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Inzynierka_Gwiazdy/machine_learning/Lib/site-packages/scipy/interpolate/_bsplines.py

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Machine learning
2023-09-20 19:46:58 +02:00
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 scipy.optimize import minimize_scalar
from . import _bspl
from . import _fitpack_impl
from scipy._lib._util import prod
from scipy.sparse import csr_array
from scipy.special import poch
from itertools import combinations
__all__ = ["BSpline", "make_interp_spline", "make_lsq_spline",
"make_smoothing_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
>>> import numpy as np
>>> 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:
>>> import numpy as np
>>> 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 = [0, 1, 1, 2]
>>> b = BSpline.basis_element(t)
>>> 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, extrapolate=False):
"""
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.
extrapolate : bool or 'periodic', optional
Whether to extrapolate based on the first and last intervals
or raise an error. If 'periodic', periodic extrapolation is used.
Default is False.
.. versionadded:: 1.10.0
Returns
-------
design_matrix : `csr_array` object
Sparse matrix in CSR format where each row contains all the basis
elements of the input row (first row = basis elements of x[0],
..., last row = basis elements x[-1]).
Examples
--------
Construct a design matrix for a B-spline
>>> from scipy.interpolate import make_interp_spline, BSpline
>>> import numpy as np
>>> 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 extrapolate != 'periodic':
extrapolate = bool(extrapolate)
if k < 0:
raise ValueError("Spline order cannot be negative.")
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
# elements 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}.")
if extrapolate == 'periodic':
# With periodic extrapolation we map x to the segment
# [t[k], t[n]].
n = t.size - k - 1
x = t[k] + (x - t[k]) % (t[n] - t[k])
extrapolate = False
elif not extrapolate and (
(min(x) < t[k]) or (max(x) > t[t.shape[0] - k - 1])
):
# Checks from `find_interval` function
raise ValueError(f'Out of bounds w/ x = {x}.')
# Compute number of non-zeros of final CSR array in order to determine
# the dtype of indices and indptr of the CSR array.
n = x.shape[0]
nnz = n * (k + 1)
if nnz < np.iinfo(np.int32).max:
int_dtype = np.int32
else:
int_dtype = np.int64
# Preallocate indptr and indices
indices = np.empty(n * (k + 1), dtype=int_dtype)
indptr = np.arange(0, (n + 1) * (k + 1), k + 1, dtype=int_dtype)
# indptr is not passed to Cython as it is already fully computed
data, indices = _bspl._make_design_matrix(
x, t, k, extrapolate, indices
)
return csr_array(
(data, indices, indptr),
shape=(x.shape[0], t.shape[0] - 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).
integral = _fitpack_impl.splint(a, b, self.tck)
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 data points 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 (``deriv_l``) sets the boundary conditions at ``x[0]`` and
the second element (``deriv_r``) 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:
>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> 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
>>> 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
>>> 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
# sanity check the input
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")
if x.size != y.shape[0]:
raise ValueError('Shapes of x {} and y {} are incompatible'
.format(x.shape, y.shape))
if np.any(x[1:] == x[:-1]):
raise ValueError("Expect x to not have duplicates")
if x.ndim != 1 or np.any(x[1:] < x[:-1]):
raise ValueError("Expect x to be a 1D strictly increasing sequence.")
# 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 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 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))
# bail out if the `y` array is zero-sized
if y.size == 0:
c = np.zeros((nt,) + y.shape[1:], dtype=float)
return BSpline.construct_fast(t, c, k, axis=axis)
# 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 matrix 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 (Least SQuared) based
fitting 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 (m,), 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:
>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> 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:
>>> 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)
#############################
# Smoothing spline helpers #
#############################
def _compute_optimal_gcv_parameter(X, wE, y, w):
"""
Returns an optimal regularization parameter from the GCV criteria [1].
