gpt4 book ai didi

python - 给定一组点 xi=(xi,yi) 在 Python 中拟合一个椭圆

转载 作者:行者123 更新时间:2023-11-28 16:46:03 26 4
gpt4 key购买 nike

我正在从二维点 (x,y) 计算一系列索引。一个指标是短轴和长轴之间的比率。为了适应椭圆,我使用以下 post

当我运行这些函数时,最终结果看起来很奇怪,因为中心和轴长度与二维点不成比例

center =  [  560415.53298363+0.j  6368878.84576771+0.j]
angle of rotation = (-0.0528033467597-5.55111512313e-17j)
axes = [0.00000000-557.21553487j 6817.76933256 +0.j]

在此先感谢您的帮助

import numpy as np
from numpy.linalg import eig, inv

def fitEllipse(x,y):
x = x[:,np.newaxis]
y = y[:,np.newaxis]
D = np.hstack((x*x, x*y, y*y, x, y, np.ones_like(x)))
S = np.dot(D.T,D)
C = np.zeros([6,6])
C[0,2] = C[2,0] = 2; C[1,1] = -1
E, V = eig(np.dot(inv(S), C))
n = np.argmax(np.abs(E))
a = V[:,n]
return a

def ellipse_center(a):
b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0]
num = b*b-a*c
x0=(c*d-b*f)/num
y0=(a*f-b*d)/num
return np.array([x0,y0])

def ellipse_angle_of_rotation( a ):
b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0]
return 0.5*np.arctan(2*b/(a-c))

def ellipse_axis_length( a ):
b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0]
up = 2*(a*f*f+c*d*d+g*b*b-2*b*d*f-a*c*g)
down1=(b*b-a*c)*( (c-a)*np.sqrt(1+4*b*b/((a-c)*(a-c)))-(c+a))
down2=(b*b-a*c)*( (a-c)*np.sqrt(1+4*b*b/((a-c)*(a-c)))-(c+a))
res1=np.sqrt(up/down1)
res2=np.sqrt(up/down2)
return np.array([res1, res2])

if __name__ == '__main__':

points = [(560036.4495758876, 6362071.890493258),
(560036.4495758876, 6362070.890493258),
(560036.9495758876, 6362070.890493258),
(560036.9495758876, 6362070.390493258),
(560037.4495758876, 6362070.390493258),
(560037.4495758876, 6362064.890493258),
(560036.4495758876, 6362064.890493258),
(560036.4495758876, 6362063.390493258),
(560035.4495758876, 6362063.390493258),
(560035.4495758876, 6362062.390493258),
(560034.9495758876, 6362062.390493258),
(560034.9495758876, 6362061.390493258),
(560032.9495758876, 6362061.390493258),
(560032.9495758876, 6362061.890493258),
(560030.4495758876, 6362061.890493258),
(560030.4495758876, 6362061.390493258),
(560029.9495758876, 6362061.390493258),
(560029.9495758876, 6362060.390493258),
(560029.4495758876, 6362060.390493258),
(560029.4495758876, 6362059.890493258),
(560028.9495758876, 6362059.890493258),
(560028.9495758876, 6362059.390493258),
(560028.4495758876, 6362059.390493258),
(560028.4495758876, 6362058.890493258),
(560027.4495758876, 6362058.890493258),
(560027.4495758876, 6362058.390493258),
(560026.9495758876, 6362058.390493258),
(560026.9495758876, 6362057.890493258),
(560025.4495758876, 6362057.890493258),
(560025.4495758876, 6362057.390493258),
(560023.4495758876, 6362057.390493258),
(560023.4495758876, 6362060.390493258),
(560023.9495758876, 6362060.390493258),
(560023.9495758876, 6362061.890493258),
(560024.4495758876, 6362061.890493258),
(560024.4495758876, 6362063.390493258),
(560024.9495758876, 6362063.390493258),
(560024.9495758876, 6362064.390493258),
(560025.4495758876, 6362064.390493258),
(560025.4495758876, 6362065.390493258),
(560025.9495758876, 6362065.390493258),
(560025.9495758876, 6362065.890493258),
(560026.4495758876, 6362065.890493258),
(560026.4495758876, 6362066.890493258),
(560026.9495758876, 6362066.890493258),
(560026.9495758876, 6362068.390493258),
(560027.4495758876, 6362068.390493258),
(560027.4495758876, 6362068.890493258),
(560027.9495758876, 6362068.890493258),
(560027.9495758876, 6362069.390493258),
(560028.4495758876, 6362069.390493258),
(560028.4495758876, 6362069.890493258),
(560033.4495758876, 6362069.890493258),
(560033.4495758876, 6362070.390493258),
(560033.9495758876, 6362070.390493258),
(560033.9495758876, 6362070.890493258),
(560034.4495758876, 6362070.890493258),
(560034.4495758876, 6362071.390493258),
(560034.9495758876, 6362071.390493258),
(560034.9495758876, 6362071.890493258),
(560036.4495758876, 6362071.890493258)]


