python - 有没有办法通过分散的 X、Y 坐标绘制 'tight' 最佳拟合线？

Question

我在尝试通过我的数据拟合平均曲线以求出长度时遇到了问题。我在一个 Pandas 数据框中有很多 X、Y 点，看起来像这样:

x = np.asarray([731501.13, 731430.24, 731360.29, 731289.36, 731909.72, 731827.89,
   731742.  , 731657.74, 731577.95, 731502.64, 731430.39, 731359.12,
   731287.3 , 731214.21, 732015.59, 731966.88, 731902.67, 731826.31,
   731743.79, 731660.94, 731581.29, 731505.4 , 731431.95, 732048.71,
   732026.66, 731995.46, 731952.18, 731894.29, 731823.58, 731745.16,
   732149.61, 732091.53, 732052.98, 732026.82, 732005.17, 731977.63,
   732691.84, 732596.62, 732499.45, 732401.62, 732306.18, 732218.35,
   732141.82, 732080.91, 732038.21, 732009.08, 733023.08, 732951.99,
   732873.32, 732787.51])

y = np.asarray([7873771.69, 7873705.34, 7873638.03, 7873571.73, 7874082.33,
   7874027.2 , 7873976.22, 7873923.58, 7873866.35, 7873804.53,
   7873739.58, 7873673.62, 7873608.23, 7873544.15, 7874286.21,
   7874197.15, 7874123.96, 7874063.21, 7874008.78, 7873954.69,
   7873897.31, 7873836.09, 7873772.38, 7874564.62, 7874448.23,
   7874341.23, 7874246.59, 7874166.93, 7874100.4 , 7874041.77,
   7874912.56, 7874833.09, 7874733.62, 7874621.43, 7874504.65,
   7874393.89, 7875225.26, 7875183.85, 7875144.42, 7875105.69,
   7875064.49, 7875015.5 , 7874954.94, 7874878.36, 7874783.13,
   7874674.  , 7875476.18, 7875410.05, 7875351.67, 7875300.61])

x 和 y 是 map View 坐标，我想计算长度。我可以对欧几里得距离进行编码，但是因为这些点是分散的并且不是一个接一个的点，所以我在尝试通过它拟合一条移动线时遇到了麻烦。我试过 polyfit 但这主要产生一条直线，即使是更高的度数，例如:

from numpy.polynomial.polynomial import polyfit
import numpy as np
import matplotlib.pyplot as plt
z = np.polyfit(x,y,10) 
p = np.poly1d(z)
plt.scatter(x,y, marker='x')
plt.scatter(x, p(x), marker='.')

plt.show()

这是为了演示我的意思 1

如有任何帮助，我们将不胜感激!

Answer 1

这将是一个适合您的数据的经验函数:

import matplotlib.pyplot as plt
import numpy as np
from scipy.optimize import curve_fit


x = np.asarray([731501.13, 731430.24, 731360.29, 731289.36, 731909.72, 731827.89,
   731742.  , 731657.74, 731577.95, 731502.64, 731430.39, 731359.12,
   731287.3 , 731214.21, 732015.59, 731966.88, 731902.67, 731826.31,
   731743.79, 731660.94, 731581.29, 731505.4 , 731431.95, 732048.71,
   732026.66, 731995.46, 731952.18, 731894.29, 731823.58, 731745.16,
   732149.61, 732091.53, 732052.98, 732026.82, 732005.17, 731977.63,
   732691.84, 732596.62, 732499.45, 732401.62, 732306.18, 732218.35,
   732141.82, 732080.91, 732038.21, 732009.08, 733023.08, 732951.99,
   732873.32, 732787.51])/732 -1000

y = np.asarray([7873771.69, 7873705.34, 7873638.03, 7873571.73, 7874082.33,
   7874027.2 , 7873976.22, 7873923.58, 7873866.35, 7873804.53,
   7873739.58, 7873673.62, 7873608.23, 7873544.15, 7874286.21,
   7874197.15, 7874123.96, 7874063.21, 7874008.78, 7873954.69,
   7873897.31, 7873836.09, 7873772.38, 7874564.62, 7874448.23,
   7874341.23, 7874246.59, 7874166.93, 7874100.4 , 7874041.77,
   7874912.56, 7874833.09, 7874733.62, 7874621.43, 7874504.65,
   7874393.89, 7875225.26, 7875183.85, 7875144.42, 7875105.69,
   7875064.49, 7875015.5 , 7874954.94, 7874878.36, 7874783.13,
   7874674.  , 7875476.18, 7875410.05, 7875351.67, 7875300.61])/7873 - 1000

def my_func( x, x0, y0, a, b, c, t, s):
    xs = x-x0
    p = a * xs**3 + b * xs**2 + c * xs + y0
    t = t * np.tanh( s * xs )
    return p + t

xth = np.linspace( -1.15, 1.5, 50 )
yth = my_func( xth, 0.03, 0.18, .01, 0, 0.05, .05 , 10)

sol, err = curve_fit( my_func, x, y, p0=[0.03, 0.18, .01, 0, 0.05, .05 , 10] ) 
print sol 
fig = plt.figure()
ax = fig.add_subplot( 1, 1, 1 )
ax.scatter( x, y )
ax.plot( xth, yth )
ax.plot( xth, my_func( xth, *sol) )
plt.show()

给予

>>[ 2.86281016e-02  1.95292660e-01  9.62290944e-03 -1.26304655e-02 5.11281073e-02  4.63955967e-02  1.02260568e+01]

和