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kktt / ccc-model-manager python
Repository URL to install this package:
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Version:
1.11.0 ▾
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import operator
import numpy as np
from numpy.core.multiarray import normalize_axis_index
from scipy.linalg import (get_lapack_funcs, LinAlgError,
cholesky_banded, cho_solve_banded,
solve, solve_banded)
from . import _bspl
from . import _fitpack_impl
from . import _fitpack as _dierckx
from scipy._lib._util import prod
from scipy.special import poch
from scipy.sparse import csr_array
from itertools import combinations
__all__ = ["BSpline", "make_interp_spline", "make_lsq_spline"]
def _get_dtype(dtype):
"""Return np.complex128 for complex dtypes, np.float64 otherwise."""
if np.issubdtype(dtype, np.complexfloating):
return np.complex_
else:
return np.float_
def _as_float_array(x, check_finite=False):
"""Convert the input into a C contiguous float array.
NB: Upcasts half- and single-precision floats to double precision.
"""
x = np.ascontiguousarray(x)
dtyp = _get_dtype(x.dtype)
x = x.astype(dtyp, copy=False)
if check_finite and not np.isfinite(x).all():
raise ValueError("Array must not contain infs or nans.")
return x
def _dual_poly(j, k, t, y):
"""
Dual polynomial of the B-spline B_{j,k,t} -
polynomial which is associated with B_{j,k,t}:
$p_{j,k}(y) = (y - t_{j+1})(y - t_{j+2})...(y - t_{j+k})$
"""
if k == 0:
return 1
return np.prod([(y - t[j + i]) for i in range(1, k + 1)])
def _diff_dual_poly(j, k, y, d, t):
"""
d-th derivative of the dual polynomial $p_{j,k}(y)$
"""
if d == 0:
return _dual_poly(j, k, t, y)
if d == k:
return poch(1, k)
comb = list(combinations(range(j + 1, j + k + 1), d))
res = 0
for i in range(len(comb) * len(comb[0])):
res += np.prod([(y - t[j + p]) for p in range(1, k + 1)
if (j + p) not in comb[i//d]])
return res
class BSpline:
r"""Univariate spline in the B-spline basis.
.. math::
S(x) = \sum_{j=0}^{n-1} c_j B_{j, k; t}(x)
where :math:`B_{j, k; t}` are B-spline basis functions of degree `k`
and knots `t`.
Parameters
----------
t : ndarray, shape (n+k+1,)
knots
c : ndarray, shape (>=n, ...)
spline coefficients
k : int
B-spline degree
extrapolate : bool or 'periodic', optional
whether to extrapolate beyond the base interval, ``t[k] .. t[n]``,
or to return nans.
If True, extrapolates the first and last polynomial pieces of b-spline
functions active on the base interval.
If 'periodic', periodic extrapolation is used.
Default is True.
axis : int, optional
Interpolation axis. Default is zero.
Attributes
----------
t : ndarray
knot vector
c : ndarray
spline coefficients
k : int
spline degree
extrapolate : bool
If True, extrapolates the first and last polynomial pieces of b-spline
functions active on the base interval.
axis : int
Interpolation axis.
tck : tuple
A read-only equivalent of ``(self.t, self.c, self.k)``
Methods
-------
__call__
basis_element
derivative
antiderivative
integrate
construct_fast
design_matrix
from_power_basis
Notes
-----
B-spline basis elements are defined via
.. math::
B_{i, 0}(x) = 1, \textrm{if $t_i \le x < t_{i+1}$, otherwise $0$,}
B_{i, k}(x) = \frac{x - t_i}{t_{i+k} - t_i} B_{i, k-1}(x)
+ \frac{t_{i+k+1} - x}{t_{i+k+1} - t_{i+1}} B_{i+1, k-1}(x)
**Implementation details**
- At least ``k+1`` coefficients are required for a spline of degree `k`,
so that ``n >= k+1``. Additional coefficients, ``c[j]`` with
``j > n``, are ignored.
- B-spline basis elements of degree `k` form a partition of unity on the
*base interval*, ``t[k] <= x <= t[n]``.
Examples
--------
Translating the recursive definition of B-splines into Python code, we have:
>>> def B(x, k, i, t):
... if k == 0:
... return 1.0 if t[i] <= x < t[i+1] else 0.0
... if t[i+k] == t[i]:
... c1 = 0.0
... else:
... c1 = (x - t[i])/(t[i+k] - t[i]) * B(x, k-1, i, t)
... if t[i+k+1] == t[i+1]:
... c2 = 0.0
... else:
... c2 = (t[i+k+1] - x)/(t[i+k+1] - t[i+1]) * B(x, k-1, i+1, t)
... return c1 + c2
>>> def bspline(x, t, c, k):
... n = len(t) - k - 1
... assert (n >= k+1) and (len(c) >= n)
... return sum(c[i] * B(x, k, i, t) for i in range(n))
Note that this is an inefficient (if straightforward) way to
evaluate B-splines --- this spline class does it in an equivalent,
but much more efficient way.
Here we construct a quadratic spline function on the base interval
``2 <= x <= 4`` and compare with the naive way of evaluating the spline:
>>> from scipy.interpolate import BSpline
>>> k = 2
>>> t = [0, 1, 2, 3, 4, 5, 6]
>>> c = [-1, 2, 0, -1]
>>> spl = BSpline(t, c, k)
>>> spl(2.5)
array(1.375)
>>> bspline(2.5, t, c, k)
1.375
Note that outside of the base interval results differ. This is because
`BSpline` extrapolates the first and last polynomial pieces of B-spline
functions active on the base interval.
>>> import matplotlib.pyplot as plt
>>> fig, ax = plt.subplots()
>>> xx = np.linspace(1.5, 4.5, 50)
>>> ax.plot(xx, [bspline(x, t, c ,k) for x in xx], 'r-', lw=3, label='naive')
>>> ax.plot(xx, spl(xx), 'b-', lw=4, alpha=0.7, label='BSpline')
>>> ax.grid(True)
>>> ax.legend(loc='best')
>>> plt.show()
References
----------
.. [1] Tom Lyche and Knut Morken, Spline methods,
http://www.uio.no/studier/emner/matnat/ifi/INF-MAT5340/v05/undervisningsmateriale/
.. [2] Carl de Boor, A practical guide to splines, Springer, 2001.