a_points = np.array(points)
x = a_points[:, 0]
y = a_points[:, 1]
from pylab import *
plot(x,y)
show()
a = fitEllipse(x,y)
center = ellipse_center(a)
phi = ellipse_angle_of_rotation(a)
axes = ellipse_axis_length(a)

print "center = ", center
print "angle of rotation = ", phi
print "axes = ", axes

from pylab import *
plot(x,y)
plot(center[0:1],center[1:], color = 'red')
show()

每个顶点是一个 xi,y,i 点 enter image description here

二维点和拟合椭圆中心的图 enter image description here

使用 OpenCV 我得到以下结果:

import cv

PointArray2D32f = cv.CreateMat(1, len(points), cv.CV_32FC2)
for (i, (x, y)) in enumerate(points):
PointArray2D32f[0, i] = (x, y)
# Fits ellipse to current contour.
(center, size, angle) = cv.FitEllipse2(PointArray2D32f)

(center, size, angle)
((560030.625, 6362066.5),(10.480490684509277, 17.20206642150879),144.34889221191406)

最佳答案

fitEllipse 的计算返回虚假结果,因为 xy 的值与值之间的差异相比非常大。例如,如果您尝试打印特征值 E,您会看到

array([  0.00000000e+00 +0.00000000e+00j,
0.00000000e+00 +0.00000000e+00j,
0.00000000e+00 +0.00000000e+00j,
-1.36159790e-12 +8.15049878e-12j,
-1.36159790e-12 -8.15049878e-12j, 1.18685632e-11 +0.00000000e+00j])

它们几乎都是零!显然这里存在某种数值上的不准确。

您可以通过将数据的均值移近零来解决此问题,以便值的大小更“正常”并且数字之间的差异变得更显着。

x = a_points[:, 0]
y = a_points[:, 1]
xmean = x.mean()
ymean = y.mean()
x = x-xmean
y = y-ymean

然后就可以成功找到圆心、phi和坐标轴,然后将圆心重新平移回(xmean, ymean):

center = ellipse_center(a)
center[0] += xmean
center[1] += ymean

import numpy as np
import numpy.linalg as linalg
import matplotlib.pyplot as plt

def fitEllipse(x,y):
x = x[:,np.newaxis]
y = y[:,np.newaxis]
D = np.hstack((x*x, x*y, y*y, x, y, np.ones_like(x)))
S = np.dot(D.T,D)
C = np.zeros([6,6])
C[0,2] = C[2,0] = 2; C[1,1] = -1
E, V = linalg.eig(np.dot(linalg.inv(S), C))
n = np.argmax(np.abs(E))
a = V[:,n]
return a

def ellipse_center(a):
b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0]
num = b*b-a*c
x0=(c*d-b*f)/num
y0=(a*f-b*d)/num
return np.array([x0,y0])

def ellipse_angle_of_rotation( a ):
b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0]
return 0.5*np.arctan(2*b/(a-c))

def ellipse_axis_length( a ):
b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0]
up = 2*(a*f*f+c*d*d+g*b*b-2*b*d*f-a*c*g)
down1=(b*b-a*c)*( (c-a)*np.sqrt(1+4*b*b/((a-c)*(a-c)))-(c+a))
down2=(b*b-a*c)*( (a-c)*np.sqrt(1+4*b*b/((a-c)*(a-c)))-(c+a))
res1=np.sqrt(up/down1)
res2=np.sqrt(up/down2)
return np.array([res1, res2])

