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具有多段三次贝塞尔曲线和距离以及曲率约束的近似数据

具有多段三次贝塞尔曲线和距离以及曲率约束的近似数据

我找到了满足我的要求的解决方案。解决方案是,首先找到一个B样条曲线,该样条线在最小二乘方意义上近似点,然后将该样条曲线转换为多段贝塞尔曲线。B样条线确实具有以下优点:与贝塞尔曲线相反,它们不会通过控制点,并且提供了一种方法来指定近似曲线的所需“平滑度”。生成此类样条曲线所需的功能在scipy提供python绑定的FITPACK库中实现。让我们假设我读我的数据导入列表xy的话,我可以这样做:

import matplotlib.pyplot as plt
import numpy as np
from scipy import interpolate
tck,u = interpolate.splprep([x,y],s=3)
unew = np.arange(0,1.01,0.01)
out = interpolate.splev(unew,tck)
plt.figure()
plt.plot(x,y,out[0],out[1])
plt.show()

如果希望曲线更平滑,则可以将s参数增加splprep。如果我希望近似值更接近数据,则可以减小s参数以减少平滑度。通过以s编程方式遍历多个参数,我可以找到适合给定要求的良好参数。

不过,问题是如何将结果转换为贝塞尔曲线。Zachary Pincus在此电子邮件中的答案。我将在此处复制他的解决方案,以完整回答我的问题:

def b_spline_to_bezier_series(tck, per = False):
  """Convert a parametric b-spline into a sequence of Bezier curves of the same degree.

  Inputs:
    tck : (t,c,k) tuple of b-spline knots, coefficients, and degree returned by splprep.
    per : if tck was created as a periodic spline, per *must* be true, else per *must* be false.

  Output:
    A list of Bezier curves of degree k that is equivalent to the input spline. 
    Each Bezier curve is an array of shape (k+1,d) where d is the dimension of the
    space; thus the curve includes the starting point, the k-1 internal control 
    points, and the endpoint, where each point is of d dimensions.
  """
  from fitpack import insert
  from numpy import asarray, unique, split, sum
  t,c,k = tck
  t = asarray(t)
  try:
    c[0][0]
  except:
    # I can't figure out a simple way to convert nonparametric splines to 
    # parametric splines. Oh well.
    raise TypeError("Only parametric b-splines are supported.")
  new_tck = tck
  if per:
    # ignore the leading and trailing k knots that exist to enforce periodicity 
    knots_to_consider = unique(t[k:-k])
  else:
    # the first and last k+1 knots are identical in the non-periodic case, so
    # no need to consider them when increasing the knot multiplicities below
    knots_to_consider = unique(t[k+1:-k-1])
  # For each unique knot, bring it's multiplicity up to the next multiple of k+1
  # This removes all continuity constraints between each of the original knots, 
  # creating a set of independent Bezier curves.
  desired_multiplicity = k+1
  for x in knots_to_consider:
    current_multiplicity = sum(t == x)
    remainder = current_multiplicity%desired_multiplicity
    if remainder != 0:
      # add enough knots to bring the current multiplicity up to the desired multiplicity
      number_to_insert = desired_multiplicity - remainder
      new_tck = insert(x, new_tck, number_to_insert, per)
  tt,cc,kk = new_tck
  # strip off the last k+1 knots, as they are redundant after knot insertion
  bezier_points = numpy.transpose(cc)[:-desired_multiplicity]
  if per:
    # again, ignore the leading and trailing k knots
    bezier_points = bezier_points[k:-k]
  # group the points into the desired bezier curves
  return split(bezier_points, len(bezier_points) / desired_multiplicity, axis = 0)

因此B样条曲线,FITPACK,numpy和scipy挽救了我的一天:)

其他 2022/1/1 18:29:13 有488人围观

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