09821538, -0.09353846, -0.08886154, -0.08418462, -0.07950769,\n -0.07483077, -0.07015385, -0.06547692, -0.0608 , -0.05612308,\n -0.05144615, -0.04676923, -0.04209231, -0.03741538, -0.03273846,\n -0.02806154, -0.02338462, -0.01870769, -0.01403077, -0.00935385,\n -0.00467692, 0. , 0.00467692, 0.00935385, 0.01403077,\n 0.01870769, 0.02338462, 0.02806154, 0.03273846, 0.03741538,\n 0.04209231, 0.04676923, 0.05144615, 0.05612308, 0.0608 ,\n 0.06547692, 0.07015385, 0.07483077, 0.07950769, 0.08418462,\n 0.08886154, 0.09353846, 0.09821538, 0.10289231, 0.10756923,\n 0.11224615, 0.11692308])\n\n# The dependent data \xe2\x80\x94 nominally f(x_coord)\ny = np.array([-0.0221931 , -0.02323915, -0.02414913, -0.0255389 , -0.02652465,\n -0.02888672, -0.03075954, -0.03355392, -0.03543005, -0.03839526,\n -0.040933 , -0.0456585 , -0.04849097, -0.05038776, -0.0466699 ,\n -0.04202133, -0.034239 , -0.02667525, -0.01404582, -0.00122683,\n 0.01703862, 0.03992694, 0.06704549, 0.11362071, 0.28149172,\n 0.6649422 , 1. , 0.6649422 , 0.28149172, 0.11362071,\n 0.06704549, 0.03992694, 0.01703862, -0.00122683, -0.01404582,\n -0.02667525, -0.034239 , -0.04202133, -0.0466699 , -0.05038776,\n -0.04849097, -0.0456585 , -0.040933 , -0.03839526, -0.03543005,\n -0.03355392, -0.03075954, -0.02888672, -0.02652465, -0.0255389 ,\n -0.02414913, -0.02323915])\n\n# define a gaussian function to fit the data\ndef gaussian(x, a, b, c):\n val = a * np.exp(-(x - b)**2 / c**2)\n return val\n\n# fit the data \npopt, pcov = optimize.curve_fit(gaussian, x_coord, y, sigma = np.array([0.01] * len(x_coord)))\n\n# plot the data and the fitting curve\nplt.plot(x_coord, y, \'b-\', x_coord, gaussian(x_coord, popt[0], popt[1], popt[2]), \'r:\')\nRun Code Online (Sandbox Code Playgroud)\n\n\n\n我应该怎样做才能获得良好的拟合曲线?
\n这实际上是一个非常好的问题,它说明找到right(局部)最优值可能非常困难。
通过p0参数,您可以给优化例程一个提示,大约在您期望的最佳位置。
如果您从最初的猜测开始[1,0,0.1]:
# fit the data
sigma = np.array([0.01] * len(x_coord))
popt, pcov = optimize.curve_fit(gaussian, x_coord, y, sigma=sigma, p0=[1,0,0.1])
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你得到以下结果:
一些注意事项:您被迫curve_fit拟合没有常数项的钟形曲线。这让事情变得有些尴尬。
如果允许 offset d,您将得到:
# define a gaussian function to fit the data
def gaussian(x, a, b, c, d):
val = a* np.exp(-(x - b)**2 / c**2) + d
return val
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并得到如下结果:
# fit the data
popt, pcov = optimize.curve_fit(gaussian, x_coord, y)
# plot the data and the fitting curve
plt.plot(x_coord, y, 'b-', x_coord, gaussian(x_coord, *popt), 'r:')
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这看起来更像是一个合理的选择。虽然看起来高斯与数据的拟合不太好。
非常尖的形状表明拉普拉斯算子可能更适合:
# define a laplacian function to fit the data
def laplacian(x, a, b, c, d):
val = a* np.exp(-np.abs(x - b) / c) + d
return val
# fit the data
popt, pcov = optimize.curve_fit(laplacian, x_coord, y, p0=[1,0,0.01,-0.1])
# plot the data and the fitting curve
plt.plot(x_coord, y, 'b-', x_coord, laplacian(x_coord, *popt), 'r:')
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这是结果:
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