【发布时间】:2020-05-10 01:47:51
【问题描述】:
来自here:
-
splrep可以根据路径和平滑因子计算 B-Spline 节点、系数和度数 -
splev使用生成的 B 样条启用插值 -
BSpline可以直接从节点、系数和度数构建样条
那么,我应该被允许表演:
import numpy as np
from scipy.interpolate import splev, splprep, BSpline
path = [(2077.0, 712.0, 1136.6176470588234), (2077.0004154771536, 974.630482962754, 1313.735294117647), (2077.1630960823995, 1302.460574562254, 1490.8529411764707), (2078.1944091179635, 1674.693193015173, 1667.9705882352941), (2080.5096120056783, 2086.976611915444, 1845.0882352941176), (2085.1051468332066, 2711.054258877495, 2022.2058823529412), (1477.0846185328733, 2803.6223679691457, 2199.323529411765), (948.4693105162195, 2802.0390667447105, 2376.4411764705883), (383.8615403256207, 2804.843424134807, 2553.5588235294117), (-41.6669725172834, 2497.067373170676, 2730.676470588235), (-37.94311919744064, 1970.5155845437525, 2907.794117647059), (-35.97395938535092, 1576.713103381243, 3084.9117647058824), (-35.125016151504795, 1214.2319876178394, 3262.029411764706), (-35.000550767864524, 893.3910350913443, 3439.1470588235297), (-35.0, 631.2108462417168, 3616.264705882353), (-35.0, 365.60545190581837, 3793.3823529411766), (-35.0, 100.00005756991993, 3970.5)]
p = [[x for x,y,z in path], [y for x,y,z in path], [z for x,y,z in path]]
tck, u = splprep(p, k=3)
t, c0, k = tck
sp = BSpline(t, k, c0)
目标是能够调整 B 样条。但是BSpline 对我的论点不满意:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/home/lalebarde/anaconda3/lib/python3.7/site-packages/scipy/interpolate/_bsplines.py", line 184, in __init__
self.k = operator.index(k)
TypeError: 'list' object cannot be interpreted as an integer
如果我检查变量的形状和类型:
type(t)
<class 'numpy.ndarray'>
type(c0)
<class 'list'>
type(k)
<class 'int'>
t.shape
(21,)
np.array(c0).shape
(3, 17)
我对 BSpline 的使用失败,来自 documentation:
class scipy.interpolate.BSpline(t, c, k, extrapolate=True, axis=0)
t: ndarray, shape (n+k+1,) --> 结
c: ndarray, shape (>=n, …) --> spline coefficients - k 次样条曲线至少需要 k+1 个系数,因此 n >= k+1。 j > n 的附加系数 c[j] 将被忽略。
k: int --> B样条顺序
除了系数c 应该是与我的路径p 长度相同的一维向量。
例如,sp = BSpline(t, c0[0], k) 执行时没有错误,与 c0[1] 或 c0[2] 一样,但当然,我希望使用由 splprep 计算的所有系数。
来自here,看来 scipy interpolate 手册令人困惑:
tck[1]:重定位控制点的x和y坐标
手册说:
(t,c,k) 包含节点向量、B 样条系数和样条度数的元组
最终,我误用了 BSpline,错误解释了它的 样条系数参数。
那么,如何从splprep 与BSpline 或其他函数返回的结点和系数构建 BSpline?
【问题讨论】:
标签: python scipy interpolation bspline