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fr/fr_env/lib/python3.8/site-packages/scipy/interpolate/_bsplines.py

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installed required libs
2021-02-17 12:26:31 +05:30
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)
from . import _bspl
from . import _fitpack_impl
from . import _fitpack as _dierckx
from scipy._lib._util import prod
__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
class BSpline(object):
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
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(BSpline, self).__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.rollaxis(self.c, axis)
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+1,)
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)
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:])
#################################
# 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 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``.
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()
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':
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)
# 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.")
x = _as_float_array(x, check_finite)
t = np.r_[x, x[-1]]
c = np.asarray(y)
c = np.rollaxis(c, axis)
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.")
x = _as_float_array(x, check_finite)
t = np.r_[x[0], x, x[-1]]
c = np.asarray(y)
c = np.rollaxis(c, axis)
c = np.ascontiguousarray(c, dtype=_get_dtype(c.dtype))
return BSpline.construct_fast(t, c, k, axis=axis)
x = _as_float_array(x, check_finite)
y = _as_float_array(y, check_finite)
k = operator.index(k)
# come up with a sensible knot vector, if needed
if t is None:
if deriv_l is None and deriv_r is None:
if 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)
y = np.rollaxis(y, axis) # now internally interp axis is zero
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)
# 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:
>>> x = np.linspace(-3, 3, 50)
>>> y = np.exp(-x**2) + 0.1 * np.random.randn(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.rollaxis(y, axis) # 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)