Parameters
----------
X : array, shape (5, n)
5 bands of the design matrix ``X`` stored in LAPACK banded storage.
wE : array, shape (5, n)
5 bands of the penalty matrix :math:`W^{-1} E` stored in LAPACK banded
storage.
y : array, shape (n,)
Ordinates.
w : array, shape (n,)
Vector of weights.
Returns
-------
lam : float
An optimal from the GCV criteria point of view regularization
parameter.
Notes
-----
No checks are performed.
References
----------
.. [1] G. Wahba, "Estimating the smoothing parameter" in Spline models
for observational data, Philadelphia, Pennsylvania: Society for
Industrial and Applied Mathematics, 1990, pp. 45-65.
:doi:`10.1137/1.9781611970128`
"""
def compute_banded_symmetric_XT_W_Y(X, w, Y):
"""
Assuming that the product :math:`X^T W Y` is symmetric and both ``X``
and ``Y`` are 5-banded, compute the unique bands of the product.
Parameters
----------
X : array, shape (5, n)
5 bands of the matrix ``X`` stored in LAPACK banded storage.
w : array, shape (n,)
Array of weights
Y : array, shape (5, n)
5 bands of the matrix ``Y`` stored in LAPACK banded storage.
Returns
-------
res : array, shape (4, n)
The result of the product :math:`X^T Y` stored in the banded way.
Notes
-----
As far as the matrices ``X`` and ``Y`` are 5-banded, their product
:math:`X^T W Y` is 7-banded. It is also symmetric, so we can store only
unique diagonals.
"""
# compute W Y
W_Y = np.copy(Y)
W_Y[2] *= w
for i in range(2):
W_Y[i, 2 - i:] *= w[:-2 + i]
W_Y[3 + i, :-1 - i] *= w[1 + i:]
n = X.shape[1]
res = np.zeros((4, n))
for i in range(n):
for j in range(min(n-i, 4)):
res[-j-1, i + j] = sum(X[j:, i] * W_Y[:5-j, i + j])
return res
def compute_b_inv(A):
"""
Inverse 3 central bands of matrix :math:`A=U^T D^{-1} U` assuming that
``U`` is a unit upper triangular banded matrix using an algorithm
proposed in [1].
Parameters
----------
A : array, shape (4, n)
Matrix to inverse, stored in LAPACK banded storage.
Returns
-------
B : array, shape (4, n)
3 unique bands of the symmetric matrix that is an inverse to ``A``.
The first row is filled with zeros.
Notes
-----
The algorithm is based on the cholesky decomposition and, therefore,
in case matrix ``A`` is close to not positive defined, the function
raises LinalgError.
Both matrices ``A`` and ``B`` are stored in LAPACK banded storage.
References
----------
.. [1] M. F. Hutchinson and F. R. de Hoog, "Smoothing noisy data with
spline functions," Numerische Mathematik, vol. 47, no. 1,
pp. 99-106, 1985.
:doi:`10.1007/BF01389878`
"""
def find_b_inv_elem(i, j, U, D, B):
rng = min(3, n - i - 1)
rng_sum = 0.
if j == 0:
# use 2-nd formula from [1]
for k in range(1, rng + 1):
rng_sum -= U[-k - 1, i + k] * B[-k - 1, i + k]
rng_sum += D[i]
B[-1, i] = rng_sum
else:
# use 1-st formula from [1]
for k in range(1, rng + 1):
diag = abs(k - j)
ind = i + min(k, j)
rng_sum -= U[-k - 1, i + k] * B[-diag - 1, ind + diag]
B[-j - 1, i + j] = rng_sum
U = cholesky_banded(A)
for i in range(2, 5):
U[-i, i-1:] /= U[-1, :-i+1]
D = 1. / (U[-1])**2
U[-1] /= U[-1]
n = U.shape[1]
B = np.zeros(shape=(4, n))
for i in range(n - 1, -1, -1):
for j in range(min(3, n - i - 1), -1, -1):
find_b_inv_elem(i, j, U, D, B)
# the first row contains garbage and should be removed
B[0] = [0.] * n
return B
def _gcv(lam, X, XtWX, wE, XtE):
r"""
Computes the generalized cross-validation criteria [1].
Parameters
----------
lam : float, (:math:`\lambda \geq 0`)
Regularization parameter.
X : array, shape (5, n)
Matrix is stored in LAPACK banded storage.
XtWX : array, shape (4, n)
Product :math:`X^T W X` stored in LAPACK banded storage.
wE : array, shape (5, n)
Matrix :math:`W^{-1} E` stored in LAPACK banded storage.
XtE : array, shape (4, n)
Product :math:`X^T E` stored in LAPACK banded storage.
Returns
-------
res : float
Value of the GCV criteria with the regularization parameter
:math:`\lambda`.
Notes
-----
Criteria is computed from the formula (1.3.2) [3]:
.. math:
GCV(\lambda) = \dfrac{1}{n} \sum\limits_{k = 1}^{n} \dfrac{ \left(
y_k - f_{\lambda}(x_k) \right)^2}{\left( 1 - \Tr{A}/n\right)^2}$.
The criteria is discussed in section 1.3 [3].
The numerator is computed using (2.2.4) [3] and the denominator is
computed using an algorithm from [2] (see in the ``compute_b_inv``
function).
References
----------
.. [1] G. Wahba, "Estimating the smoothing parameter" in Spline models
for observational data, Philadelphia, Pennsylvania: Society for
Industrial and Applied Mathematics, 1990, pp. 45-65.
:doi:`10.1137/1.9781611970128`
.. [2] M. F. Hutchinson and F. R. de Hoog, "Smoothing noisy data with
spline functions," Numerische Mathematik, vol. 47, no. 1,
pp. 99-106, 1985.
:doi:`10.1007/BF01389878`
.. [3] E. Zemlyanoy, "Generalized cross-validation smoothing splines",
BSc thesis, 2022. Might be available (in Russian)
`here <https://www.hse.ru/ba/am/students/diplomas/620910604>`_
"""
# Compute the numerator from (2.2.4) [3]
n = X.shape[1]
c = solve_banded((2, 2), X + lam * wE, y)
res = np.zeros(n)
# compute ``W^{-1} E c`` with respect to banded-storage of ``E``
tmp = wE * c
for i in range(n):
for j in range(max(0, i - n + 3), min(5, i + 3)):
res[i] += tmp[j, i + 2 - j]
numer = np.linalg.norm(lam * res)**2 / n
# compute the denominator
lhs = XtWX + lam * XtE
try:
b_banded = compute_b_inv(lhs)
# compute the trace of the product b_banded @ XtX
tr = b_banded * XtWX
tr[:-1] *= 2
# find the denominator
denom = (1 - sum(sum(tr)) / n)**2
except LinAlgError:
# cholesky decomposition cannot be performed
raise ValueError('Seems like the problem is ill-posed')
res = numer / denom
return res
n = X.shape[1]
XtWX = compute_banded_symmetric_XT_W_Y(X, w, X)
XtE = compute_banded_symmetric_XT_W_Y(X, w, wE)
def fun(lam):
return _gcv(lam, X, XtWX, wE, XtE)
gcv_est = minimize_scalar(fun, bounds=(0, n), method='Bounded')
if gcv_est.success:
return gcv_est.x
raise ValueError(f"Unable to find minimum of the GCV "
f"function: {gcv_est.message}")
def _coeff_of_divided_diff(x):
"""
Returns the coefficients of the divided difference.
Parameters
----------
x : array, shape (n,)
Array which is used for the computation of divided difference.
Returns
-------
res : array_like, shape (n,)
Coefficients of the divided difference.
Notes
-----
Vector ``x`` should have unique elements, otherwise an error division by
zero might be raised.
