如何在python中使用最小二乘拟合找到圆心？
mim*_*lea 5 python geometry scipy
我正在尝试拟合一些数据点以找到圆的中心.以下所有点都是围绕圆周的噪声数据点:
data = [(2.2176383052987667, 4.218574252410221),
(3.3041214516913033, 5.223500807396272),
(4.280815855023374, 6.461487709813785),
(4.946375258539319, 7.606952538212697),
(5.382428804463699, 9.045717060494576),
(5.752578028217334, 10.613667377465823),
(5.547729017414035, 11.92662513852466),
(5.260208374620305, 13.57722448066025),
(4.642126672822957, 14.88238955729078),
(3.820310290976751, 16.10605425390148),
(2.8099420132544024, 17.225880123445773),
(1.5731539516426183, 18.17052077121059),
(0.31752822350872545, 18.75261434891438),
(-1.2408437559671106, 19.119355580780265),
(-2.680901948575409, 19.15018791257732),
(-4.190406775175328, 19.001321726517297),
(-5.533990404926917, 18.64857428377178),
(-6.903383826792998, 17.730112542165955),
(-8.082883753215347, 16.928080323602334),
(-9.138397388219254, 15.84088004983959),
(-9.92610373064812, 14.380575762984085),
(-10.358670204629814, 13.018017342781242),
(-10.600053524240247, 11.387283417089911),
(-10.463673966507077, 10.107554951600699),
(-10.179820255235496, 8.429558128401448),
(-9.572153386953028, 7.1976672709797676),
(-8.641475289758178, 5.8312286526738175),
(-7.665976739804268, 4.782663065707469),
(-6.493033077746997, 3.8549965442534684),
(-5.092340806635571, 3.384419909199452),
(-3.6530364510489073, 2.992272643733981),
(-2.1522365767310796, 3.020780664301393),
(-0.6855406924835704, 3.0767643753777447),
(0.7848958776292426, 3.6196842530995332),
(2.0614188482646947, 4.32795711960546),
(3.2705467984691508, 5.295836809444288),
(4.359297538484424, 6.378324784240816),
(4.981264502955681, 7.823851404553242)]
我试图使用像Scipy http://wiki.scipy.org/Cookbook/Least_Squares_Circle这样的库,但我在使用可用功能时遇到了麻烦.
有例如:
# == METHOD 2 ==
from scipy import optimize
method_2 = "leastsq"
def calc_R(xc, yc):
""" calculate the distance of each 2D points from the center (xc, yc) """
return sqrt((x-xc)**2 + (y-yc)**2)
def f_2(c):
""" calculate the algebraic distance between the data points and the mean circle centered at c=(xc, yc) """
Ri = calc_R(*c)
return Ri - Ri.mean()
center_estimate = x_m, y_m
center_2, ier = optimize.leastsq(f_2, center_estimate)
xc_2, yc_2 = center_2
Ri_2 = calc_R(*center_2)
R_2 = Ri_2.mean()
residu_2 = sum((Ri_2 - R_2)**2)
但这似乎是使用单个xy？有关如何将此功能插入我的数据示例的任何想法？
您的数据点看起来相当干净，我看不到异常值，因此许多圆拟合算法都可以使用。
我建议您从 Coope 方法开始，该方法通过神奇地线性化问题来工作：
(X-Xc)² + (Y-Yc)² = R² 被改写为
2 Xc X + 2 Yc Y + R² - Xc² - Yc² = X² + Y²， 然后
A X + B Y + C = X² + Y²，由线性最小二乘法求解。
- 看起来 @MadPhysicist 最终确实添加了它。您可以在这里找到该函数：https://github.com/madphysicalist/scikit-guess/blob/master/src/skg/nsphere.py。哦，它记录在这里：https://scikit-guess.readthedocs.io/en/latest/ generated/skg.nsphere.html (3认同)
作为 Bas Swinckels 帖子的后续内容，我想我应该发布实现拟合椭圆的 Halir 和 Flusser 方法的代码
https://github.com/bdhammel/least-squares-ellipse-fitting
pip install lsq-ellipse
现在使用 import numpy as np data=np.array(data) 将数据放入 numpy 数组中
使用上面的代码，您可以通过以下方法找到中心。
from ellipse import LsqEllipse
import numpy as np
import matplotlib.pyplot as plt
import statistics
from statistics import mean
from matplotlib.patches import Ellipse
lsqe = LsqEllipse()
lsqe.fit(data)
center, width, height, phi = lsqe.as_parameters()
plt.close('all')
fig = plt.figure(figsize=(6,6))
ax = fig.add_subplot(111)
ax.axis('equal')
ax.plot(data[:,0], data[:,1], 'ro', label='test data', zorder=1)
ellipse = Ellipse(xy=center, width=2*width, height=2*height, angle=np.rad2deg(phi),
edgecolor='b', fc='None', lw=2, label='Fit', zorder = 2)
ax.add_patch(ellipse)
print('center of fitted circle =',center, '\n','radius =', mean([width,height]),'+/- stddev=',statistics.stdev([width,height]))
plt.legend()
plt.show()
这里我们只是取这个椭圆的高度和宽度的平均值作为圆形拟合的半径，并将它们的标准差作为误差。这个可以修改