"""
def __init__(self, t, c, k, extrapolate=True, axis=0):
super().__init__()
self.k = operator.index(k)
self.c = np.asarray(c)
self.t = np.ascontiguousarray(t, dtype=np.float64)
if extrapolate == 'periodic':
self.extrapolate = extrapolate
else:
self.extrapolate = bool(extrapolate)
n = self.t.shape[0] - self.k - 1
axis = normalize_axis_index(axis, self.c.ndim)
# Note that the normalized axis is stored in the object.
self.axis = axis
if axis != 0:
# roll the interpolation axis to be the first one in self.c
# More specifically, the target shape for self.c is (n, ...),
# and axis !=0 means that we have c.shape (..., n, ...)
# ^
# axis
self.c = np.moveaxis(self.c, axis, 0)
if k < 0:
raise ValueError("Spline order cannot be negative.")
if self.t.ndim != 1:
raise ValueError("Knot vector must be one-dimensional.")
if n < self.k + 1:
raise ValueError("Need at least %d knots for degree %d" %
(2*k + 2, k))
if (np.diff(self.t) < 0).any():
raise ValueError("Knots must be in a non-decreasing order.")
if len(np.unique(self.t[k:n+1])) < 2:
raise ValueError("Need at least two internal knots.")
if not np.isfinite(self.t).all():
raise ValueError("Knots should not have nans or infs.")
if self.c.ndim < 1:
raise ValueError("Coefficients must be at least 1-dimensional.")
if self.c.shape[0] < n:
raise ValueError("Knots, coefficients and degree are inconsistent.")
dt = _get_dtype(self.c.dtype)
self.c = np.ascontiguousarray(self.c, dtype=dt)
@classmethod
def construct_fast(cls, t, c, k, extrapolate=True, axis=0):
"""Construct a spline without making checks.
Accepts same parameters as the regular constructor. Input arrays
`t` and `c` must of correct shape and dtype.
"""
self = object.__new__(cls)
self.t, self.c, self.k = t, c, k
self.extrapolate = extrapolate
self.axis = axis
return self
@property
def tck(self):
"""Equivalent to ``(self.t, self.c, self.k)`` (read-only).
"""
return self.t, self.c, self.k
@classmethod
def basis_element(cls, t, extrapolate=True):
"""Return a B-spline basis element ``B(x | t[0], ..., t[k+1])``.
Parameters
----------
t : ndarray, shape (k+2,)
internal knots
extrapolate : bool or 'periodic', optional
whether to extrapolate beyond the base interval, ``t[0] .. t[k+1]``,
or to return nans.
If 'periodic', periodic extrapolation is used.
Default is True.
Returns
-------
basis_element : callable
A callable representing a B-spline basis element for the knot
vector `t`.
Notes
-----
The degree of the B-spline, `k`, is inferred from the length of `t` as
``len(t)-2``. The knot vector is constructed by appending and prepending
``k+1`` elements to internal knots `t`.
Examples
--------
Construct a cubic B-spline:
>>> from scipy.interpolate import BSpline
>>> b = BSpline.basis_element([0, 1, 2, 3, 4])
>>> k = b.k
>>> b.t[k:-k]
array([ 0., 1., 2., 3., 4.])
>>> k
3
Construct a quadratic B-spline on ``[0, 1, 1, 2]``, and compare
to its explicit form:
>>> t = [-1, 0, 1, 1, 2]
>>> b = BSpline.basis_element(t[1:])
>>> def f(x):
... return np.where(x < 1, x*x, (2. - x)**2)
>>> import matplotlib.pyplot as plt
>>> fig, ax = plt.subplots()
>>> x = np.linspace(0, 2, 51)
>>> ax.plot(x, b(x), 'g', lw=3)
>>> ax.plot(x, f(x), 'r', lw=8, alpha=0.4)
>>> ax.grid(True)
>>> plt.show()
"""
k = len(t) - 2
t = _as_float_array(t)
t = np.r_[(t[0]-1,) * k, t, (t[-1]+1,) * k]
c = np.zeros_like(t)
c[k] = 1.
return cls.construct_fast(t, c, k, extrapolate)
@classmethod
def design_matrix(cls, x, t, k):
"""
Returns a design matrix as a CSR format sparse array.
Parameters
----------
x : array_like, shape (n,)
Points to evaluate the spline at.
t : array_like, shape (nt,)
Sorted 1D array of knots.
k : int
B-spline degree.
Returns
-------
design_matrix : `csr_array` object
Sparse matrix in CSR format where in each row all the basis
elements are evaluated at the certain point (first row - x[0],
..., last row - x[-1]).
Examples
--------
Construct a design matrix for a B-spline
>>> from scipy.interpolate import make_interp_spline, BSpline
>>> x = np.linspace(0, np.pi * 2, 4)
>>> y = np.sin(x)
>>> k = 3
>>> bspl = make_interp_spline(x, y, k=k)
>>> design_matrix = bspl.design_matrix(x, bspl.t, k)
>>> design_matrix.toarray()
[[1. , 0. , 0. , 0. ],
[0.2962963 , 0.44444444, 0.22222222, 0.03703704],
[0.03703704, 0.22222222, 0.44444444, 0.2962963 ],
[0. , 0. , 0. , 1. ]]
Construct a design matrix for some vector of knots
>>> k = 2
>>> t = [-1, 0, 1, 2, 3, 4, 5, 6]
>>> x = [1, 2, 3, 4]
>>> design_matrix = BSpline.design_matrix(x, t, k).toarray()
>>> design_matrix
[[0.5, 0.5, 0. , 0. , 0. ],
[0. , 0.5, 0.5, 0. , 0. ],
[0. , 0. , 0.5, 0.5, 0. ],
[0. , 0. , 0. , 0.5, 0.5]]
This result is equivalent to the one created in the sparse format
>>> c = np.eye(len(t) - k - 1)
>>> design_matrix_gh = BSpline(t, c, k)(x)
>>> np.allclose(design_matrix, design_matrix_gh, atol=1e-14)
True
Notes
-----
.. versionadded:: 1.8.0
In each row of the design matrix all the basis elements are evaluated
at the certain point (first row - x[0], ..., last row - x[-1]).
`nt` is a length of the vector of knots: as far as there are
`nt - k - 1` basis elements, `nt` should be not less than `2 * k + 2`
to have at least `k + 1` basis element.
Out of bounds `x` raises a ValueError.
