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更新时间：2023-02-26 16:50:49
我能够重新创建我问过的Mathematica示例在上一篇文章中使用Python / scipy。结果如下：
I was able to recreate the Mathematica example I asked about in the previous post using Python/scipy. Here's the result:
诀窍是截取系数，即元素的1 scipy.interpolate.splrep 返回的元组，并在将它们交给 scipy.interpolate.splev $ c之前将其替换为控制点值$ c>，或者，如果您自己创建节点很好，也可以不用 splrep 自己创建整个元组。
The trick was to either intercept the coefficients, i.e. element 1 of the tuple returned by scipy.interpolate.splrep, and to replace them with the control point values before handing them to scipy.interpolate.splev, or, if you are fine with creating the knots yourself, you can also do without splrep and create the entire tuple yourself.
这一切奇怪的是，根据手册， splrep 返回（和 splev 期望一个元组，其中包含一个样条系数向量，每个结具有一个系数。但是，根据我发现的所有资料，将样条线定义为 N_control_points 基本样条线的加权和，因此我希望系数向量具有与控制点一样多的元素，而不是结点。
What is strange about this all, though, is that, according to the manual, splrep returns (and splev expects) a tuple containing, among others, a spline coefficients vector with one coefficient per knot. However, according to all sources I found, a spline is defined as the weighted sum of the N_control_points basis splines, so I would expect the coefficients vector to have as many elements as control points, not knot positions.
实际上，当提供 splrep 的结果元组时，系数向量已如上所述修改为 scipy.interpolate.splev ，事实证明该向量的第一个 N_control_points 实际上是 N_control_points 基本样条的期望系数。该向量的最后一个 degree + 1 元素似乎无效。我很困惑为什么要这样做。如果有人可以澄清这一点，那就太好了。这是生成上述图的来源：
In fact, when supplying splrep's result tuple with the coefficients vector modified as described above to scipy.interpolate.splev, it turns out that the first N_control_points of that vector actually are the expected coefficients for the N_control_points basis splines. The last degree + 1 elements of that vector seem to have no effect. I'm stumped as to why it's done this way. If anyone can clarify that, that would be great. Here's the source that generates the above plots:
import numpy as np
import matplotlib.pyplot as plt
import scipy.interpolate as si
points = [[0, 0], [0, 2], [2, 3], [4, 0], [6, 3], [8, 2], [8, 0]];
points = np.array(points)
x = points[:,0]
y = points[:,1]
t = range(len(points))
ipl_t = np.linspace(0.0, len(points) - 1, 100)
x_tup = si.splrep(t, x, k=3)
y_tup = si.splrep(t, y, k=3)
x_list = list(x_tup)
xl = x.tolist()
x_list[1] = xl + [0.0, 0.0, 0.0, 0.0]
y_list = list(y_tup)
yl = y.tolist()
y_list[1] = yl + [0.0, 0.0, 0.0, 0.0]
x_i = si.splev(ipl_t, x_list)
y_i = si.splev(ipl_t, y_list)
#==============================================================================
# Plot
#==============================================================================
fig = plt.figure()
ax = fig.add_subplot(231)
plt.plot(t, x, '-og')
plt.plot(ipl_t, x_i, 'r')
plt.xlim([0.0, max(t)])
plt.title('Splined x(t)')
ax = fig.add_subplot(232)
plt.plot(t, y, '-og')
plt.plot(ipl_t, y_i, 'r')
plt.xlim([0.0, max(t)])
plt.title('Splined y(t)')
ax = fig.add_subplot(233)
plt.plot(x, y, '-og')
plt.plot(x_i, y_i, 'r')
plt.xlim([min(x) - 0.3, max(x) + 0.3])
plt.ylim([min(y) - 0.3, max(y) + 0.3])
plt.title('Splined f(x(t), y(t))')
ax = fig.add_subplot(234)
for i in range(7):
vec = np.zeros(11)
vec[i] = 1.0
x_list = list(x_tup)
x_list[1] = vec.tolist()
x_i = si.splev(ipl_t, x_list)
plt.plot(ipl_t, x_i)
plt.xlim([0.0, max(t)])
plt.title('Basis splines')
plt.show()
现在为了创建如下闭合曲线，这是可以在网上找到的另一个Mathematica示例，
Now in order to create a closed curve like the following, which is another Mathematica example that can be found on the web,
如果使用的话，必须在 splrep 调用中设置 per 参数。如图所示，在最后用 degree + 1 值填充控制点列表之后，这似乎效果很好。
it is necessary to set the per parameter in the splrep call, if you use that. After padding the list of control points with degree+1 values at the end, this seems to work well enough, as the images show.
但是，这里的下一个特点是系数向量中的第一个和最后一个 degree 元素无效，这意味着控制必须将点从第二个位置（即位置1）开始放入向量中。对于度数k = 4和k = 5，该位置甚至变为位置2。
The next peculiarity here, however, is that the first and the last degree elements in the coefficients vector have no effect, meaning that the control points must be put in the vector starting at the second position, i.e. position 1. Only then are the results ok. For degrees k=4 and k=5, that position even changes to position 2.
