任意轮廓的平滑样条表示,f(长度) - > X,Y

时间:2013-01-15 18:11:30

标签: numpy scipy curve-fitting bezier spline

假设我有一组x,y坐标,用于标记沿轮廓的点。有没有办法可以构建轮廓的样条曲线表示,我可以在其长度上的特定位置进行求值并恢复插值的x,y坐标?

通常情况下,X和Y值之间不存在1:1的对应关系,因此单变量样条对我没有好处。双变量样条曲线会很好,但据我所知,所有用于评估scipy.interpolate中的二变量样条函数的函数取x,y值并返回z,而我需要给出z并返回x,y(因为x ,y是一条线上的点,每个z映射到唯一的x,y)。

以下是我希望能够做到的草图:

import numpy as np
from matplotlib.pyplot import plot

# x,y coordinates of contour points, not monotonically increasing
x = np.array([ 2.,  1.,  1.,  2.,  2.,  4.,  4.,  3.])
y = np.array([ 1.,  2.,  3.,  4.,  2.,  3.,  2.,  1.])

# f: X --> Y might not be a 1:1 correspondence
plot(x,y,'-o')

# get the cumulative distance along the contour
dist = [0]
for ii in xrange(x.size-1):
    dist.append(np.sqrt((x[ii+1]-x[ii])**2 + (y[ii+1]-y[ii])**2))
d = np.array(dist)

# build a spline representation of the contour
spl = ContourSpline(x,y,d)

# resample it at smaller distance intervals
interp_d = np.linspace(d[0],d[-1],1000)
interp_x,interp_y = spl(interp_d)

3 个答案:

答案 0 :(得分:25)

您希望使用参数化样条,而不是从y值中插入x,而是设置新参数t,并插入yx的值t,并使用单变量样条函数。如何为每个点分配t值会影响结果,并且如您的问题所示,使用距离可能是一个好主意:

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

x = np.array([ 2.,  1.,  1.,  2.,  2.,  4.,  4.,  3.])
y = np.array([ 1.,  2.,  3.,  4.,  2.,  3.,  2.,  1.])
plt.plot(x,y, label='poly')

t = np.arange(x.shape[0], dtype=float)
t /= t[-1]
nt = np.linspace(0, 1, 100)
x1 = scipy.interpolate.spline(t, x, nt)
y1 = scipy.interpolate.spline(t, y, nt)
plt.plot(x1, y1, label='range_spline')

t = np.zeros(x.shape)
t[1:] = np.sqrt((x[1:] - x[:-1])**2 + (y[1:] - y[:-1])**2)
t = np.cumsum(t)
t /= t[-1]
x2 = scipy.interpolate.spline(t, x, nt)
y2 = scipy.interpolate.spline(t, y, nt)
plt.plot(x2, y2, label='dist_spline')

plt.legend(loc='best')
plt.show()

enter image description here

答案 1 :(得分:2)

以下是使用splprepsplev的示例:

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

# x,y coordinates of contour points, not monotonically increasing
x = np.array([2.,  1.,  1.,  2.,  2.,  4.,  4.,  3.])
y = np.array([1.,  2.,  3.,  4.,  2.,  3.,  2.,  1.])

# f: X --> Y might not be a 1:1 correspondence
plot(x, y, '-o')

# get the cumulative distance along the contour
dist = np.sqrt((x[:-1] - x[1:])**2 + (y[:-1] - y[1:])**2)
dist_along = np.concatenate(([0], dist.cumsum()))

# build a spline representation of the contour
spline, u = scipy.interpolate.splprep([x, y], u=dist_along, s=0)

# resample it at smaller distance intervals
interp_d = np.linspace(dist_along[0], dist_along[-1], 50)
interp_x, interp_y = scipy.interpolate.splev(interp_d, spline)
plot(interp_x, interp_y, '-o')

Parametric spline example

答案 2 :(得分:1)

您可以使用splprep和splev,看看scipy cookbook,例如与您的问题非常相似。

http://wiki.scipy.org/Cookbook/Interpolation#head-34818696f8d7066bb3188495567dd776a451cf11