No checks are performed.
"""
n = x.shape[0]
res = np.zeros(n)
for i in range(n):
pp = 1.
for k in range(n):
if k != i:
pp *= (x[i] - x[k])
res[i] = 1. / pp
return res
def make_smoothing_spline(x, y, w=None, lam=None):
r"""
Compute the (coefficients of) smoothing cubic spline function using
``lam`` to control the tradeoff between the amount of smoothness of the
curve and its proximity to the data. In case ``lam`` is None, using the
GCV criteria [1] to find it.
A smoothing spline is found as a solution to the regularized weighted
linear regression problem:
.. math::
\sum\limits_{i=1}^n w_i\lvert y_i - f(x_i) \rvert^2 +
\lambda\int\limits_{x_1}^{x_n} (f^{(2)}(u))^2 d u
where :math:`f` is a spline function, :math:`w` is a vector of weights and
:math:`\lambda` is a regularization parameter.
If ``lam`` is None, we use the GCV criteria to find an optimal
regularization parameter, otherwise we solve the regularized weighted
linear regression problem with given parameter. The parameter controls
the tradeoff in the following way: the larger the parameter becomes, the
smoother the function gets.
Parameters
----------
x : array_like, shape (n,)
Abscissas.
y : array_like, shape (n,)
Ordinates.
w : array_like, shape (n,), optional
Vector of weights. Default is ``np.ones_like(x)``.
lam : float, (:math:`\lambda \geq 0`), optional
Regularization parameter. If ``lam`` is None, then it is found from
the GCV criteria. Default is None.
Returns
-------
func : a BSpline object.
A callable representing a spline in the B-spline basis
as a solution of the problem of smoothing splines using
the GCV criteria [1] in case ``lam`` is None, otherwise using the
given parameter ``lam``.
Notes
-----
This algorithm is a clean room reimplementation of the algorithm
introduced by Woltring in FORTRAN [2]. The original version cannot be used
in SciPy source code because of the license issues. The details of the
reimplementation are discussed here (available only in Russian) [4].
If the vector of weights ``w`` is None, we assume that all the points are
equal in terms of weights, and vector of weights is vector of ones.
Note that in weighted residual sum of squares, weights are not squared:
:math:`\sum\limits_{i=1}^n w_i\lvert y_i - f(x_i) \rvert^2` while in
``splrep`` the sum is built from the squared weights.
In cases when the initial problem is ill-posed (for example, the product
:math:`X^T W X` where :math:`X` is a design matrix is not a positive
defined matrix) a ValueError is raised.
References
----------
.. [1] G. Wahba, "Estimating the smoothing parameter" in Spline models for
observational data, Philadelphia, Pennsylvania: Society for Industrial
and Applied Mathematics, 1990, pp. 45-65.
:doi:`10.1137/1.9781611970128`
.. [2] H. J. Woltring, A Fortran package for generalized, cross-validatory
spline smoothing and differentiation, Advances in Engineering
Software, vol. 8, no. 2, pp. 104-113, 1986.
:doi:`10.1016/0141-1195(86)90098-7`
.. [3] T. Hastie, J. Friedman, and R. Tisbshirani, "Smoothing Splines" in
The elements of Statistical Learning: Data Mining, Inference, and
prediction, New York: Springer, 2017, pp. 241-249.
:doi:`10.1007/978-0-387-84858-7`
.. [4] E. Zemlyanoy, "Generalized cross-validation smoothing splines",
BSc thesis, 2022.