"""
x = _as_float_array(x, True)
t = _as_float_array(t, True)
if t.ndim != 1 or np.any(t[1:] < t[:-1]):
raise ValueError(f"Expect t to be a 1-D sorted array_like, but "
f"got t={t}.")
# There are `nt - k - 1` basis elements in a BSpline built on the
# vector of knots with length `nt`, so to have at least `k + 1` basis
# element we need to have at least `2 * k + 2` elements in the vector
# of knots.
if len(t) < 2 * k + 2:
raise ValueError(f"Length t is not enough for k={k}.")
# Checks from `find_interval` function
if (min(x) < t[k]) or (max(x) > t[t.shape[0] - k - 1]):
raise ValueError(f'Out of bounds w/ x = {x}.')
n, nt = x.shape[0], t.shape[0]
data, idx = _bspl._make_design_matrix(x, t, k)
return csr_array((data, idx), (n, nt - k - 1))
def __call__(self, x, nu=0, extrapolate=None):
"""
Evaluate a spline function.
Parameters
----------
x : array_like
points to evaluate the spline at.
nu : int, optional
derivative to evaluate (default is 0).
extrapolate : bool or 'periodic', optional
whether to extrapolate based on the first and last intervals
or return nans. If 'periodic', periodic extrapolation is used.
Default is `self.extrapolate`.
Returns
-------
y : array_like
Shape is determined by replacing the interpolation axis
in the coefficient array with the shape of `x`.
"""
if extrapolate is None:
extrapolate = self.extrapolate
x = np.asarray(x)
x_shape, x_ndim = x.shape, x.ndim
x = np.ascontiguousarray(x.ravel(), dtype=np.float_)
# With periodic extrapolation we map x to the segment
# [self.t[k], self.t[n]].
if extrapolate == 'periodic':
n = self.t.size - self.k - 1
x = self.t[self.k] + (x - self.t[self.k]) % (self.t[n] -
self.t[self.k])
extrapolate = False
out = np.empty((len(x), prod(self.c.shape[1:])), dtype=self.c.dtype)
self._ensure_c_contiguous()
self._evaluate(x, nu, extrapolate, out)
out = out.reshape(x_shape + self.c.shape[1:])
if self.axis != 0:
# transpose to move the calculated values to the interpolation axis
l = list(range(out.ndim))
l = l[x_ndim:x_ndim+self.axis] + l[:x_ndim] + l[x_ndim+self.axis:]
out = out.transpose(l)
return out
def _evaluate(self, xp, nu, extrapolate, out):
_bspl.evaluate_spline(self.t, self.c.reshape(self.c.shape[0], -1),
self.k, xp, nu, extrapolate, out)
def _ensure_c_contiguous(self):
"""
c and t may be modified by the user. The Cython code expects
that they are C contiguous.
"""
if not self.t.flags.c_contiguous:
self.t = self.t.copy()
if not self.c.flags.c_contiguous:
self.c = self.c.copy()
def derivative(self, nu=1):
"""Return a B-spline representing the derivative.
Parameters
----------
nu : int, optional
Derivative order.
Default is 1.
Returns
-------
b : BSpline object
A new instance representing the derivative.
See Also
--------
splder, splantider
"""
c = self.c
# pad the c array if needed
ct = len(self.t) - len(c)
if ct > 0:
c = np.r_[c, np.zeros((ct,) + c.shape[1:])]
tck = _fitpack_impl.splder((self.t, c, self.k), nu)
return self.construct_fast(*tck, extrapolate=self.extrapolate,
axis=self.axis)
def antiderivative(self, nu=1):
"""Return a B-spline representing the antiderivative.
Parameters
----------
nu : int, optional
Antiderivative order. Default is 1.
Returns
-------
b : BSpline object
A new instance representing the antiderivative.
Notes
-----
If antiderivative is computed and ``self.extrapolate='periodic'``,
it will be set to False for the returned instance. This is done because
the antiderivative is no longer periodic and its correct evaluation
outside of the initially given x interval is difficult.
See Also
--------
splder, splantider
"""
c = self.c
# pad the c array if needed
ct = len(self.t) - len(c)
if ct > 0:
c = np.r_[c, np.zeros((ct,) + c.shape[1:])]
tck = _fitpack_impl.splantider((self.t, c, self.k), nu)
if self.extrapolate == 'periodic':
extrapolate = False
else:
extrapolate = self.extrapolate
return self.construct_fast(*tck, extrapolate=extrapolate,
axis=self.axis)
def integrate(self, a, b, extrapolate=None):
"""Compute a definite integral of the spline.
Parameters
----------
a : float
Lower limit of integration.
b : float
Upper limit of integration.
extrapolate : bool or 'periodic', optional
whether to extrapolate beyond the base interval,
``t[k] .. t[-k-1]``, or take the spline to be zero outside of the
base interval. If 'periodic', periodic extrapolation is used.
If None (default), use `self.extrapolate`.
Returns
-------
I : array_like
Definite integral of the spline over the interval ``[a, b]``.
Examples
--------
Construct the linear spline ``x if x < 1 else 2 - x`` on the base
interval :math:`[0, 2]`, and integrate it
>>> from scipy.interpolate import BSpline
>>> b = BSpline.basis_element([0, 1, 2])
>>> b.integrate(0, 1)
array(0.5)
If the integration limits are outside of the base interval, the result
is controlled by the `extrapolate` parameter
>>> b.integrate(-1, 1)
array(0.0)
>>> b.integrate(-1, 1, extrapolate=False)
array(0.5)
>>> import matplotlib.pyplot as plt
>>> fig, ax = plt.subplots()
>>> ax.grid(True)
>>> ax.axvline(0, c='r', lw=5, alpha=0.5) # base interval
>>> ax.axvline(2, c='r', lw=5, alpha=0.5)
>>> xx = [-1, 1, 2]
>>> ax.plot(xx, b(xx))
>>> plt.show()
"""
if extrapolate is None:
extrapolate = self.extrapolate
# Prepare self.t and self.c.
self._ensure_c_contiguous()
# Swap integration bounds if needed.
sign = 1
if b < a:
a, b = b, a
sign = -1
n = self.t.size - self.k - 1
if extrapolate != "periodic" and not extrapolate:
# Shrink the integration interval, if needed.
a = max(a, self.t[self.k])
b = min(b, self.t[n])
if self.c.ndim == 1:
# Fast path: use FITPACK's routine
# (cf _fitpack_impl.splint).
t, c, k = self.tck
integral, wrk = _dierckx._splint(t, c, k, a, b)
return integral * sign
out = np.empty((2, prod(self.c.shape[1:])), dtype=self.c.dtype)
# Compute the antiderivative.
c = self.c
ct = len(self.t) - len(c)
if ct > 0:
c = np.r_[c, np.zeros((ct,) + c.shape[1:])]
ta, ca, ka = _fitpack_impl.splantider((self.t, c, self.k), 1)
if extrapolate == 'periodic':
# Split the integral into the part over period (can be several
# of them) and the remaining part.
ts, te = self.t[self.k], self.t[n]
period = te - ts
interval = b - a
n_periods, left = divmod(interval, period)
if n_periods > 0:
# Evaluate the difference of antiderivatives.
x = np.asarray([ts, te], dtype=np.float_)
_bspl.evaluate_spline(ta, ca.reshape(ca.shape[0], -1),
ka, x, 0, False, out)
integral = out[1] - out[0]
integral *= n_periods
else:
integral = np.zeros((1, prod(self.c.shape[1:])),
dtype=self.c.dtype)
# Map a to [ts, te], b is always a + left.
a = ts + (a - ts) % period
b = a + left
# If b <= te then we need to integrate over [a, b], otherwise
# over [a, te] and from xs to what is remained.
if b <= te:
x = np.asarray([a, b], dtype=np.float_)
_bspl.evaluate_spline(ta, ca.reshape(ca.shape[0], -1),
ka, x, 0, False, out)
integral += out[1] - out[0]
else:
x = np.asarray([a, te], dtype=np.float_)
_bspl.evaluate_spline(ta, ca.reshape(ca.shape[0], -1),
ka, x, 0, False, out)
integral += out[1] - out[0]
x = np.asarray([ts, ts + b - te], dtype=np.float_)
_bspl.evaluate_spline(ta, ca.reshape(ca.shape[0], -1),
ka, x, 0, False, out)
integral += out[1] - out[0]
else:
# Evaluate the difference of antiderivatives.
x = np.asarray([a, b], dtype=np.float_)
_bspl.evaluate_spline(ta, ca.reshape(ca.shape[0], -1),
ka, x, 0, extrapolate, out)
integral = out[1] - out[0]
integral *= sign
return integral.reshape(ca.shape[1:])
@classmethod
def from_power_basis(cls, pp, bc_type='not-a-knot'):
r"""
Construct a polynomial in the B-spline basis
from a piecewise polynomial in the power basis.
For now, accepts ``CubicSpline`` instances only.
Parameters
----------
pp : CubicSpline
A piecewise polynomial in the power basis, as created
by ``CubicSpline``
bc_type : string, optional
Boundary condition type as in ``CubicSpline``: one of the
``not-a-knot``, ``natural``, ``clamped``, or ``periodic``.
Necessary for construction an instance of ``BSpline`` class.
Default is ``not-a-knot``.
Returns
-------
b : BSpline object
A new instance representing the initial polynomial
in the B-spline basis.
Notes
-----
.. versionadded:: 1.8.0
Accepts only ``CubicSpline`` instances for now.
The algorithm follows from differentiation
the Marsden's identity [1]: each of coefficients of spline
interpolation function in the B-spline basis is computed as follows:
.. math::
c_j = \sum_{m=0}^{k} \frac{(k-m)!}{k!}
c_{m,i} (-1)^{k-m} D^m p_{j,k}(x_i)
:math:`c_{m, i}` - a coefficient of CubicSpline,
:math:`D^m p_{j, k}(x_i)` - an m-th defivative of a dual polynomial
in :math:`x_i`.
``k`` always equals 3 for now.
First ``n - 2`` coefficients are computed in :math:`x_i = x_j`, e.g.
.. math::
c_1 = \sum_{m=0}^{k} \frac{(k-1)!}{k!} c_{m,1} D^m p_{j,3}(x_1)
Last ``nod + 2`` coefficients are computed in ``x[-2]``,
``nod`` - number of derivatives at the ends.
For example, consider :math:`x = [0, 1, 2, 3, 4]`,
:math:`y = [1, 1, 1, 1, 1]` and bc_type = ``natural``
The coefficients of CubicSpline in the power basis:
:math:`[[0, 0, 0, 0, 0], [0, 0, 0, 0, 0],
[0, 0, 0, 0, 0], [1, 1, 1, 1, 1]]`
The knot vector: :math:`t = [0, 0, 0, 0, 1, 2, 3, 4, 4, 4, 4]`
In this case
.. math::
c_j = \frac{0!}{k!} c_{3, i} k! = c_{3, i} = 1,~j = 0, ..., 6
References
----------
.. [1] Tom Lyche and Knut Morken, Spline Methods, 2005, Section 3.1.2
"""
from ._cubic import CubicSpline
if not isinstance(pp, CubicSpline):
raise NotImplementedError("Only CubicSpline objects are accepted"
"for now. Got %s instead." % type(pp))
x = pp.x
coef = pp.c
k = pp.c.shape[0] - 1
n = x.shape[0]
if bc_type == 'not-a-knot':
t = _not_a_knot(x, k)
elif bc_type == 'natural' or bc_type == 'clamped':
t = _augknt(x, k)
elif bc_type == 'periodic':
t = _periodic_knots(x, k)
else:
raise TypeError('Unknown boundary condition: %s' % bc_type)
nod = t.shape[0] - (n + k + 1) # number of derivatives at the ends
c = np.zeros(n + nod, dtype=pp.c.dtype)
for m in range(k + 1):
for i in range(n - 2):
c[i] += poch(k + 1, -m) * coef[m, i]\
* np.power(-1, k - m)\
* _diff_dual_poly(i, k, x[i], m, t)
for j in range(n - 2, n + nod):
c[j] += poch(k + 1, -m) * coef[m, n - 2]\
* np.power(-1, k - m)\
* _diff_dual_poly(j, k, x[n - 2], m, t)
return cls.construct_fast(t, c, k, pp.extrapolate, pp.axis)
#################################
# Interpolating spline helpers #
#################################
def _not_a_knot(x, k):
"""Given data x, construct the knot vector w/ not-a-knot BC.
cf de Boor, XIII(12)."""
x = np.asarray(x)
if k % 2 != 1:
raise ValueError("Odd degree for now only. Got %s." % k)
m = (k - 1) // 2
t = x[m+1:-m-1]
t = np.r_[(x[0],)*(k+1), t, (x[-1],)*(k+1)]
return t
def _augknt(x, k):
"""Construct a knot vector appropriate for the order-k interpolation."""
return np.r_[(x[0],)*k, x, (x[-1],)*k]
def _convert_string_aliases(deriv, target_shape):
if isinstance(deriv, str):
if deriv == "clamped":
deriv = [(1, np.zeros(target_shape))]
elif deriv == "natural":
deriv = [(2, np.zeros(target_shape))]
else:
raise ValueError("Unknown boundary condition : %s" % deriv)
return deriv
def _process_deriv_spec(deriv):
if deriv is not None:
try:
ords, vals = zip(*deriv)
except TypeError as e:
msg = ("Derivatives, `bc_type`, should be specified as a pair of "
"iterables of pairs of (order, value).")
raise ValueError(msg) from e
else:
ords, vals = [], []
return np.atleast_1d(ords, vals)
def _woodbury_algorithm(A, ur, ll, b, k):
'''
Solve a cyclic banded linear system with upper right
and lower blocks of size ``(k-1) / 2`` using
the Woodbury formula
Parameters
----------
A : 2-D array, shape(k, n)
Matrix of diagonals of original matrix(see
``solve_banded`` documentation).
ur : 2-D array, shape(bs, bs)
Upper right block matrix.
ll : 2-D array, shape(bs, bs)
Lower left block matrix.
b : 1-D array, shape(n,)
Vector of constant terms of the system of linear equations.
k : int
B-spline degree.
Returns
-------
c : 1-D array, shape(n,)
Solution of the original system of linear equations.
Notes
-----
This algorithm works only for systems with banded matrix A plus
a correction term U @ V.T, where the matrix U @ V.T gives upper right
and lower left block of A
The system is solved with the following steps:
1. New systems of linear equations are constructed:
A @ z_i = u_i,
u_i - columnn vector of U,
i = 1, ..., k - 1
2. Matrix Z is formed from vectors z_i:
Z = [ z_1 | z_2 | ... | z_{k - 1} ]
3. Matrix H = (1 + V.T @ Z)^{-1}
4. The system A' @ y = b is solved
5. x = y - Z @ (H @ V.T @ y)
Also, ``n`` should be greater than ``k``, otherwise corner block
elements will intersect with diagonals.
Examples
--------
Consider the case of n = 8, k = 5 (size of blocks - 2 x 2).
The matrix of a system: U: V:
x x x * * a b a b 0 0 0 0 1 0
x x x x * * c 0 c 0 0 0 0 0 1
x x x x x * * 0 0 0 0 0 0 0 0
* x x x x x * 0 0 0 0 0 0 0 0
* * x x x x x 0 0 0 0 0 0 0 0
d * * x x x x 0 0 d 0 1 0 0 0
e f * * x x x 0 0 e f 0 1 0 0
References
----------
.. [1] William H. Press, Saul A. Teukolsky, William T. Vetterling
and Brian P. Flannery, Numerical Recipes, 2007, Section 2.7.3
'''
k_mod = k - k % 2
bs = int((k - 1) / 2) + (k + 1) % 2
n = A.shape[1] + 1
U = np.zeros((n - 1, k_mod))
VT = np.zeros((k_mod, n - 1)) # V transpose
# upper right block
U[:bs, :bs] = ur
VT[np.arange(bs), np.arange(bs) - bs] = 1
# lower left block
U[-bs:, -bs:] = ll
VT[np.arange(bs) - bs, np.arange(bs)] = 1
Z = solve_banded((bs, bs), A, U)
H = solve(np.identity(k_mod) + VT @ Z, np.identity(k_mod))
y = solve_banded((bs, bs), A, b)
c = y - Z @ (H @ (VT @ y))
return c
def _periodic_knots(x, k):
'''
returns vector of nodes on circle
'''
xc = np.copy(x)
n = len(xc)
if k % 2 == 0:
dx = np.diff(xc)
xc[1: -1] -= dx[:-1] / 2
dx = np.diff(xc)
t = np.zeros(n + 2 * k)
t[k: -k] = xc
for i in range(0, k):
# filling first `k` elements in descending order
t[k - i - 1] = t[k - i] - dx[-(i % (n - 1)) - 1]
# filling last `k` elements in ascending order
t[-k + i] = t[-k + i - 1] + dx[i % (n - 1)]
return t
def _make_interp_per_full_matr(x, y, t, k):
'''
Returns a solution of a system for B-spline interpolation with periodic
boundary conditions. First ``k - 1`` rows of matrix are condtions of
periodicity (continuity of ``k - 1`` derivatives at the boundary points).
Last ``n`` rows are interpolation conditions.
RHS is ``k - 1`` zeros and ``n`` ordinates in this case.
Parameters
----------
x : 1-D array, shape (n,)
Values of x - coordinate of a given set of points.
y : 1-D array, shape (n,)
Values of y - coordinate of a given set of points.
t : 1-D array, shape(n+2*k,)
Vector of knots.
k : int
The maximum degree of spline
Returns
-------
c : 1-D array, shape (n+k-1,)
B-spline coefficients
Notes
-----
``t`` is supposed to be taken on circle.
'''
x, y, t = map(np.asarray, (x, y, t))
n = x.size
# LHS: the collocation matrix + derivatives at edges
matr = np.zeros((n + k - 1, n + k - 1))
# derivatives at x[0] and x[-1]:
for i in range(k - 1):
bb = _bspl.evaluate_all_bspl(t, k, x[0], k, nu=i + 1)
matr[i, : k + 1] += bb
bb = _bspl.evaluate_all_bspl(t, k, x[-1], n + k - 1, nu=i + 1)[:-1]
matr[i, -k:] -= bb
# collocation matrix
for i in range(n):
xval = x[i]
# find interval
if xval == t[k]:
left = k
else:
left = np.searchsorted(t, xval) - 1
# fill a row
bb = _bspl.evaluate_all_bspl(t, k, xval, left)
matr[i + k - 1, left-k:left+1] = bb
# RHS
b = np.r_[[0] * (k - 1), y]
c = solve(matr, b)
return c
def _make_periodic_spline(x, y, t, k, axis):
'''
Compute the (coefficients of) interpolating B-spline with periodic
boundary conditions.
Parameters
----------
x : array_like, shape (n,)
Abscissas.
y : array_like, shape (n,)
Ordinates.
k : int
B-spline degree.
t : array_like, shape (n + 2 * k,).
Knots taken on a circle, ``k`` on the left and ``k`` on the right
of the vector ``x``.
Returns
-------
b : a BSpline object of the degree ``k`` and with knots ``t``.
Notes
-----
The original system is formed by ``n + k - 1`` equations where the first
``k - 1`` of them stand for the ``k - 1`` derivatives continuity on the
edges while the other equations correspond to an interpolating case
(matching all the input points). Due to a special form of knot vector, it
can be proved that in the original system the first and last ``k``
coefficients of a spline function are the same, respectively. It follows
from the fact that all ``k - 1`` derivatives are equal term by term at ends
and that the matrix of the original system of linear equations is
non-degenerate. So, we can reduce the number of equations to ``n - 1``
(first ``k - 1`` equations could be reduced). Another trick of this
implementation is cyclic shift of values of B-splines due to equality of
``k`` unknown coefficients. With this we can receive matrix of the system
with upper right and lower left blocks, and ``k`` diagonals. It allows
to use Woodbury formula to optimize the computations.
'''
n = y.shape[0]
extradim = prod(y.shape[1:])
y_new = y.reshape(n, extradim)
c = np.zeros((n + k - 1, extradim))
# n <= k case is solved with full matrix
if n <= k:
for i in range(extradim):
c[:, i] = _make_interp_per_full_matr(x, y_new[:, i], t, k)
c = np.ascontiguousarray(c.reshape((n + k - 1,) + y.shape[1:]))
return BSpline.construct_fast(t, c, k, extrapolate='periodic', axis=axis)
nt = len(t) - k - 1
# size of block elements
kul = int(k / 2)
# kl = ku = k
ab = np.zeros((3 * k + 1, nt), dtype=np.float_, order='F')
# upper right and lower left blocks
ur = np.zeros((kul, kul))
ll = np.zeros_like(ur)
# `offset` is made to shift all the non-zero elements to the end of the
# matrix
_bspl._colloc(x, t, k, ab, offset=k)
# remove zeros before the matrix
ab = ab[-k - (k + 1) % 2:, :]
# The least elements in rows (except repetitions) are diagonals
# of block matrices. Upper right matrix is an upper triangular
# matrix while lower left is a lower triangular one.
for i in range(kul):
ur += np.diag(ab[-i - 1, i: kul], k=i)
ll += np.diag(ab[i, -kul - (k % 2): n - 1 + 2 * kul - i], k=-i)
# remove elements that occur in the last point
# (first and last points are equivalent)
A = ab[:, kul: -k + kul]
for i in range(extradim):
cc = _woodbury_algorithm(A, ur, ll, y_new[:, i][:-1], k)
c[:, i] = np.concatenate((cc[-kul:], cc, cc[:kul + k % 2]))
c = np.ascontiguousarray(c.reshape((n + k - 1,) + y.shape[1:]))
return BSpline.construct_fast(t, c, k, extrapolate='periodic', axis=axis)
def make_interp_spline(x, y, k=3, t=None, bc_type=None, axis=0,
check_finite=True):
"""Compute the (coefficients of) interpolating B-spline.
Parameters
----------
x : array_like, shape (n,)
Abscissas.
y : array_like, shape (n, ...)
Ordinates.
k : int, optional
B-spline degree. Default is cubic, k=3.
t : array_like, shape (nt + k + 1,), optional.
Knots.
The number of knots needs to agree with the number of datapoints and
the number of derivatives at the edges. Specifically, ``nt - n`` must
equal ``len(deriv_l) + len(deriv_r)``.
bc_type : 2-tuple or None
Boundary conditions.
Default is None, which means choosing the boundary conditions
automatically. Otherwise, it must be a length-two tuple where the first
element sets the boundary conditions at ``x[0]`` and the second
element sets the boundary conditions at ``x[-1]``. Each of these must
be an iterable of pairs ``(order, value)`` which gives the values of
derivatives of specified orders at the given edge of the interpolation
interval.
Alternatively, the following string aliases are recognized:
* ``"clamped"``: The first derivatives at the ends are zero. This is
equivalent to ``bc_type=([(1, 0.0)], [(1, 0.0)])``.
* ``"natural"``: The second derivatives at ends are zero. This is
equivalent to ``bc_type=([(2, 0.0)], [(2, 0.0)])``.
* ``"not-a-knot"`` (default): The first and second segments are the
same polynomial. This is equivalent to having ``bc_type=None``.
* ``"periodic"``: The values and the first ``k-1`` derivatives at the
ends are equivalent.
axis : int, optional
Interpolation axis. Default is 0.
check_finite : bool, optional
Whether to check that the input arrays contain only finite numbers.
Disabling may give a performance gain, but may result in problems
(crashes, non-termination) if the inputs do contain infinities or NaNs.
Default is True.
Returns
-------
b : a BSpline object of the degree ``k`` and with knots ``t``.
