Catmull-Rom splines in python

Question

Is there a library or function in python to compute Catmull-Rom spline from three points ?

What I need in the end are the x,y coordinates of points along the spline, provided that they are always equidistant of a given amount t along the spline (say, the spline curve is 3 units long and I want the x,y coordinates at spline length 0, 1, 2 and 3)

Nothing really exciting. I am writing it by myself, but if you find something nice, It would be great for testing (or to save time)

Answer 1

3 points ? Catmull-Rom is defined for 4 points, say p_1 p0 p1 p2; a cubic curve goes from p0 to p1, and outer points p_1 and p2 determine the slopes at p0 and p1. To draw a curve through some points in an array P, do something like this:

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

One can use piecewise quadratic curves through 3 points -- see Dodgson, Quadratic Interpolation for Image Resampling. What do you really want to do ?

Answer 2

As mentioned previously you do need 4 points for catmull-rom, and the endpoints are an issue. I was looking at applying these myself instead of natural cubic splines (which the potential overshoots beyond the known data range is impractical in my application). Similar to @denis's code, here is something that might help you (notice a few things. 1) That code just randomly generates points, I'm sure you can use the commented out examples for how to use your own data. 2) I create extended endpoints, preserving the slope between the first/last two points - using an arbitrary distance of 1% of the domain. 3)I include uniform, centripetal, and chordial knot parameterization for comparison):

# 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]: