在Matplotlib中模拟Excel的“平滑曲线散射”样条函数3个点

Question

我正在尝试模仿Excel的

插入＆gt; Scatter＆gt; Scatter，带有平滑的线条和标记

Matplotlib中的

命令

scipy函数 interpolate 会产生类似的效果，其中有一些很好的示例，说明如何在此处简单实现： How to draw cubic spline in matplotlib

然而，Excel的样条算法也能够通过三个点生成平滑曲线（例如x = [0,1,2] y = [4,2,1]）;并且不可能使用三次样条函数来完成此操作。

我见过的讨论表明Excel算法使用Catmull-Rom样条曲线;但是不要真正理解这些，或者它们如何适应Matplotlib： http://answers.microsoft.com/en-us/office/forum/office_2007-excel/how-does-excel-plot-smooth-curves/c751e8ff-9f99-4ac7-a74a-fba41ac80300

是否有一种简单的方法可以使用插入库修改上述示例以通过三个或更多点实现平滑曲线？

非常感谢

Answer 1

到目前为止，您可能已找到Centripetal Catmull-Rom spline的维基百科页面，但如果您还没有，则包含以下示例代码：

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

很好地计算n >= 4点的插值，如下所示：

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')

生成此matplotlib图片：

更新

或者，BarycentricInterpolator有一个scipy.interpolate功能，似乎可以执行您正在寻找的功能。它非常简单易用，适用于只有3个数据点的情况。

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)

更新2

scipy中的另一个选项是通过Akima1DInterpolator进行akima插值。它与Barycentric一样容易实现，但其优点是避免了数据集边缘的大振荡。这里列出了一些测试用例，它们展示了您目前所要求的所有标准。

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')

Answer 2

@Lanery：回复：更新2：最好的变得更好！

不得不将列表x2，y2，x3，y3重新定义为numpy数组，以使您的示例在我的系统上运行（Spyder / Python 2.7）：

x2 = np.array([0,2,3,6,12])
y2 = np.array([100,50,30,18,14])
x3 = np.array([4, 6, 8])
y3 = np.array([60, 80, 40])

但现在就像梦一样！非常感谢您的专业知识和明确的解释。