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Emulating Excel's "scatter with smooth curve" spline function in Matplotlib for 3 points


I'm trying to emulate Excel's

Insert>Scatter>Scatter with smooth lines and markers

command in Matplotlib

The scipy function interpolate creates a similar effect, with some nice examples of how to simply implement this here: How to draw cubic spline in matplotlib

However Excel's spline algorithm is also able to generate a smooth curve through just three points (e.g. x = [0,1,2] y = [4,2,1]); and it isn't possible to do this with cubic splines.

I have seen discussions that suggest that the Excel algorithm uses Catmull-Rom splines; but don't really understand these, or how they could be adapted to Matplotlib: http://answers.microsoft.com/en-us/office/forum/office_2007-excel/how-does-excel-plot-smooth-curves/c751e8ff-9f99-4ac7-a74a-fba41ac80300

Is there a simple way of modifying the above examples to achieve smooth curves through three or more points using the interpolate library?

Many thanks


Solution

  • By now you may have found the Wikipedia page for the Centripetal Catmull-Rom spline, but in case you haven't, it includes this sample code:

    import numpy
    import matplotlib.pyplot as plt
    
    def CatmullRomSpline(P0, P1, P2, P3, 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 = 0.5
      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):
      """
      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])
        C.extend(c)
    
      return C
    

    which nicely computes the interpolation for n >= 4 points like so:

    points = [[0,1.5],[2,2],[3,1],[4,0.5],[5,1],[6,2],[7,3]]
    c = CatmullRomChain(points)
    px, py = zip(*points)
    x, y = zip(*c)
    
    plt.plot(x, y)
    plt.plot(px, py, 'or')
    

    resulting in this matplotlib image:

    enter image description here

    Update:

    Alternatively, there is a scipy.interpolate function for BarycentricInterpolator that appears to do what you're looking for. It is rather straightforward to use and works for cases in which you have only 3 data points.

    from scipy.interpolate import BarycentricInterpolator
    
    # create some data points
    points1 = [[0, 2], [1, 4], [2, -2], [3, 6], [4, 2]]
    points2 = [[1, 1], [2, 5], [3, -1]]
    
    # put data into x, y tuples
    x1, y1 =zip(*points1)
    x2, y2 = zip(*points2)
    
    # create the interpolator
    bci1 = BarycentricInterpolator(x1, y1)
    bci2 = BarycentricInterpolator(x2, y2)
    
    # define dense x-axis for interpolating over
    x1_new = np.linspace(min(x1), max(x1), 1000)
    x2_new = np.linspace(min(x2), max(x2), 1000)
    
    # plot it all
    plt.plot(x1, y1, 'o')
    plt.plot(x2, y2, 'o')
    plt.plot(x1_new, bci1(x1_new))
    plt.plot(x2_new, bci2(x2_new))
    plt.xlim(-1, 5)
    

    enter image description here

    Update 2

    Another option within scipy is akima interpolation via Akima1DInterpolator. It is as easy to implement as Barycentric, but has the advantage that it avoids large oscillations at the edge of a data set. Here's a few test cases that exhibit all the criteria you've asked for so far.

    from scipy.interpolate import Akima1DInterpolator
    
    x1, y1 = np.arange(13), np.random.randint(-10, 10, 13)
    x2, y2 = [0,2,3,6,12], [100,50,30,18,14]
    x3, y3 = [4, 6, 8], [60, 80, 40]
    
    akima1 = Akima1DInterpolator(x1, y1)
    akima2 = Akima1DInterpolator(x2, y2)
    akima3 = Akima1DInterpolator(x3, y3)
    
    x1_new = np.linspace(min(x1), max(x1), 1000)
    x2_new = np.linspace(min(x2), max(x2), 1000)
    x3_new = np.linspace(min(x3), max(x3), 1000)
    
    plt.plot(x1, y1, 'bo')
    plt.plot(x2, y2, 'go')
    plt.plot(x3, y3, 'ro')
    plt.plot(x1_new, akima1(x1_new), 'b', label='random points')
    plt.plot(x2_new, akima2(x2_new), 'g', label='exponential')
    plt.plot(x3_new, akima3(x3_new), 'r', label='3 points')
    plt.xlim(-1, 15)
    plt.ylim(-10, 110)
    plt.legend(loc='best')
    

    enter image description here