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首页 > 解决方案 > 使用 scipy.interpolate BSpline 的困难：“TypeError：'list' 对象不能解释为整数”

python - 使用 scipy.interpolate BSpline 的困难：“TypeError：'list' 对象不能解释为整数”

问题描述

从这里：

然后，我应该被允许执行：

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-Spline。但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 的使用失败了，来自文档：

class scipy.interpolate.BSpline(t, c, k, extrapolate=True, axis=0)

t: ndarray, 形状 (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c应该是与我的路径长度相同的一维向量的系数p。

例如，sp = BSpline(t, c0[0], k)执行时没有错误，与c0[1]or一样c0[2]，但当然，我希望splprep使用计算的所有系数。

从这里，scipy interpolate 手册似乎令人困惑：

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，错误地解释了它的spline coefficients 参数。

那么，如何从splprepwithBSpline或 with 另一个函数返回的结点和系数构建 BSpline 呢？

标签： pythonscipyinterpolationbspline

解决方案

BSpline(t, k, c0)应该BSpline(t, c0, k)

编辑。其实还有一个问题： splprep 返回的数组列表与BSpline.

注意 splrep 和 spl p rep之间的区别：

基本上，splrep/splev 是一致的，splrep/BSpline 是一致的，但是 spl p rep/BSpline 不是。这是一个已知的疣，不能以向后兼容的方式修复。

如果要将它们一起使用，则需要转置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]])

这个答案基于https://github.com/scipy/scipy/issues/10389。那里的一般建议适用：如果你想要插值，更make_interp_spline喜欢splrepand splprep。如果您想要平滑，目前只有 FITPACK，或者 splrep（与 BSpline 兼容）或 splprep（您需要手动转置）。

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