Получение контрольных точек и манипуляторов кубической кривой Безье из серии точек

Ответом для моего случая была функция Безье best fit, которая принимает ввод значений точек, подгоняет точки к кубическому сплайну и выводит ручки Безье кривой, находя их коэффициенты.

Вот один из таких скриптов, fitCurves, который можно использовать следующим образом:

import numpy as np
from fitCurve import fitCurve
import matplotlib.pyplot as plt

y = [0.0,
            0.03771681353260319,
            0.20421680080883106,
            0.49896111463402026,
            0.7183501026981503,
            0.8481517096346528,
            0.9256128196832564,
            0.9705404287079152,
            0.9933297674379904,
            1.0]

x = np.linspace(0, 1, len(y))
pts = np.array([x,y]).T
bezier_handles = fitCurve(points=pts , maxError=20)
x_bez = []
y_bez = []
for bez in bezier_handles:
    for pt in bez:
      x_bez.append(pt[0])
      y_bez.append(pt[1])

plt.plot(pts[:,0], pts[:,1], 'bo-', label='Points')
plt.plot(x_bez[:2], y_bez[:2], 'ro--', label='Handle') # handle 1
plt.plot(x_bez[2:4], y_bez[2:4], 'ro--') # handle 2
plt.legend()
plt.show()

FitCurve.py

from numpy import *

""" Python implementation of
    Algorithm for Automatically Fitting Digitized Curves
    by Philip J. Schneider
    "Graphics Gems", Academic Press, 1990
"""

# evaluates cubic bezier at t, return point
def q(ctrlPoly, t):
    return (1.0-t)**3 * ctrlPoly[0] + 3*(1.0-t)**2 * t * ctrlPoly[1] + 3*(1.0-t)* t**2 * ctrlPoly[2] + t**3 * ctrlPoly[3]


# evaluates cubic bezier first derivative at t, return point
def qprime(ctrlPoly, t):
    return 3*(1.0-t)**2 * (ctrlPoly[1]-ctrlPoly[0]) + 6*(1.0-t) * t * (ctrlPoly[2]-ctrlPoly[1]) + 3*t**2 * (ctrlPoly[3]-ctrlPoly[2])


# evaluates cubic bezier second derivative at t, return point
def qprimeprime(ctrlPoly, t):
    return 6*(1.0-t) * (ctrlPoly[2]-2*ctrlPoly[1]+ctrlPoly[0]) + 6*(t) * (ctrlPoly[3]-2*ctrlPoly[2]+ctrlPoly[1])

# Fit one (ore more) Bezier curves to a set of points
def fitCurve(points, maxError):
    leftTangent = normalize(points[1] - points[0])
    rightTangent = normalize(points[-2] - points[-1])
    return fitCubic(points, leftTangent, rightTangent, maxError)


def fitCubic(points, leftTangent, rightTangent, error):
    # Use heuristic if region only has two points in it
    if (len(points) == 2):
        dist = linalg.norm(points[0] - points[1]) / 3.0
        bezCurve = [points[0], points[0] + leftTangent * dist, points[1] + rightTangent * dist, points[1]]
        return [bezCurve]

    # Parameterize points, and attempt to fit curve
    u = chordLengthParameterize(points)
    bezCurve = generateBezier(points, u, leftTangent, rightTangent)
    # Find max deviation of points to fitted curve
    maxError, splitPoint = computeMaxError(points, bezCurve, u)
    if maxError < error:
        return [bezCurve]

    # If error not too large, try some reparameterization and iteration
    if maxError < error**2:
        for i in range(20):
            uPrime = reparameterize(bezCurve, points, u)
            bezCurve = generateBezier(points, uPrime, leftTangent, rightTangent)
            maxError, splitPoint = computeMaxError(points, bezCurve, uPrime)
            if maxError < error:
                return [bezCurve]
            u = uPrime

    # Fitting failed -- split at max error point and fit recursively
    beziers = []
    centerTangent = normalize(points[splitPoint-1] - points[splitPoint+1])
    beziers += fitCubic(points[:splitPoint+1], leftTangent, centerTangent, error)
    beziers += fitCubic(points[splitPoint:], -centerTangent, rightTangent, error)

    return beziers


def generateBezier(points, parameters, leftTangent, rightTangent):
    bezCurve = [points[0], None, None, points[-1]]

    # compute the A's
    A = zeros((len(parameters), 2, 2))
    for i, u in enumerate(parameters):
        A[i][0] = leftTangent  * 3*(1-u)**2 * u
        A[i][1] = rightTangent * 3*(1-u)    * u**2

    # Create the C and X matrices
    C = zeros((2, 2))
    X = zeros(2)

    for i, (point, u) in enumerate(zip(points, parameters)):
        C[0][0] += dot(A[i][0], A[i][0])
        C[0][1] += dot(A[i][0], A[i][1])
        C[1][0] += dot(A[i][0], A[i][1])
        C[1][1] += dot(A[i][1], A[i][1])

        tmp = point - q([points[0], points[0], points[-1], points[-1]], u)

        X[0] += dot(A[i][0], tmp)
        X[1] += dot(A[i][1], tmp)

    # Compute the determinants of C and X
    det_C0_C1 = C[0][0] * C[1][1] - C[1][0] * C[0][1]
    det_C0_X  = C[0][0] * X[1] - C[1][0] * X[0]
    det_X_C1  = X[0] * C[1][1] - X[1] * C[0][1]

    # Finally, derive alpha values
    alpha_l = 0.0 if det_C0_C1 == 0 else det_X_C1 / det_C0_C1
    alpha_r = 0.0 if det_C0_C1 == 0 else det_C0_X / det_C0_C1

    # If alpha negative, use the Wu/Barsky heuristic (see text) */
    # (if alpha is 0, you get coincident control points that lead to
    # divide by zero in any subsequent NewtonRaphsonRootFind() call. */
    segLength = linalg.norm(points[0] - points[-1])
    epsilon = 1.0e-6 * segLength
    if alpha_l < epsilon or alpha_r < epsilon:
        # fall back on standard (probably inaccurate) formula, and subdivide further if needed.
        bezCurve[1] = bezCurve[0] + leftTangent * (segLength / 3.0)
        bezCurve[2] = bezCurve[3] + rightTangent * (segLength / 3.0)

    else:
        # First and last control points of the Bezier curve are
        # positioned exactly at the first and last data points
        # Control points 1 and 2 are positioned an alpha distance out
        # on the tangent vectors, left and right, respectively
        bezCurve[1] = bezCurve[0] + leftTangent * alpha_l
        bezCurve[2] = bezCurve[3] + rightTangent * alpha_r

    return bezCurve


def reparameterize(bezier, points, parameters):
    return [newtonRaphsonRootFind(bezier, point, u) for point, u in zip(points, parameters)]


def newtonRaphsonRootFind(bez, point, u):
    """
       Newton's root finding algorithm calculates f(x)=0 by reiterating
       x_n+1 = x_n - f(x_n)/f'(x_n)

       We are trying to find curve parameter u for some point p that minimizes
       the distance from that point to the curve. Distance point to curve is d=q(u)-p.
       At minimum distance the point is perpendicular to the curve.
       We are solving
       f = q(u)-p * q'(u) = 0
       with
       f' = q'(u) * q'(u) + q(u)-p * q''(u)

       gives
       u_n+1 = u_n - |q(u_n)-p * q'(u_n)| / |q'(u_n)**2 + q(u_n)-p * q''(u_n)|
    """
    d = q(bez, u)-point
    numerator = (d * qprime(bez, u)).sum()
    denominator = (qprime(bez, u)**2 + d * qprimeprime(bez, u)).sum()

    if denominator == 0.0:
        return u
    else:
        return u - numerator/denominator


def chordLengthParameterize(points):
    u = [0.0]
    for i in range(1, len(points)):
        u.append(u[i-1] + linalg.norm(points[i] - points[i-1]))

    for i, _ in enumerate(u):
        u[i] = u[i] / u[-1]

    return u


def computeMaxError(points, bez, parameters):
    maxDist = 0.0
    splitPoint = len(points)/2
    for i, (point, u) in enumerate(zip(points, parameters)):
        dist = linalg.norm(q(bez, u)-point)**2
        if dist > maxDist:
            maxDist = dist
            splitPoint = i

    return maxDist, splitPoint


def normalize(v):
    return v / linalg.norm(v)