我试图用高斯曲线拟合我的数据。这是我的代码:
import numpy as np
from scipy import optimize
# The independent variable where the data is measured
x_coord = np.array([-0.1216 , -0.11692308, -0.11224615, -0.10756923, -0.10289231,
-0.09821538, -0.09353846, -0.08886154, -0.08418462, -0.07950769,
-0.07483077, -0.07015385, -0.06547692, -0.0608 , -0.05612308,
-0.05144615, -0.04676923, -0.04209231, -0.03741538, -0.03273846,
-0.02806154, -0.02338462, -0.01870769, -0.01403077, -0.00935385,
-0.00467692, 0. , 0.00467692, 0.00935385, 0.01403077,
0.01870769, 0.02338462, 0.02806154, 0.03273846, 0.03741538,
0.04209231, 0.04676923, 0.05144615, 0.05612308, 0.0608 ,
0.06547692, 0.07015385, 0.07483077, 0.07950769, 0.08418462,
0.08886154, 0.09353846, 0.09821538, 0.10289231, 0.10756923,
0.11224615, 0.11692308])
# The dependent data — nominally f(x_coord)
y = np.array([-0.0221931 , -0.02323915, -0.02414913, -0.0255389 , -0.02652465,
-0.02888672, -0.03075954, -0.03355392, -0.03543005, -0.03839526,
-0.040933 , -0.0456585 , -0.04849097, -0.05038776, -0.0466699 ,
-0.04202133, -0.034239 , -0.02667525, -0.01404582, -0.00122683,
0.01703862, 0.03992694, 0.06704549, 0.11362071, 0.28149172,
0.6649422 , 1. , 0.6649422 , 0.28149172, 0.11362071,
0.06704549, 0.03992694, 0.01703862, -0.00122683, -0.01404582,
-0.02667525, -0.034239 , -0.04202133, -0.0466699 , -0.05038776,
-0.04849097, -0.0456585 , -0.040933 , -0.03839526, -0.03543005,
-0.03355392, -0.03075954, -0.02888672, -0.02652465, -0.0255389 ,
-0.02414913, -0.02323915])
# define a gaussian function to fit the data
def gaussian(x, a, b, c):
val = a * np.exp(-(x - b)**2 / c**2)
return val
# fit the data
popt, pcov = optimize.curve_fit(gaussian, x_coord, y, sigma = np.array([0.01] * len(x_coord)))
# plot the data and the fitting curve
plt.plot(x_coord, y, 'b-', x_coord, gaussian(x_coord, popt[0], popt[1], popt[2]), 'r:')
我该怎样做才能获得合适的曲线?
答案 0 :(得分:4)
这实际上是一个非常好的问题,表明找到right
(本地)最优可能非常困难。
通过p0
参数,您可以为优化例程提供一个提示,大概可以看出最佳值。
如果您从[1,0,0.1]
的初始猜测开始:
# 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])
您得到以下结果:
一些注释:你强迫curve_fit
在没有恒定项的情况下拟合钟形曲线。这让事情变得有些尴尬。
如果您允许偏移d
,则会得到:
# 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
获得以下结果:
# 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:')
看起来更合适。虽然高斯似乎不适合数据。
非常尖锐的形状表明拉普拉斯可能更适合:
# 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:')
结果如下: