[SciPy-User] B-spline basis functions?
Charles R Harris
Tue Jul 31 10:46:14 CDT 2012
On Tue, Jul 31, 2012 at 9:37 AM, Nathaniel Smith <firstname.lastname@example.org> wrote:
> Hi all,
> I'd like to be able to do spline regression in patsy, which means
> that I need to be able to compute b-spline basis functions. I am not
> an initiate into the mysteries of practical spline computations, but I
> *think* the stuff in scipy.signal is not quite usable as is, because
> it's focused on doing interpolation directly rather than exposing the
> basis functions themselves?
> Specifically, to achieve feature parity with R , I need to be able to
> - an arbitrary order
> - an arbitrary collection of knot positions (which may be irregularly
> - a vector x of points at which to evaluate the basis functions
> and spit out the value of each spline basis function evaluated at each
> point in the x vector.
> It looks like scipy.signal.bspline *might* be useful, but I can't
> quite tell? Or alternatively someone might have some code lying around
> to do this already?
> Basically I have a copy of Schumaker here and I'm hoping someone will
> save me from having to read it :-).
I have this floating around
def splvander(x, deg, knots):
"""Vandermonde type matrix for splines.
Returns a matrix whose columns are the values of the b-splines of deg
`deg` associated with the knot sequence `knots` evaluated at the points
x : array_like
Points at which to evaluate the b-splines.
deg : int
Degree of the splines.
knots : array_like
List of knots. The convention here is that the interior knots have
been extended at both ends by ``deg + 1`` extra knots.
vander : ndarray
Vandermonde like matrix of shape (m,n), where ``m = len(x)`` and
``m = len(knots) - deg - 1``
The knots exending the interior points are usually taken to be the same
as the endpoints of the interval on which the spline will be evaluated.
m = len(knots) - deg - 1
v = np.zeros((m, len(x)))
d = np.eye(m, len(knots))
for i in range(m):
v[i] = spl.splev(x, (knots, d[i], deg))
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