How to draw cubic spline in matplotlib

Question

I want to connect the following points using smooth line, say cubic spline

points = [(3.28,0.00),(4.00,0.50),(4.40,1.0),(4.60,1.52),(5.00,2.5),(5.00,3.34),(4.70,3.8)]
points = points + [(4.50,3.96),(4.20,4.0),(3.70,3.90),(3.00,3.5),(2.00,2.9)]

and finally get orange line like this (this one is created using a vector plotting language Asymptote)

I'm wondering how to do it in matplotlib in a simple way. I already had a look at similar question, e.g. Generating smooth line graph using matplotlib, but direct use of that method produces figure like this

which is of course not what I want.

Answer 1

You need to take a parametric approach, like this:

import numpy as np
import matplotlib.pyplot as plt
from scipy import interpolate

points = [(3.28,0.00),(4.00,0.50),(4.40,1.0),(4.60,1.52),(5.00,2.5),(5.00,3.34),(4.70,3.8)]
points = points + [(4.50,3.96),(4.20,4.0),(3.70,3.90),(3.00,3.5),(2.00,2.9)]
data = np.array(points)

tck,u = interpolate.splprep(data.transpose(), s=0)
unew = np.arange(0, 1.01, 0.01)
out = interpolate.splev(unew, tck)

plt.figure()
plt.plot(out[0], out[1], color='orange')
plt.plot(data[:,0], data[:,1], 'ob')
plt.show()

This is basically just reworked from the last example in the section here.

Answer 2

This is pretty much following the circle example here.

import numpy as np
import matplotlib.pyplot as plt
from scipy import interpolate

def annotate_points(ax, A, B):
    for xy in zip(A, B):
        ax.annotate('(%s, %s)' % xy, xy=xy, textcoords='offset points')

points = [(3.28,0.00),(4.00,0.50),(4.40,1.0),(4.60,1.52),(5.00,2.5),(5.00,3.34),(4.70,3.8)]
points = points + [(4.50,3.96),(4.20,4.0),(3.70,3.90),(3.00,3.5),(2.00,2.9)]
x, y = zip(*points)

fig = plt.figure()
ax = fig.add_subplot(111)
plt.scatter(x, y, color='black')
annotate_points(ax, x, y)

tck,u = interpolate.splprep([x, y], s=0)
unew = np.arange(0, 1.01, 0.01)
out = interpolate.splev(unew, tck)

plt.plot(x, y, 'orange', out[0], out[1])
plt.legend(['connect the dots', 'cubic spline'])

plt.show()