`<https://www.hse.ru/ba/am/students/diplomas/620910604>`_ (in
Russian)
Examples
--------
Generate some noisy data
>>> import numpy as np
>>> np.random.seed(1234)
>>> n = 200
>>> def func(x):
... return x**3 + x**2 * np.sin(4 * x)
>>> x = np.sort(np.random.random_sample(n) * 4 - 2)
>>> y = func(x) + np.random.normal(scale=1.5, size=n)
Make a smoothing spline function
>>> from scipy.interpolate import make_smoothing_spline
>>> spl = make_smoothing_spline(x, y)
Plot both
>>> import matplotlib.pyplot as plt
>>> grid = np.linspace(x[0], x[-1], 400)
>>> plt.plot(grid, spl(grid), label='Spline')
>>> plt.plot(grid, func(grid), label='Original function')
>>> plt.scatter(x, y, marker='.')
>>> plt.legend(loc='best')
>>> plt.show()
"""
x = np.ascontiguousarray(x, dtype=float)
y = np.ascontiguousarray(y, dtype=float)
if any(x[1:] - x[:-1] <= 0):
raise ValueError('``x`` should be an ascending array')
if x.ndim != 1 or y.ndim != 1 or x.shape[0] != y.shape[0]:
raise ValueError('``x`` and ``y`` should be one dimensional and the'
' same size')
if w is None:
w = np.ones(len(x))
else:
w = np.ascontiguousarray(w)
if any(w <= 0):
raise ValueError('Invalid vector of weights')
t = np.r_[[x[0]] * 3, x, [x[-1]] * 3]
n = x.shape[0]
# It is known that the solution to the stated minimization problem exists
# and is a natural cubic spline with vector of knots equal to the unique
# elements of ``x`` [3], so we will solve the problem in the basis of
# natural splines.
# create design matrix in the B-spline basis
X_bspl = BSpline.design_matrix(x, t, 3)
# move from B-spline basis to the basis of natural splines using equations
# (2.1.7) [4]
# central elements
X = np.zeros((5, n))
for i in range(1, 4):
X[i, 2: -2] = X_bspl[i: i - 4, 3: -3][np.diag_indices(n - 4)]
# first elements
X[1, 1] = X_bspl[0, 0]
X[2, :2] = ((x[2] + x[1] - 2 * x[0]) * X_bspl[0, 0],
X_bspl[1, 1] + X_bspl[1, 2])
X[3, :2] = ((x[2] - x[0]) * X_bspl[1, 1], X_bspl[2, 2])
# last elements
X[1, -2:] = (X_bspl[-3, -3], (x[-1] - x[-3]) * X_bspl[-2, -2])
X[2, -2:] = (X_bspl[-2, -3] + X_bspl[-2, -2],
(2 * x[-1] - x[-2] - x[-3]) * X_bspl[-1, -1])
X[3, -2] = X_bspl[-1, -1]
# create penalty matrix and divide it by vector of weights: W^{-1} E
wE = np.zeros((5, n))
wE[2:, 0] = _coeff_of_divided_diff(x[:3]) / w[:3]
wE[1:, 1] = _coeff_of_divided_diff(x[:4]) / w[:4]
for j in range(2, n - 2):
wE[:, j] = (x[j+2] - x[j-2]) * _coeff_of_divided_diff(x[j-2:j+3])\
/ w[j-2: j+3]
wE[:-1, -2] = -_coeff_of_divided_diff(x[-4:]) / w[-4:]
wE[:-2, -1] = _coeff_of_divided_diff(x[-3:]) / w[-3:]
wE *= 6
if lam is None:
lam = _compute_optimal_gcv_parameter(X, wE, y, w)
elif lam < 0.:
raise ValueError('Regularization parameter should be non-negative')
# solve the initial problem in the basis of natural splines
c = solve_banded((2, 2), X + lam * wE, y)
# move back to B-spline basis using equations (2.2.10) [4]
c_ = np.r_[c[0] * (t[5] + t[4] - 2 * t[3]) + c[1],
c[0] * (t[5] - t[3]) + c[1],
c[1: -1],
c[-1] * (t[-4] - t[-6]) + c[-2],
c[-1] * (2 * t[-4] - t[-5] - t[-6]) + c[-2]]
return BSpline.construct_fast(t, c_, 3)
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