Interpolating a closed curve using scipy

2020-02-01 04:25发布

I'm writing a python script to interpolate a given set of points with splines. The points are defined by their [x, y] coordinates.

I tried to use this code:

x = np.array([23, 24, 24, 25, 25])
y = np.array([13, 12, 13, 12, 13])
tck, u = scipy.interpolate.splprep([x,y], s=0)
unew = np.arange(0, 1.00, 0.005)
out = scipy.interpolate.splev(unew, tck) 

which gives me a curve like this:

Bad Interpolation

However, I need to have a smooth closed curve - on the picture above the derivatives at one of the points are obviously not the same. How can I achieve this?

1条回答
手持菜刀,她持情操
2楼-- · 2020-02-01 04:30

Your closed path can be considered as a parametric curve, x=f(u), y=g(u) where u is distance along the curve, bounded on the interval [0, 1). You can use scipy.interpolate.splprep with per=True to treat your x and y points as periodic, then evaluate the fitted splines using scipy.interpolate.splev:

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

x = np.array([23, 24, 24, 25, 25])
y = np.array([13, 12, 13, 12, 13])

# append the starting x,y coordinates
x = np.r_[x, x[0]]
y = np.r_[y, y[0]]

# fit splines to x=f(u) and y=g(u), treating both as periodic. also note that s=0
# is needed in order to force the spline fit to pass through all the input points.
tck, u = interpolate.splprep([x, y], s=0, per=True)

# evaluate the spline fits for 1000 evenly spaced distance values
xi, yi = interpolate.splev(np.linspace(0, 1, 1000), tck)

# plot the result
fig, ax = plt.subplots(1, 1)
ax.plot(x, y, 'or')
ax.plot(xi, yi, '-b')

enter image description here

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