Examples
--------
Use cubic interpolation on Chebyshev nodes:
>>> def cheb_nodes(N):
... jj = 2.*np.arange(N) + 1
... x = np.cos(np.pi * jj / 2 / N)[::-1]
... return x
>>> x = cheb_nodes(20)
>>> y = np.sqrt(1 - x**2)
>>> from scipy.interpolate import BSpline, make_interp_spline
>>> b = make_interp_spline(x, y)
>>> np.allclose(b(x), y)
True
Note that the default is a cubic spline with a not-a-knot boundary condition
>>> b.k
3
Here we use a 'natural' spline, with zero 2nd derivatives at edges:
>>> l, r = [(2, 0.0)], [(2, 0.0)]
>>> b_n = make_interp_spline(x, y, bc_type=(l, r)) # or, bc_type="natural"
>>> np.allclose(b_n(x), y)
True
>>> x0, x1 = x[0], x[-1]
>>> np.allclose([b_n(x0, 2), b_n(x1, 2)], [0, 0])
True
Interpolation of parametric curves is also supported. As an example, we
compute a discretization of a snail curve in polar coordinates
>>> phi = np.linspace(0, 2.*np.pi, 40)
>>> r = 0.3 + np.cos(phi)
>>> x, y = r*np.cos(phi), r*np.sin(phi) # convert to Cartesian coordinates
Build an interpolating curve, parameterizing it by the angle
>>> from scipy.interpolate import make_interp_spline
>>> spl = make_interp_spline(phi, np.c_[x, y])
Evaluate the interpolant on a finer grid (note that we transpose the result
to unpack it into a pair of x- and y-arrays)
>>> phi_new = np.linspace(0, 2.*np.pi, 100)
>>> x_new, y_new = spl(phi_new).T
Plot the result
>>> import matplotlib.pyplot as plt
>>> plt.plot(x, y, 'o')
>>> plt.plot(x_new, y_new, '-')
>>> plt.show()
Build a B-spline curve with 2 dimensional y
>>> x = np.linspace(0, 2*np.pi, 10)
>>> y = np.array([np.sin(x), np.cos(x)])
Periodic condition is satisfied because y coordinates of points on the ends
are equivalent
>>> ax = plt.axes(projection='3d')
>>> xx = np.linspace(0, 2*np.pi, 100)
>>> bspl = make_interp_spline(x, y, k=5, bc_type='periodic', axis=1)
>>> ax.plot3D(xx, *bspl(xx))
>>> ax.scatter3D(x, *y, color='red')
>>> plt.show()
See Also
--------
BSpline : base class representing the B-spline objects
CubicSpline : a cubic spline in the polynomial basis
make_lsq_spline : a similar factory function for spline fitting
UnivariateSpline : a wrapper over FITPACK spline fitting routines
splrep : a wrapper over FITPACK spline fitting routines
"""
# convert string aliases for the boundary conditions
if bc_type is None or bc_type == 'not-a-knot' or bc_type == 'periodic':
deriv_l, deriv_r = None, None
elif isinstance(bc_type, str):
deriv_l, deriv_r = bc_type, bc_type
else:
try:
deriv_l, deriv_r = bc_type
except TypeError as e:
raise ValueError("Unknown boundary condition: %s" % bc_type) from e
y = np.asarray(y)
axis = normalize_axis_index(axis, y.ndim)
x = _as_float_array(x, check_finite)
y = _as_float_array(y, check_finite)
y = np.moveaxis(y, axis, 0) # now internally interp axis is zero
if bc_type == 'periodic' and not np.allclose(y[0], y[-1], atol=1e-15):
raise ValueError("First and last points does not match while "
"periodic case expected")
# special-case k=0 right away
if k == 0:
if any(_ is not None for _ in (t, deriv_l, deriv_r)):
raise ValueError("Too much info for k=0: t and bc_type can only "
"be None.")
t = np.r_[x, x[-1]]
c = np.asarray(y)
c = np.ascontiguousarray(c, dtype=_get_dtype(c.dtype))
return BSpline.construct_fast(t, c, k, axis=axis)
# special-case k=1 (e.g., Lyche and Morken, Eq.(2.16))
if k == 1 and t is None:
if not (deriv_l is None and deriv_r is None):
raise ValueError("Too much info for k=1: bc_type can only be None.")
t = np.r_[x[0], x, x[-1]]
c = np.asarray(y)
c = np.ascontiguousarray(c, dtype=_get_dtype(c.dtype))
return BSpline.construct_fast(t, c, k, axis=axis)
k = operator.index(k)
if bc_type == 'periodic' and t is not None:
raise NotImplementedError("For periodic case t is constructed "
"automatically and can not be passed manually")
# come up with a sensible knot vector, if needed
if t is None:
if deriv_l is None and deriv_r is None:
if bc_type == 'periodic':
t = _periodic_knots(x, k)
elif k == 2:
# OK, it's a bit ad hoc: Greville sites + omit
# 2nd and 2nd-to-last points, a la not-a-knot
t = (x[1:] + x[:-1]) / 2.
t = np.r_[(x[0],)*(k+1),
t[1:-1],
(x[-1],)*(k+1)]
else:
t = _not_a_knot(x, k)
else:
t = _augknt(x, k)
t = _as_float_array(t, check_finite)
if x.ndim != 1 or np.any(x[1:] < x[:-1]):
raise ValueError("Expect x to be a 1-D sorted array_like.")
if np.any(x[1:] == x[:-1]):
raise ValueError("Expect x to not have duplicates")
if k < 0:
raise ValueError("Expect non-negative k.")
if t.ndim != 1 or np.any(t[1:] < t[:-1]):
raise ValueError("Expect t to be a 1-D sorted array_like.")
if x.size != y.shape[0]:
raise ValueError('Shapes of x {} and y {} are incompatible'
.format(x.shape, y.shape))
if t.size < x.size + k + 1:
raise ValueError('Got %d knots, need at least %d.' %
(t.size, x.size + k + 1))
if (x[0] < t[k]) or (x[-1] > t[-k]):
raise ValueError('Out of bounds w/ x = %s.' % x)
if bc_type == 'periodic':
return _make_periodic_spline(x, y, t, k, axis)
# Here : deriv_l, r = [(nu, value), ...]
deriv_l = _convert_string_aliases(deriv_l, y.shape[1:])
deriv_l_ords, deriv_l_vals = _process_deriv_spec(deriv_l)
nleft = deriv_l_ords.shape[0]
deriv_r = _convert_string_aliases(deriv_r, y.shape[1:])
deriv_r_ords, deriv_r_vals = _process_deriv_spec(deriv_r)
nright = deriv_r_ords.shape[0]
# have `n` conditions for `nt` coefficients; need nt-n derivatives
n = x.size
nt = t.size - k - 1
if nt - n != nleft + nright:
raise ValueError("The number of derivatives at boundaries does not "
"match: expected %s, got %s+%s" % (nt-n, nleft, nright))
# set up the LHS: the collocation matrix + derivatives at boundaries
kl = ku = k
ab = np.zeros((2*kl + ku + 1, nt), dtype=np.float_, order='F')
_bspl._colloc(x, t, k, ab, offset=nleft)
if nleft > 0:
_bspl._handle_lhs_derivatives(t, k, x[0], ab, kl, ku, deriv_l_ords)
if nright > 0:
_bspl._handle_lhs_derivatives(t, k, x[-1], ab, kl, ku, deriv_r_ords,
offset=nt-nright)
# set up the RHS: values to interpolate (+ derivative values, if any)
extradim = prod(y.shape[1:])
rhs = np.empty((nt, extradim), dtype=y.dtype)
if nleft > 0:
rhs[:nleft] = deriv_l_vals.reshape(-1, extradim)
rhs[nleft:nt - nright] = y.reshape(-1, extradim)
if nright > 0:
rhs[nt - nright:] = deriv_r_vals.reshape(-1, extradim)
# solve Ab @ x = rhs; this is the relevant part of linalg.solve_banded
if check_finite:
ab, rhs = map(np.asarray_chkfinite, (ab, rhs))
gbsv, = get_lapack_funcs(('gbsv',), (ab, rhs))
lu, piv, c, info = gbsv(kl, ku, ab, rhs,
overwrite_ab=True, overwrite_b=True)
if info > 0:
raise LinAlgError("Collocation matix is singular.")
elif info < 0:
raise ValueError('illegal value in %d-th argument of internal gbsv' % -info)
c = np.ascontiguousarray(c.reshape((nt,) + y.shape[1:]))
return BSpline.construct_fast(t, c, k, axis=axis)
def make_lsq_spline(x, y, t, k=3, w=None, axis=0, check_finite=True):
r"""Compute the (coefficients of) an LSQ B-spline.
