科学中的样条插值系数 [英] coefficients of spline interpolation in scipy

问题描述

我想通过scipy计算样条插值的系数. 在MATLAB中:

I want to calculate the coefficients of a spline interpolation by scipy. In MATLAB:

x=[0:3];
y=[0,1,4,0];
spl=spline(x,y);
disp(spl.coefs);

，它将返回:

ans =

   -1.5000    5.5000   -3.0000         0
   -1.5000    1.0000    3.5000    1.0000
   -1.5000   -3.5000    1.0000    4.0000

但是我不能通过scipy中的interpolate.splrep来做到这一点.你能告诉我怎么计算吗?

But i can't do that by interpolate.splrep in scipy. Can you tell me how to calc it?

推荐答案

我不确定是否有任何方法可以从scipy中准确获得这些系数. scipy.interpolate.splrep给您的是b样条的结系数. Matlab的样条曲线为您提供的似乎是描述连接传递点的三次方程的偏多项式系数，这使我相信Matlab样条曲线是基于控制点的样条曲线，例如Hermite或Catmull-Rom而不是b样条.

I'm not sure there is any way to get exactly those coefficients from scipy. What scipy.interpolate.splrep gives you is the coefficients for the knots for a b-spline. What Matlab's spline gives you appears to be the partial polynomial coefficients describing the cubic equations connecting the points you pass in, which leads me to believe that the Matlab spline is a control-point based spline such as a Hermite or Catmull-Rom instead of a b-spline.

但是，scipy.interpolate.interpolate.spltopp确实提供了一种获取b样条的部分多项式系数的方法.不幸的是，它似乎不能很好地工作.

However, scipy.interpolate.interpolate.spltopp does provide a way to get the partial polynomial coefficients of a b-spline. Unfortunately, it doesn't seem to work very well.

>>> import scipy.interpolate
>>> x = [0, 1, 2, 3]
>>> y = [0, 1, 4, 0]
>>> tck = scipy.interpolate.splrep(x, y)
>>> tck
Out: 
    (array([ 0.,  0.,  0.,  0.,  3.,  3.,  3.,  3.]),
    array([  3.19142761e-16,  -3.00000000e+00,   1.05000000e+01,
        0.00000000e+00,   0.00000000e+00,   0.00000000e+00,
        0.00000000e+00,   0.00000000e+00]),
    3)

>>> pp = scipy.interpolate.interpolate.spltopp(tck[0][1:-1], tck[1], tck[2])

>>> pp.coeffs.T
Out: 
    array([[ -4.54540394e-322,   0.00000000e+000,   0.00000000e+000,
           0.00000000e+000],
        [ -4.54540394e-322,   0.00000000e+000,   0.00000000e+000,
           0.00000000e+000],
        [ -4.54540394e-322,   0.00000000e+000,   0.00000000e+000,
           0.00000000e+000],
        [  0.00000000e+000,   0.00000000e+000,   0.00000000e+000,
           0.00000000e+000],
        [  0.00000000e+000,   0.00000000e+000,   0.00000000e+000,
           0.00000000e+000]])

请注意，每个结有一组系数，而不是每个传入的原始点都有一个系数.而且，将系数乘以b样条基矩阵似乎没有什么帮助.

Note that there is one set of coefficients per knot, not one for each of the original points passed in. Also, multiplying the coefficients by the b-spline basis matrix doesn't seem to be very helpful.

>>> bsbm = array([[-1,  3, -3,  1], [ 3, -6,  3,  0], [-3,  0,  3,  0], 
                 [ 1,  4,  1,  0]]) * 1.0/6
Out: 
    array([[-0.16666667,  0.5       , -0.5       ,  0.16666667],
        [ 0.5       , -1.        ,  0.5       ,  0.        ],
        [-0.5       ,  0.        ,  0.5       ,  0.        ],
        [ 0.16666667,  0.66666667,  0.16666667,  0.        ]])

>>> dot(pp.coeffs.T, bsbm)
Out: 
    array([[  7.41098469e-323,  -2.27270197e-322,   2.27270197e-322,
           -7.41098469e-323],
        [  7.41098469e-323,  -2.27270197e-322,   2.27270197e-322,
           -7.41098469e-323],
        [  7.41098469e-323,  -2.27270197e-322,   2.27270197e-322,
           -7.41098469e-323],
        [  0.00000000e+000,   0.00000000e+000,   0.00000000e+000,
           0.00000000e+000],
        [  0.00000000e+000,   0.00000000e+000,   0.00000000e+000,
           0.00000000e+000]])

FORTRAN分段多项式软件包 PPPack 具有命令bsplpp从B样条转换为分段多项式形式，可以满足您的需求.不幸的是，目前没有用于PPPack的Python包装器.

The FORTRAN Piecewise Polynomial Package, PPPack, has a command bsplpp that converts from B-spline to piecewise polynomial form, which may serve your needs. Unfortunately, there isn't a Python wrapper for PPPack at this time.

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