我正在使用UnivariateSpline为我拥有的某些数据构建分段多项式.然后我想在其他程序中使用这些样条(在C或FORTRAN中),因此我想了解生成样条函数背后的等式.
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')
这是由此产生的情节.我相信每个区间都有有效的样条线,但看起来我的样条方程不正确…我找不到任何关于scipy文档中应该是什么的引用.有人知道吗?谢谢 !
解决方法:
关于如何获取系数并手动生成样条曲线,scipy documentation没有任何说法.但是,有可能从现有的B样条文献中找出如何做到这一点.以下函数bspleval显示了如何构造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,可以很容易地生成样条曲线:
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.这将获取输入数据并使用Scipy的功能以及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结点,并使用这些作为我们手动计算的例程bspleval的输入,我们重现它们所做的相同曲线.请注意,手动评估曲线与Scipy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()