The result is a linear combination
.. math::
S(x) = \sum_j c_j B_j(x; t)
of the B-spline basis elements, :math:`B_j(x; t)`, which minimizes
.. math::
\sum_{j} \left( w_j \times (S(x_j) - y_j) \right)^2
Parameters
----------
x : array_like, shape (m,)
Abscissas.
y : array_like, shape (m, ...)
Ordinates.
t : array_like, shape (n + k + 1,).
Knots.
Knots and data points must satisfy Schoenberg-Whitney conditions.
k : int, optional
B-spline degree. Default is cubic, k=3.
w : array_like, shape (n,), optional
Weights for spline fitting. Must be positive. If ``None``,
then weights are all equal.
Default is ``None``.
axis : int, optional
Interpolation axis. Default is zero.
check_finite : bool, optional
Whether to check that the input arrays contain only finite numbers.
Disabling may give a performance gain, but may result in problems
(crashes, non-termination) if the inputs do contain infinities or NaNs.
Default is True.
Returns
-------
b : a BSpline object of the degree `k` with knots `t`.
Notes
-----
The number of data points must be larger than the spline degree `k`.
Knots `t` must satisfy the Schoenberg-Whitney conditions,
i.e., there must be a subset of data points ``x[j]`` such that
``t[j] < x[j] < t[j+k+1]``, for ``j=0, 1,...,n-k-2``.
Examples
--------
Generate some noisy data:
>>> rng = np.random.default_rng()
>>> x = np.linspace(-3, 3, 50)
>>> y = np.exp(-x**2) + 0.1 * rng.standard_normal(50)
Now fit a smoothing cubic spline with a pre-defined internal knots.
Here we make the knot vector (k+1)-regular by adding boundary knots:
>>> from scipy.interpolate import make_lsq_spline, BSpline
>>> t = [-1, 0, 1]
>>> k = 3
>>> t = np.r_[(x[0],)*(k+1),
... t,
... (x[-1],)*(k+1)]
>>> spl = make_lsq_spline(x, y, t, k)
For comparison, we also construct an interpolating spline for the same
set of data:
>>> from scipy.interpolate import make_interp_spline
>>> spl_i = make_interp_spline(x, y)
Plot both:
>>> import matplotlib.pyplot as plt
>>> xs = np.linspace(-3, 3, 100)
>>> plt.plot(x, y, 'ro', ms=5)
>>> plt.plot(xs, spl(xs), 'g-', lw=3, label='LSQ spline')
>>> plt.plot(xs, spl_i(xs), 'b-', lw=3, alpha=0.7, label='interp spline')
>>> plt.legend(loc='best')
>>> plt.show()
**NaN handling**: If the input arrays contain ``nan`` values, the result is
not useful since the underlying spline fitting routines cannot deal with
``nan``. A workaround is to use zero weights for not-a-number data points:
>>> y[8] = np.nan
>>> w = np.isnan(y)
>>> y[w] = 0.
>>> tck = make_lsq_spline(x, y, t, w=~w)
Notice the need to replace a ``nan`` by a numerical value (precise value
does not matter as long as the corresponding weight is zero.)
See Also
--------
BSpline : base class representing the B-spline objects
make_interp_spline : a similar factory function for interpolating splines
LSQUnivariateSpline : a FITPACK-based spline fitting routine
splrep : a FITPACK-based fitting routine
"""
x = _as_float_array(x, check_finite)
y = _as_float_array(y, check_finite)
t = _as_float_array(t, check_finite)
if w is not None:
w = _as_float_array(w, check_finite)
else:
w = np.ones_like(x)
k = operator.index(k)
axis = normalize_axis_index(axis, y.ndim)
y = np.moveaxis(y, axis, 0) # now internally interp axis is zero
if x.ndim != 1 or np.any(x[1:] - x[:-1] <= 0):
raise ValueError("Expect x to be a 1-D sorted array_like.")
if x.shape[0] < k+1:
raise ValueError("Need more x points.")
if k < 0:
raise ValueError("Expect non-negative k.")
if t.ndim != 1 or np.any(t[1:] - t[:-1] < 0):
raise ValueError("Expect t to be a 1-D sorted array_like.")
if x.size != y.shape[0]:
raise ValueError('Shapes of x {} and y {} are incompatible'
.format(x.shape, y.shape))
if k > 0 and np.any((x < t[k]) | (x > t[-k])):
raise ValueError('Out of bounds w/ x = %s.' % x)
if x.size != w.size:
raise ValueError('Shapes of x {} and w {} are incompatible'
.format(x.shape, w.shape))
# number of coefficients
n = t.size - k - 1
# construct A.T @ A and rhs with A the collocation matrix, and
# rhs = A.T @ y for solving the LSQ problem ``A.T @ A @ c = A.T @ y``
lower = True
extradim = prod(y.shape[1:])
ab = np.zeros((k+1, n), dtype=np.float_, order='F')
rhs = np.zeros((n, extradim), dtype=y.dtype, order='F')
_bspl._norm_eq_lsq(x, t, k,
y.reshape(-1, extradim),
w,
ab, rhs)
rhs = rhs.reshape((n,) + y.shape[1:])
# have observation matrix & rhs, can solve the LSQ problem
cho_decomp = cholesky_banded(ab, overwrite_ab=True, lower=lower,
check_finite=check_finite)
c = cho_solve_banded((cho_decomp, lower), rhs, overwrite_b=True,
check_finite=check_finite)
c = np.ascontiguousarray(c)
return BSpline.construct_fast(t, c, k, axis=axis)