Difficulties to use scipy.interpolate BSpline: "TypeError: 'list' object cannot be interpreted as an integer"

Question

来自 :

splrep 可以根据路径和平滑因子计算 B 样条结、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，应该让我表演：

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 样条曲线。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,) --> knots

c: ndarray, shape (>=n, …) --> spline coefficients - At least k+1 coefficients are required for a spline of degree k, so that n >= k+1. Additional coefficients, c[j] with j > n, are ignored.

k: int --> B-spline order

除了系数 c，它应该是与我的路径 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 splprep 计算的所有系数.

从 here 看来，scipy 插值手册令人困惑：

tck[1]: x and y coordinates of the relocated control points

手册说：

(t,c,k) a tuple containing the vector of knots, the B-spline coefficients, and the degree of the spline

最终，我错误地解释了 BSpline 的 样条系数参数 。

那么，我如何使用 BSpline 或其他函数从 splprep 返回的结和系数构建 BSpline？

Answer 1

BSpline(t, k, c0) 应该是 BSpline(t, c0, k)

编辑。事实上，还有一个问题：splprep returns 数组列表与 BSpline.

不一致

注意splrep和splprep的区别：

基本上，splrep/splev一致，splrep/BSpline一致，但是splprep/BSpline不一致。这是一个已知的疣，无法以向后兼容的方式修复。

如果你想一起使用它们，你需要转置 c 数组。基于您的 OP 示例：

In [1]: import numpy as np
   ...: from scipy.interpolate import splev, splprep, BSpline
   ...: path =  [(2077.0, 712.0, 1136.6176470588234), (2077.0004154771536, 974.6
   ...: 30482962754, 1313.735294117647), (2077.1630960823995, 1302.460574562254,
   ...:  1490.8529411764707), (2078.1944091179635, 1674.693193015173, 1667.97058
   ...: 82352941), (2080.5096120056783, 2086.976611915444, 1845.0882352941176), 
   ...: (2085.1051468332066, 2711.054258877495, 2022.2058823529412), (1477.08461
   ...: 85328733, 2803.6223679691457, 2199.323529411765), (948.4693105162195, 28
   ...: 02.0390667447105, 2376.4411764705883), (383.8615403256207, 2804.84342413
   ...: 4807, 2553.5588235294117), (-41.6669725172834, 2497.067373170676, 2730.6
   ...: 76470588235), (-37.94311919744064, 1970.5155845437525, 2907.794117647059
   ...: ), (-35.97395938535092, 1576.713103381243, 3084.9117647058824), (-35.125
   ...: 016151504795, 1214.2319876178394, 3262.029411764706), (-35.0005507678645
   ...: 24, 893.3910350913443, 3439.1470588235297), (-35.0, 631.2108462417168, 3
   ...: 616.264705882353), (-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, s=0)      # ADDED s=0 for clarity
   ...: 

In [2]: t, c, k = tck

In [3]: c1 = np.asarray(c)

In [4]: spl = BSpline(t, c1.T, k)         # Note the transpose

In [5]: spl(u) - path                     # these should match, and they do
Out[5]: 
array([[ -4.54747351e-13,  -1.13686838e-13,  -4.54747351e-13],
       [  0.00000000e+00,  -1.13686838e-13,   0.00000000e+00],
       [ -4.54747351e-13,   0.00000000e+00,   0.00000000e+00],
       [  0.00000000e+00,  -2.27373675e-13,  -2.27373675e-13],
       [ -4.54747351e-13,   0.00000000e+00,   4.54747351e-13],
       [ -4.54747351e-13,   0.00000000e+00,  -6.82121026e-13],
       [  2.27373675e-13,   0.00000000e+00,   0.00000000e+00],
       [ -1.13686838e-13,  -4.54747351e-13,  -4.54747351e-13],
       [  0.00000000e+00,   0.00000000e+00,   0.00000000e+00],
       [  4.26325641e-14,  -9.09494702e-13,   0.00000000e+00],
       [  1.42108547e-14,  -4.54747351e-13,   0.00000000e+00],
       [  0.00000000e+00,   0.00000000e+00,   0.00000000e+00],
       [  7.10542736e-15,   0.00000000e+00,  -4.54747351e-13],
       [  0.00000000e+00,  -3.41060513e-13,   0.00000000e+00],
       [ -7.10542736e-15,  -1.13686838e-13,   0.00000000e+00],
       [  0.00000000e+00,   0.00000000e+00,   0.00000000e+00],
       [  0.00000000e+00,   0.00000000e+00,   0.00000000e+00]])

此答案基于 https://github.com/scipy/scipy/issues/10389。那里的一般建议适用：如果你想要插值，更喜欢 make_interp_spline 而不是 splrep 和 splprep。如果你想要平滑，目前只有 FITPACK，splrep（与 BSpline 兼容）或 splprep（你需要手动转置）。

Difficulties to use scipy.interpolate BSpline: "TypeError: 'list' object cannot be interpreted as an integer"

Difficulties to use scipy.interpolate BSpline: "TypeError: 'list' object cannot be interpreted as an integer"

python

interpolation

scipy

bspline