python - 从 UnivariateSpline 对象获取样条方程

Question

我正在使用 UnivariateSpline 为我拥有的一些数据构造分段多项式。然后我想在其他程序(C 或 FORTRAN 中)中使用这些样条，因此我想了解生成的样条背后的方程式。

这是我的代码:

import numpy as np
import scipy as sp
from  scipy.interpolate import UnivariateSpline
import matplotlib.pyplot as plt
import bisect

data = np.loadtxt('test_C12H26.dat')
Tmid = 800.0
print "Tmid", Tmid
nmid = bisect.bisect(data[:,0],Tmid)
fig = plt.figure()
plt.plot(data[:,0], data[:,7],ls='',marker='o',markevery=20)
npts = len(data[:,0])
#print "npts", npts
w = np.ones(npts)
w[0] = 100
w[nmid] = 100
w[npts-1] = 100
spline1 = UnivariateSpline(data[:nmid,0],data[:nmid,7],s=1,w=w[:nmid])
coeffs = spline1.get_coeffs()
print coeffs
print spline1.get_knots()
print spline1.get_residual()
print coeffs[0] + coeffs[1] * (data[0,0] - data[0,0]) \
                + coeffs[2] * (data[0,0] - data[0,0])**2 \
                + coeffs[3] * (data[0,0] - data[0,0])**3, \
      data[0,7]
print coeffs[0] + coeffs[1] * (data[nmid,0] - data[0,0]) \
                + coeffs[2] * (data[nmid,0] - data[0,0])**2 \
                + coeffs[3] * (data[nmid,0] - data[0,0])**3, \
      data[nmid,7]

print Tmid,data[-1,0]
spline2 = UnivariateSpline(data[nmid-1:,0],data[nmid-1:,7],s=1,w=w[nmid-1:])
print spline2.get_coeffs()
print spline2.get_knots()
print spline2.get_residual()
plt.plot(data[:,0],spline1(data[:,0]))
plt.plot(data[:,0],spline2(data[:,0]))
plt.savefig('test.png')

这是结果图。我相信我对每个间隔都有有效的样条曲线，但看起来我的样条曲线方程不正确......我在 scipy 文档中找不到它应该是什么的任何引用。有人知道吗？谢谢!

Answer 1

scipy documentation 没有说明如何获取系数并手动生成样条曲线。但是，可以从有关 B 样条的现有文献中找出如何做到这一点。下面的函数 bspleval 展示了如何构造 B 样条基函数(代码中的矩阵 B)，从中可以很容易地通过乘以系数生成样条曲线具有最高阶基函数和求和:

def bspleval(x, knots, coeffs, order, debug=False):
    '''
    Evaluate a B-spline at a set of points.

    Parameters
    ----------
    x : list or ndarray
        The set of points at which to evaluate the spline.
    knots : list or ndarray
        The set of knots used to define the spline.
    coeffs : list of ndarray
        The set of spline coefficients.
    order : int
        The order of the spline.

    Returns
    -------
    y : ndarray
        The value of the spline at each point in x.
    '''

    k = order
    t = knots
    m = alen(t)
    npts = alen(x)
    B = zeros((m-1,k+1,npts))

    if debug:
        print('k=%i, m=%i, npts=%i' % (k, m, npts))
        print('t=', t)
        print('coeffs=', coeffs)

    ## Create the zero-order B-spline basis functions.
    for i in range(m-1):
        B[i,0,:] = float64(logical_and(x >= t[i], x < t[i+1]))

    if (k == 0):
        B[m-2,0,-1] = 1.0

    ## Next iteratively define the higher-order basis functions, working from lower order to higher.
    for j in range(1,k+1):
        for i in range(m-j-1):
            if (t[i+j] - t[i] == 0.0):
                first_term = 0.0
            else:
                first_term = ((x - t[i]) / (t[i+j] - t[i])) * B[i,j-1,:]

            if (t[i+j+1] - t[i+1] == 0.0):
                second_term = 0.0
            else:
                second_term = ((t[i+j+1] - x) / (t[i+j+1] - t[i+1])) * B[i+1,j-1,:]

            B[i,j,:] = first_term + second_term
        B[m-j-2,j,-1] = 1.0

    if debug:
        plt.figure()
        for i in range(m-1):
            plt.plot(x, B[i,k,:])
        plt.title('B-spline basis functions')

    ## Evaluate the spline by multiplying the coefficients with the highest-order basis functions.
    y = zeros(npts)
    for i in range(m-k-1):
        y += coeffs[i] * B[i,k,:]

    if debug:
        plt.figure()
        plt.plot(x, y)
        plt.title('spline curve')
        plt.show()

    return(y)

为了举例说明如何将其与 Scipy 现有的单变量样条函数一起使用，下面是一个示例脚本。这需要输入数据并使用 Scipy 的功能及其面向对象的样条拟合方法。从两者中的任何一个获取系数和结点，并将它们用作我们手动计算的例程 bspleval 的输入，我们再现了它们所做的相同曲线。请注意，手动评估的曲线与 Scipy 的评估方法之间的差异非常小，几乎可以肯定是浮点噪声。

x = array([-273.0, -176.4, -79.8, 16.9, 113.5, 210.1, 306.8, 403.4, 500.0])
y = array([2.25927498e-53, 2.56028619e-03, 8.64512988e-01, 6.27456769e+00, 1.73894734e+01,
        3.29052124e+01, 5.14612316e+01, 7.20531200e+01, 9.40718450e+01])

x_nodes = array([-273.0, -263.5, -234.8, -187.1, -120.3, -34.4, 70.6, 194.6, 337.8, 500.0])
y_nodes = array([2.25927498e-53, 3.83520726e-46, 8.46685318e-11, 6.10568083e-04, 1.82380809e-01,
                2.66344008e+00, 1.18164677e+01, 3.01811501e+01, 5.78812583e+01, 9.40718450e+01])

## Now get scipy's spline fit.
k = 3
tck = splrep(x_nodes, y_nodes, k=k, s=0)
knots = tck[0]
coeffs = tck[1]
print('knot points=', knots)
print('coefficients=', coeffs)

## Now try scipy's object-oriented version. The result is exactly the same as "tck": the knots are the
## same and the coeffs are the same, they are just queried in a different way.
uspline = UnivariateSpline(x_nodes, y_nodes, s=0)
uspline_knots = uspline.get_knots()
uspline_coeffs = uspline.get_coeffs()

## Here are scipy's native spline evaluation methods. Again, "ytck" and "y_uspline" are exactly equal.
ytck = splev(x, tck)
y_uspline = uspline(x)
y_knots = uspline(knots)

## Now let's try our manually-calculated evaluation function.
y_eval = bspleval(x, knots, coeffs, k, debug=False)

plt.plot(x, ytck, label='tck')
plt.plot(x, y_uspline, label='uspline')
plt.plot(x, y_eval, label='manual')

## Next plot the knots and nodes.
plt.plot(x_nodes, y_nodes, 'ko', markersize=7, label='input nodes')            ## nodes
plt.plot(knots, y_knots, 'mo', markersize=5, label='tck knots')    ## knots
plt.xlim((-300.0,530.0))
plt.legend(loc='best', prop={'size':14})

plt.figure()
plt.title('difference')
plt.plot(x, ytck-y_uspline, label='tck-uspl')
plt.plot(x, ytck-y_eval, label='tck-manual')
plt.legend(loc='best', prop={'size':14})

plt.show()