Python 給定隱式方程，找到那個方程上的點？

Question

上下文：將.iges轉換為.vtk 。

我有以下等式Ax^2+Bxy+Cy^2+Dx+Ey+F=0代表圓錐曲線。

給出了參數A~F 。 我想在圓錐截面上找到點，這樣我就可以用線把它們連接起來，然后做一個網格。

我需要這些點而不是僅僅使用matplotlib Ellipse的原因是因為我正在創建一個網格而不是 plot。

它在 3 維空間中，但我首先在 xy 平面上得到點，然后使用仿射變換將其發送到 3 維空間。

問題： How do I find points given an implicit equation?

Answer 1

為了避免在這方面花費太多時間，我編寫了一些似乎可以處理一般省略號的代碼。 根據需要，它可以擴展為其他圓錐曲線。 該代碼采用橢圓的一般二次方程的系數和要在橢圓上生成的多個所需點，並在橢圓上生成一組點。

import numpy as np

def equation(conic, points):
     '''
     equation of a conic with coefficients 'conic' 
     applied to a matrix number_of_points x 3 whose each row is the coordinates
     of each point
     '''
     c = np.array(conic)
     x = np.array([points[:,0]**2, points[:, 0]*points[:,1], points[:,1]**2, points[:,0], points[:,1], np.ones(points.shape[0])])
     return c.dot(x)
 
def equation_to_matrix(eq):
    '''
    eq[0]*x**2 + eq[1]*x*y + eq[2]*y**2 + eq[3]*x + eq[4]*y + eq[5] = 0
    '''
    return np.array([[2*eq[0],   eq[1],   eq[3]],
                     [  eq[1], 2*eq[2],   eq[4]],
                     [  eq[3],   eq[4], 2*eq[5]]]) / 2

def solve_quadratic(a, b, c):
    '''
    solves
    ax^2 + bx + c = 0
    '''
    D = b**2 - 4*a*c
    D = np.sqrt(D)
    return (-b-D)/(2*a), (-b+D)/(2*a)

def eigen2(S):
    '''
    solves the eigen-decomposition problem
    for a 2x2 symmetric matrix
    '''
    k1, k2 = solve_quadratic(1, -S[0,0]-S[1,1], S[0,0]*S[1,1] - S[0,1]*S[1,0])
    u1 = np.array([-S[0,1], S[0,0]-k1, 0])
    u1 = u1 / np.sqrt(u1.dot(u1))
    u2 = np.array([-u1[1], u1[0], 0])
    return np.array([k1, k2]), np.array([u1, u2, np.array([0,0,1])]).T

def center(conic_matrix):
    center = np.linalg.solve(conic_matrix, np.array([0,0,1]))
    return center/center[2]
    
def find_rotation_and_translation(conic_matrix):
    '''
    conic = c[0]x^2 + c[1]*xy + c[2]*y^2 + c[3]*x + c[4]*y + c[5] = 0
    the result is rotation U such that U.T C U = diag
    '''
    k, U = eigen2(conic_matrix)
    U[:,2] = center(conic_matrix)
    return U, k

def find_transform(conic):
    C = equation_to_matrix(conic)
    U, k = find_rotation_and_translation(C)
    C = (U.T).dot(C.dot(U))
    C = - C / C[2,2]
    k = np.array([1/np.sqrt(C[0,0]), 1/np.sqrt(C[1,1]), 1])
    return U.dot(np.diag(k))

def generate_points_on(conic, num_points):
    '''
    conic = [c[0], c[1], c[2], c[3], c[4], c[5]]
    coefficients of the qudaratic equation:
    conic: c[0]x^2 + c[1]*xy + c[2]*y^2 + c[3]*x + c[4]*y + c[5] = 0
    result is the affine transformation (scaling, rotation, translation)
    that maps the unit circle to the ellipse defined by the coefficients
    'conic' 
    '''
    cos_ = np.cos(2*np.pi* np.arange(0, num_points)/ num_points)
    sin_ = np.sin(2*np.pi* np.arange(0, num_points)/ num_points)
    U = find_transform(conic)
    points = np.array([cos_, sin_, np.ones(num_points)])
    return ((U.dot(points)).T)[:,[0,1]]

'''
Test:
'''

'''
Ellipse with equation whose coefficients are in the list E.
The ellipse has semi-major axes 2 and 1,
it is rotated 60 deg from the horizontal,
and its center is at (1, 4)
'''   
E = [ 3.25,  -2.59807621, 1.75, -23.40192379, 6.89230485,   39.35769515]


'''
U maps points from unit circle to points on E
'''
U = find_transform(E)

print(U)

'''
the set of points on the ellipse E
'''
p = generate_points_on(E, num_points = 20)
print(p)


'''
check that the points p lie on the ellipse E
'''
print(equation(E, p).round(10))

'''
plot
'''
fig = plt.figure()
ax = fig.add_subplot()
ax.plot(p[:,0], p[:,1], 'ro')
ax.set_aspect('equal')
plt.show()

