Python中的Catmull-Rom样条

Question

有没有什么库或者函数可以用来在Python中计算Catmull-Rom样条曲线，给定三个点？

我最终想要的是沿着样条曲线的一些点的x和y坐标，要求这些点在样条曲线上的间距是固定的，比如说样条曲线的长度是3个单位，我想要在长度为0、1、2和3的地方的x和y坐标。

其实没什么特别的。我自己在写这个，但如果你找到什么好的方法，那就太好了，可以用来测试或者节省时间。

Answer 1

这里有一个链接：jj_catmull，看起来是用Python写的，也许你可以在那找到你需要的东西。

Answer 2

如之前提到的，使用Catmull-Rom插值需要4个点，而端点的问题比较棘手。我在考虑自己应用这些方法，而不是使用自然三次样条（因为在我的应用中，超出已知数据范围的情况是不实际的）。和@denis的代码类似，这里有一些可能对你有帮助的内容（注意几点：1）这段代码只是随机生成点，我相信你可以参考注释掉的示例来使用你自己的数据。2）我创建了扩展的端点，保持了第一个和最后两个点之间的斜率——使用了一个任意的距离，设为域的1%。3）我包括了均匀、向心和弦长参数化以供比较）：

# coding: utf-8

# In[1]:

import numpy
import matplotlib.pyplot as plt
get_ipython().magic(u'pylab inline')


# In[2]:

def CatmullRomSpline(P0, P1, P2, P3, a, nPoints=100):
  """
  P0, P1, P2, and P3 should be (x,y) point pairs that define the Catmull-Rom spline.
  nPoints is the number of points to include in this curve segment.
  """
  # Convert the points to numpy so that we can do array multiplication
  P0, P1, P2, P3 = map(numpy.array, [P0, P1, P2, P3])

  # Calculate t0 to t4
  alpha = a
  def tj(ti, Pi, Pj):
    xi, yi = Pi
    xj, yj = Pj
    return ( ( (xj-xi)**2 + (yj-yi)**2 )**0.5 )**alpha + ti

  t0 = 0
  t1 = tj(t0, P0, P1)
  t2 = tj(t1, P1, P2)
  t3 = tj(t2, P2, P3)

  # Only calculate points between P1 and P2
  t = numpy.linspace(t1,t2,nPoints)

  # Reshape so that we can multiply by the points P0 to P3
  # and get a point for each value of t.
  t = t.reshape(len(t),1)

  A1 = (t1-t)/(t1-t0)*P0 + (t-t0)/(t1-t0)*P1
  A2 = (t2-t)/(t2-t1)*P1 + (t-t1)/(t2-t1)*P2
  A3 = (t3-t)/(t3-t2)*P2 + (t-t2)/(t3-t2)*P3

  B1 = (t2-t)/(t2-t0)*A1 + (t-t0)/(t2-t0)*A2
  B2 = (t3-t)/(t3-t1)*A2 + (t-t1)/(t3-t1)*A3

  C  = (t2-t)/(t2-t1)*B1 + (t-t1)/(t2-t1)*B2
  return C

def CatmullRomChain(P,alpha):
  """
  Calculate Catmull Rom for a chain of points and return the combined curve.
  """
  sz = len(P)

  # The curve C will contain an array of (x,y) points.
  C = []
  for i in range(sz-3):
    c = CatmullRomSpline(P[i], P[i+1], P[i+2], P[i+3],alpha)
    C.extend(c)

  return C


# In[139]:

# Define a set of points for curve to go through
Points = numpy.random.rand(10,2)
#Points=array([array([153.01,722.67]),array([152.73,699.92]),array([152.91,683.04]),array([154.6,643.45]),
#        array([158.07,603.97])])
#Points = array([array([0,92.05330318]),
#               array([2.39580622,29.76345192]),
#               array([10.01564963,16.91470591]),
#               array([15.26219886,71.56301997]),
#               array([15.51234733,73.76834447]),
#               array([24.88468545,50.89432899]),
#               array([27.83934153,81.1341789]),
#               array([36.80443404,56.55810783]),
#               array([43.1404725,16.96946811]),
#               array([45.27824599,15.75903418]),
#               array([51.58871027,90.63583215])])

x1=Points[0][0]
x2=Points[1][0]
y1=Points[0][1]
y2=Points[1][1]
x3=Points[-2][0]
x4=Points[-1][0]
y3=Points[-2][1]
y4=Points[-1][1]
dom=max(Points[:,0])-min(Points[:,0])
rng=max(Points[:,1])-min(Points[:,1])
pctdom=1
pctdom=float(pctdom)/100
prex=x1+sign(x1-x2)*dom*pctdom
prey=(y1-y2)/(x1-x2)*(prex-x1)+y1
endx=x4+sign(x4-x3)*dom*pctdom
endy=(y4-y3)/(x4-x3)*(endx-x4)+y4
print len(Points)
Points=list(Points)
Points.insert(0,array([prex,prey]))
Points.append(array([endx,endy]))
print len(Points)


# In[140]:

#Define alpha
a=0.

# Calculate the Catmull-Rom splines through the points
c = CatmullRomChain(Points,a)

# Convert the Catmull-Rom curve points into x and y arrays and plot
x,y = zip(*c)
plt.plot(x,y,c='green',zorder=10)

a=0.5
c = CatmullRomChain(Points,a)
x,y = zip(*c)
plt.plot(x,y,c='blue')

a=1.
c = CatmullRomChain(Points,a)
x,y = zip(*c)
plt.plot(x,y,c='red')

# Plot the control points
px, py = zip(*Points)
plt.plot(px,py,'o',c='black')

plt.grid(b=True)
plt.show()


# In[141]:

Points


# In[104]:

Answer 3

3个点？Catmull-Rom曲线是由4个点定义的，分别是p_1、p0、p1和p2；这条三次曲线是从p0到p1的，而外面的点p_1和p2则决定了在p0和p1的斜率。要在一个点数组P中绘制一条曲线，可以这样做：

for j in range( 1, len(P)-2 ):  # skip the ends
    for t in range( 10 ):  # t: 0 .1 .2 .. .9
        p = spline_4p( t/10, P[j-1], P[j], P[j+1], P[j+2] )
        # draw p

def spline_4p( t, p_1, p0, p1, p2 ):
    """ Catmull-Rom
        (Ps can be numpy vectors or arrays too: colors, curves ...)
    """
        # wikipedia Catmull-Rom -> Cubic_Hermite_spline
        # 0 -> p0,  1 -> p1,  1/2 -> (- p_1 + 9 p0 + 9 p1 - p2) / 16
    # assert 0 <= t <= 1
    return (
          t*((2-t)*t - 1)   * p_1
        + (t*t*(3*t - 5) + 2) * p0
        + t*((4 - 3*t)*t + 1) * p1
        + (t-1)*t*t         * p2 ) / 2

你也可以通过3个点使用分段二次曲线——可以参考Dodgson的文章，关于图像重采样的二次插值。你到底想要做什么呢？