在Matplotlib中模拟Excel的“带平滑曲线的散点图”样条函数，针对3个点。

Question

在Matplotlib中模拟Excel的“带平滑曲线的散点图”样条函数，针对3个点。

4

我正在尝试模拟Excel中的“插入>散点图>带平滑线和标记的散点图”命令，使用Matplotlib。

Scipy函数interpolate可以创建类似的效果，并提供了一些简单实现的好例子，可以在这里找到：如何在Matplotlib中绘制三次样条曲线然而，Excel的样条算法也能够通过仅三个点（例如x = [0,1,2] y = [4,2,1]）生成平滑曲线；而使用三次样条则无法实现此功能。

我看到有讨论表明Excel算法使用Catmull-Rom样条；但我并不真正理解这些，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修改上述示例，使用interpolate库实现通过三个或更多点的平滑曲线？

非常感谢

- Dave

2个回答

0

@Lanery：回复：更新2：更好的变得更好了！

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

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])

但现在它像梦一样运行！非常感谢您的专业知识和清晰的解释。

- DaveW

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- lanery · Accepted Answer

到目前为止，您可能已经找到了Centripetal Catmull-Rom spline的维基百科页面，但是如果您还没有找到，它包括以下示例代码：

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

这个函数可以很好地计算出n >= 4个点的插值，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')

导致这个matplotlib图像的结果是：

更新：

另外，有一个 scipy.interpolate 函数 BarycentricInterpolator 可以完成你所需的功能。它非常简单易用，并且适用于只有 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，但具有避免数据集边缘处大振荡的优点。以下是一些展示到目前为止您所要求的所有标准的测试用例。

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')