Answer 2

下面的代碼處理雙曲線的情況。 它在很大程度上改編了這里的代碼

import numpy as np
import matplotlib.pyplot as plt

def equation_to_matrix(eq):
    '''
    eq[0]*x**2 + eq[1]*x*y + eq[2]*y**2 + eq[3]*x + eq[4]*y + eq[5] = 0
    '''
    return np.array([[2*eq[0],   eq[1],   eq[3]],
                      [  eq[1], 2*eq[2],   eq[4]],
                      [  eq[3],   eq[4], 2*eq[5]]]) / 2

def hyp_params_from_general(coeffs):
    # get the matrix of the quadratic equation
    Aq = equation_to_matrix(coeffs)

    # get the matrix of the quadratic form A33
    A33 = Aq[:2, :2]

    # determinant of A33
    detA33 = np.linalg.det(A33)

    if detA33 > 0:
        raise ValueError('coeffs do not represent a hyperbola: det A33 must be negative!')

    # get the center
    x0 = -np.linalg.det(np.array([Aq[:2, 2], Aq[:2, 1]]).T) / detA33
    y0 = -np.linalg.det(np.array([Aq[:2, 0], Aq[:2, 2]]).T) / detA33

    # The semi-major and semi-minor axis lengths (these are not sorted).
    # get discriminant of the conic section
    delta = np.linalg.det(Aq)

    # get the eigenvalues
    k1, k2 = np.linalg.eigvals(A33)

    k1isk2 = np.isclose(k1/k2, -1)

    ap = np.sqrt(abs(delta/k1/detA33))
    bp = np.sqrt(abs(delta/k2/detA33))

    # Eccentricity.
    fac = np.sqrt((Aq[0, 0] - Aq[1, 1])**2 + Aq[0, 1]**2)
    if delta < 0:
        nu = 1
    else:
        nu = -1
    e = np.sqrt(2*fac/(nu*(Aq[0, 0] - Aq[1, 1]) + fac))

    # slope of the asymptotes
    if Aq[0, 0] == Aq[1, 1] and k1isk2:
        m1 = 0.
        m2 = np.nan
    else:
        m1 = Aq[0, 0]/(-Aq[0, 1] - np.sqrt(-detA33))
        m2 = Aq[0, 0]/(-Aq[0, 1] + np.sqrt(-detA33))

    # Sort the semi-major and semi-minor axis lengths but keep track of
    # the original relative magnitudes of width and height.
    width_gt_height = True
    if ap < bp and not k1isk2:
        width_gt_height = False
        ap, bp = bp, ap

    # The angle of anticlockwise rotation of the major-axis from x-axis.
    if Aq[0, 1] == 0:
        phi = 0 if Aq[0, 0] < Aq[1, 1] else np.pi/2
    elif Aq[0, 0] == Aq[1, 1]:
        phi = np.pi/4 # would divide by zero and arctan(inf) -> pi/4
        if m1 > 0 and m2 > 0:
            width_gt_height = True
    else:# Aq[0, 0] > Aq[1, 1]:
        phi = np.arctan(2*Aq[0, 1]/(Aq[0, 0] - Aq[1, 1])) / 2

    if not width_gt_height:
        # Ensure that phi is the angle to rotate to the semi-major axis.
        phi += np.pi/2
    phi = phi % np.pi

    return x0, y0, ap, bp, phi, e, m1, m2, width_gt_height

def get_hyperbola_pts(params, npts=100, tmin=-1, tmax=1):
    x0, y0, ap, bp, phi, m1, m2 = params

    # A grid of the parametric variable, t.
    t = np.linspace(tmin, tmax, npts)

    # points
    x = x0 + ap * np.cosh(t) * np.cos(phi) - bp * np.sinh(t) * np.sin(phi)
    y = y0 + ap * np.cosh(t) * np.sin(phi) + bp * np.sinh(t) * np.cos(phi)

    # asymptotes
    ya1 = y0 + m1*(x - x0)
    ya2 = y0 + m2*(x - x0)

    return x, y, ya1, ya2

if __name__ == '__main__':
    coeffs = [1., 6., -2., 3., 0., 0.]

    x0, y0, ap, bp, phi, e, m1, m2, width_gt_height = hyp_params_from_general(coeffs)
    print('x0, y0, ap, bp, phi, e, m1, m2, width_gt_height = ', x0, y0, ap, bp, phi, e, m1, m2)

    x_, y_, ya1, ya2 = get_hyperbola_pts((x0, y0, ap, bp, phi, m1, m2), npts=250, tmin=-2, tmax=3)

    fig, ax = plt.subplots(figsize=(16, 9))
    ax.plot(x_, y_, marker='.', linewidth=0.5, c='r')
    ax.plot(x_, ya1, marker='.', linewidth=0.2, c='b')
    ax.plot(x_, ya2, marker='.', linewidth=0.2, c='b')
    ax.grid(True, linestyle='--')