以下是生成闭合曲线的来源：
Here's the source for generating the closed curve:
import numpy as np
import matplotlib.pyplot as plt
import scipy.interpolate as si
points = [[-2, 2], [0, 1], [-2, 0], [0, -1], [-2, -2], [-4, -4], [2, -4], [4, 0], [2, 4], [-4, 4]]
degree = 3
points = points + points[0:degree + 1]
points = np.array(points)
n_points = len(points)
x = points[:,0]
y = points[:,1]
t = range(len(x))
ipl_t = np.linspace(1.0, len(points) - degree, 1000)
x_tup = si.splrep(t, x, k=degree, per=1)
y_tup = si.splrep(t, y, k=degree, per=1)
x_list = list(x_tup)
xl = x.tolist()
x_list[1] = [0.0] + xl + [0.0, 0.0, 0.0, 0.0]
y_list = list(y_tup)
yl = y.tolist()
y_list[1] = [0.0] + yl + [0.0, 0.0, 0.0, 0.0]
x_i = si.splev(ipl_t, x_list)
y_i = si.splev(ipl_t, y_list)
#==============================================================================
# Plot
#==============================================================================
fig = plt.figure()
ax = fig.add_subplot(231)
plt.plot(t, x, '-og')
plt.plot(ipl_t, x_i, 'r')
plt.xlim([0.0, max(t)])
plt.title('Splined x(t)')
ax = fig.add_subplot(232)
plt.plot(t, y, '-og')
plt.plot(ipl_t, y_i, 'r')
plt.xlim([0.0, max(t)])
plt.title('Splined y(t)')
ax = fig.add_subplot(233)
plt.plot(x, y, '-og')
plt.plot(x_i, y_i, 'r')
plt.xlim([min(x) - 0.3, max(x) + 0.3])
plt.ylim([min(y) - 0.3, max(y) + 0.3])
plt.title('Splined f(x(t), y(t))')
ax = fig.add_subplot(234)
for i in range(n_points - degree - 1):
vec = np.zeros(11)
vec[i] = 1.0
x_list = list(x_tup)
x_list[1] = vec.tolist()
x_i = si.splev(ipl_t, x_list)
plt.plot(ipl_t, x_i)
plt.xlim([0.0, 9.0])
plt.title('Periodic basis splines')
plt.show()
最后，这也是我无法解释的效果，这是在转到5度时，样条曲线中会出现一个小的不连续点，请参见右上图，它是该鼻形的月亮。产生此问题的源代码在下面列出。
Lastly, there is an effect that I can not explain either, and this is when going to degree 5, there is a small discontinuity that appears in the splined curve, see the upper right panel, which is a close-up of that 'half-moon-with-nose-shape'. The source code that produces this is listed below.
import numpy as np
import matplotlib.pyplot as plt
import scipy.interpolate as si
points = [[-2, 2], [0, 1], [-2, 0], [0, -1], [-2, -2], [-4, -4], [2, -4], [4, 0], [2, 4], [-4, 4]]
degree = 5
points = points + points[0:degree + 1]
points = np.array(points)
n_points = len(points)
x = points[:,0]
y = points[:,1]
t = range(len(x))
ipl_t = np.linspace(1.0, len(points) - degree, 1000)
knots = np.linspace(-degree, len(points), len(points) + degree + 1).tolist()
xl = x.tolist()
coeffs_x = [0.0, 0.0] + xl + [0.0, 0.0, 0.0]
yl = y.tolist()
coeffs_y = [0.0, 0.0] + yl + [0.0, 0.0, 0.0]
x_i = si.splev(ipl_t, (knots, coeffs_x, degree))
y_i = si.splev(ipl_t, (knots, coeffs_y, degree))
#==============================================================================
# Plot
#==============================================================================
fig = plt.figure()
ax = fig.add_subplot(231)
plt.plot(t, x, '-og')
plt.plot(ipl_t, x_i, 'r')
plt.xlim([0.0, max(t)])
plt.title('Splined x(t)')
ax = fig.add_subplot(232)
plt.plot(t, y, '-og')
plt.plot(ipl_t, y_i, 'r')
plt.xlim([0.0, max(t)])
plt.title('Splined y(t)')
ax = fig.add_subplot(233)
plt.plot(x, y, '-og')
plt.plot(x_i, y_i, 'r')
plt.xlim([min(x) - 0.3, max(x) + 0.3])
plt.ylim([min(y) - 0.3, max(y) + 0.3])
plt.title('Splined f(x(t), y(t))')
ax = fig.add_subplot(234)
for i in range(n_points - degree - 1):
vec = np.zeros(11)
vec[i] = 1.0
x_i = si.splev(ipl_t, (knots, vec, degree))
plt.plot(ipl_t, x_i)
plt.xlim([0.0, 9.0])
plt.title('Periodic basis splines')
plt.show()
给出b-样条在科学界无处不在，而scipy是一个如此全面的工具箱，而且我无法在网上找到关于我所要询问的内容的很多信息，这使我相信我走错了路，或者俯视的东西。任何帮助将不胜感激。
Given that b-splines are ubiquitous in the scientific community, and that scipy is such a comprehensive toolbox, and that I have not been able to find much about what I'm asking here on the web, leads me to believe I'm on the wrong track or overlooking something. Any help would be appreciated.