def find_ellipse(x, y):
xmean = x.mean()
ymean = y.mean()
x -= xmean
y -= ymean
a = fitEllipse(x,y)
center = ellipse_center(a)
center[0] += xmean
center[1] += ymean
phi = ellipse_angle_of_rotation(a)
axes = ellipse_axis_length(a)
x += xmean
y += ymean
return center, phi, axes

if __name__ == '__main__':

points = [(560036.4495758876, 6362071.890493258),
(560036.4495758876, 6362070.890493258),
(560036.9495758876, 6362070.890493258),
(560036.9495758876, 6362070.390493258),
(560037.4495758876, 6362070.390493258),
(560037.4495758876, 6362064.890493258),
(560036.4495758876, 6362064.890493258),
(560036.4495758876, 6362063.390493258),
(560035.4495758876, 6362063.390493258),
(560035.4495758876, 6362062.390493258),
(560034.9495758876, 6362062.390493258),
(560034.9495758876, 6362061.390493258),
(560032.9495758876, 6362061.390493258),
(560032.9495758876, 6362061.890493258),
(560030.4495758876, 6362061.890493258),
(560030.4495758876, 6362061.390493258),
(560029.9495758876, 6362061.390493258),
(560029.9495758876, 6362060.390493258),
(560029.4495758876, 6362060.390493258),
(560029.4495758876, 6362059.890493258),
(560028.9495758876, 6362059.890493258),
(560028.9495758876, 6362059.390493258),
(560028.4495758876, 6362059.390493258),
(560028.4495758876, 6362058.890493258),
(560027.4495758876, 6362058.890493258),
(560027.4495758876, 6362058.390493258),
(560026.9495758876, 6362058.390493258),
(560026.9495758876, 6362057.890493258),
(560025.4495758876, 6362057.890493258),
(560025.4495758876, 6362057.390493258),
(560023.4495758876, 6362057.390493258),
(560023.4495758876, 6362060.390493258),
(560023.9495758876, 6362060.390493258),
(560023.9495758876, 6362061.890493258),
(560024.4495758876, 6362061.890493258),
(560024.4495758876, 6362063.390493258),
(560024.9495758876, 6362063.390493258),
(560024.9495758876, 6362064.390493258),
(560025.4495758876, 6362064.390493258),
(560025.4495758876, 6362065.390493258),
(560025.9495758876, 6362065.390493258),
(560025.9495758876, 6362065.890493258),
(560026.4495758876, 6362065.890493258),
(560026.4495758876, 6362066.890493258),
(560026.9495758876, 6362066.890493258),
(560026.9495758876, 6362068.390493258),
(560027.4495758876, 6362068.390493258),
(560027.4495758876, 6362068.890493258),
(560027.9495758876, 6362068.890493258),
(560027.9495758876, 6362069.390493258),
(560028.4495758876, 6362069.390493258),
(560028.4495758876, 6362069.890493258),
(560033.4495758876, 6362069.890493258),
(560033.4495758876, 6362070.390493258),
(560033.9495758876, 6362070.390493258),
(560033.9495758876, 6362070.890493258),
(560034.4495758876, 6362070.890493258),
(560034.4495758876, 6362071.390493258),
(560034.9495758876, 6362071.390493258),
(560034.9495758876, 6362071.890493258),
(560036.4495758876, 6362071.890493258)]

fig, axs = plt.subplots(2, 1, sharex = True, sharey = True)
a_points = np.array(points)
x = a_points[:, 0]
y = a_points[:, 1]
axs[0].plot(x,y)
center, phi, axes = find_ellipse(x, y)
print "center = ", center
print "angle of rotation = ", phi
print "axes = ", axes

axs[1].plot(x, y)
axs[1].scatter(center[0],center[1], color = 'red', s = 100)
axs[1].set_xlim(x.min(), x.max())
axs[1].set_ylim(y.min(), y.max())

plt.show()

enter image description here

关于python - 给定一组点 xi=(xi,yi) 在 Python 中拟合一个椭圆,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13635528/

26 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com