我正在嘗試拟合一些資料點以找到圓的中心.以下所有點都是圍繞圓周的噪聲資料點：
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)]
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)
但這似乎是使用單個xy？有關如何将此功能插入我的資料示例的任何想法？