scipy documentation 没有说明如何获取系数并手动生成样条曲线。但是，可以从现有的关于 B 样条的文献中弄清楚如何做到这一点。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），通过将系数与最高阶基函数相乘并求和，可以轻松生成样条曲线：
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，以下是一个示例脚本。这需要输入数据并使用 Scipy 的函数以及其面向对象的样条拟合方法。从两者中的任何一个中获取系数和节点并将它们用作我们手动计算的例程bspleval 的输入，我们重现了与它们相同的曲线。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()