如何将样条曲线拟合转换为分段函数？

简短的答案是，如果您对以标准幂为单位的多项式系数感兴趣，那么最好使用CubicSpline（请参见此讨论）：

cu = scipy.interpolate.CubicSpline(x_sample, d_sample)

plt.plot(x_sample, y_sample, 'ko')
for i in range(len(cu.x)-1):
    xs = np.linspace(cu.x[i], cu.x[i+1], 100)
    plt.plot(xs, np.polyval(cu.c[:,i], xs - cu.x[i]))

为了回答您的问题，您可以改为使用numpy.piecewise，的断点cu.x和的系数从此处创建一个分段函数cu.c，然后自己直接编写多项式表达式或使用numpy.polyval。例如，

cu.c[:,0]  # coeffs for 0th segment
# array([-0.01316353, -0.02680068,  0.51629024,  3.        ])

# equivalent ways to code polynomial for this segment
f0 = lambda x: cu.c[0,0]*(x-x[0])**3 + cu.c[1,0]*(x-x[0])**2 + cu.c[2,0]*(x-x[0]) + cu.c[3,0]
f0 = lambda x: [cu.c[i,0]*(x-x[0])**(3-i) for i in range(4)]

# ... or getting values directly from x's
y0 = np.polyval(cu.c[:,0], xs - cu.x[0])

更长的答案：

这里有一些潜在的混乱点：

• UnivariateSpline适合B样条曲线，因此系数与标准多项式幂基础不同
• In order to convert from B-spline, we can use PPoly.from_spline, but unfortunately UnivariateSpline returns a truncated list of knots and coefficients that won't play with this function. We can resolve this problem by accessing the internal data of the spline object, which is a little taboo.
• Also, the coefficient matrix c (whether from UnivariateSpline or CubicSpline) is in reverse degree order and assumes you are "centering" yourself, e.g. the coefficient at c[k,i] belongs to c[k,i]*(x-x[i])^(3-k).

Given your setup, note that if instead of using the UnivariateSpline wrapper, we directly fit with splrep and no smoothing (s=0), we can grab the tck (knots-coefficients-degree) tuple and send it to the PPoly.from_spline function and get the coefficients we want:

tck = scipy.interpolate.splrep(x_sample, d_sample, s=0)
tck
# (array([0.        , 0.        , 0.        , 0.        , 2.68456376,
#        4.02684564, 5.36912752, 6.7114094 , 9.39597315, 9.39597315,
#        9.39597315, 9.39597315]),
# array([3.        , 3.46200469, 4.05843704, 3.89649312, 3.33792889,
#        2.29435138, 1.65015175, 1.59021688, 0.        , 0.        ,
#        0.        , 0.        ]),
# 3)

p = scipy.interpolate.PPoly.from_spline(tck)
p.x.shape  # breakpoints in unexpected shape
# (12,)

p.c.shape  # polynomial coeffs in unexpected shape
# (4, 11)

注意怪异重复断点tck，又在p.x：这是一个FITPACK事情（算法运行这一切）。

如果我们尝试使用发送tck来自UnivariateSpline的元组(s.get_knots(), s.get_coeffs(), 3)，那么我们将丢失这些重复，因此from_spline不起作用。检查出的来源，虽然它出现在全矢量存储在self._data，所以我们可以做

s = scipy.interpolate.UnivariateSpline(x_sample, d_sample, s=0)
tck = (s._data[8], s._data[9], 3)
p = scipy.interpolate.PPoly.from_spline(tck)

和以前一样 要检查这些系数，请执行以下操作：

plt.plot(x_sample, d_sample, 'o')

for i in range(len(p.x)-1):
    xs = np.linspace(p.x[i], p.x[i+1], 10)
    plt.plot(xs, np.polyval(p.c[:,i], xs - p.x[i]))

注意numpy.polyval要使系数无效，因此我们可以按原样传递p.c。