难以使用scipy.interpolate BSpline："TypeError：'list‘对象不能解释为整数“

Question

来自这里

• splrep允许从路径和平滑因子计算B样条结、系数和度。
• splev使用生成的B样条进行插值。
• BSpline允许直接从节点、系数和度构建样条。

然后，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BSpline对我的论点并不满意：

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

类scipy.interpolate.BSpline(t，c，k，extrapolate=True，axis=0) t: ndarray，形似(n+k+1，)->节 c: ndarray，形状(>=n，…)->样条系数-- k次样条至少需要k+1系数，从而忽略了n >= k+1，忽略了j>n的附加系数. k: int ->B样条阶

除了系数c外，它应该是与我的路径p相同长度的一维向量。

例如，sp = BSpline(t, c0[0], k)执行时没有错误，就像对c0[1]或c0[2]一样，但是当然，我期望使用splprep计算的所有系数。

从这里来看，枕叶内插手册似乎令人困惑：

重新定位控制点的tck1: X和y坐标

手册上说：

(t，c，k)包含节点向量B样条系数和样条度的元组。

最后，我对BSpline的样条系数参数进行了错误的解释。

那么，如何从splprep返回的带BSpline或其他函数的节点和系数构建BSpline呢？

Answer 1

BSpline(t, k, c0)应该是BSpline(t, c0, k)

编辑。实际上，还有一个问题: splprep返回数组列表，并且它与BSpline不一致。

注意spl与splprep之间的区别：

基本上，样条/样条是一致的，样条/样条是一致的，但splprep/BSpline不是一致的。这是一个已知的疣，不能用向后兼容的方式修复。

如果您想将它们一起使用，则需要转置c数组。根据您的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]])

这个答案是基于https://github.com/scipy/scipy/issues/10389的。那里的一般建议是适用的:如果您想要插值，请选择make_interp_spline而不是splrep和splprep。如果您想要平滑，目前只有FITPACK，要么是splrep (它是BSpline兼容的)，要么是splprep (在这里您需要手动转置)。