1012 lines
34 KiB
Python
1012 lines
34 KiB
Python
import operator
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import numpy as np
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from numpy.core.multiarray import normalize_axis_index
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from scipy.linalg import (get_lapack_funcs, LinAlgError,
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cholesky_banded, cho_solve_banded)
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from . import _bspl
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from . import _fitpack_impl
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from . import _fitpack as _dierckx
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from scipy._lib._util import prod
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__all__ = ["BSpline", "make_interp_spline", "make_lsq_spline"]
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def _get_dtype(dtype):
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"""Return np.complex128 for complex dtypes, np.float64 otherwise."""
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if np.issubdtype(dtype, np.complexfloating):
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return np.complex_
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else:
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return np.float_
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def _as_float_array(x, check_finite=False):
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"""Convert the input into a C contiguous float array.
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NB: Upcasts half- and single-precision floats to double precision.
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"""
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x = np.ascontiguousarray(x)
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dtyp = _get_dtype(x.dtype)
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x = x.astype(dtyp, copy=False)
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if check_finite and not np.isfinite(x).all():
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raise ValueError("Array must not contain infs or nans.")
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return x
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class BSpline(object):
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r"""Univariate spline in the B-spline basis.
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.. math::
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S(x) = \sum_{j=0}^{n-1} c_j B_{j, k; t}(x)
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where :math:`B_{j, k; t}` are B-spline basis functions of degree `k`
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and knots `t`.
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Parameters
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----------
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t : ndarray, shape (n+k+1,)
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knots
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c : ndarray, shape (>=n, ...)
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spline coefficients
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k : int
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B-spline degree
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extrapolate : bool or 'periodic', optional
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whether to extrapolate beyond the base interval, ``t[k] .. t[n]``,
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or to return nans.
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If True, extrapolates the first and last polynomial pieces of b-spline
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functions active on the base interval.
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If 'periodic', periodic extrapolation is used.
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Default is True.
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axis : int, optional
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Interpolation axis. Default is zero.
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Attributes
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----------
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t : ndarray
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knot vector
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c : ndarray
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spline coefficients
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k : int
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spline degree
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extrapolate : bool
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If True, extrapolates the first and last polynomial pieces of b-spline
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functions active on the base interval.
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axis : int
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Interpolation axis.
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tck : tuple
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A read-only equivalent of ``(self.t, self.c, self.k)``
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Methods
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-------
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__call__
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basis_element
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derivative
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antiderivative
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integrate
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construct_fast
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Notes
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-----
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B-spline basis elements are defined via
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.. math::
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B_{i, 0}(x) = 1, \textrm{if $t_i \le x < t_{i+1}$, otherwise $0$,}
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B_{i, k}(x) = \frac{x - t_i}{t_{i+k} - t_i} B_{i, k-1}(x)
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+ \frac{t_{i+k+1} - x}{t_{i+k+1} - t_{i+1}} B_{i+1, k-1}(x)
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**Implementation details**
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- At least ``k+1`` coefficients are required for a spline of degree `k`,
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so that ``n >= k+1``. Additional coefficients, ``c[j]`` with
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``j > n``, are ignored.
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- B-spline basis elements of degree `k` form a partition of unity on the
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*base interval*, ``t[k] <= x <= t[n]``.
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Examples
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--------
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Translating the recursive definition of B-splines into Python code, we have:
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>>> def B(x, k, i, t):
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... if k == 0:
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... return 1.0 if t[i] <= x < t[i+1] else 0.0
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... if t[i+k] == t[i]:
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... c1 = 0.0
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... else:
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... c1 = (x - t[i])/(t[i+k] - t[i]) * B(x, k-1, i, t)
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... if t[i+k+1] == t[i+1]:
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... c2 = 0.0
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... else:
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... c2 = (t[i+k+1] - x)/(t[i+k+1] - t[i+1]) * B(x, k-1, i+1, t)
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... return c1 + c2
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>>> def bspline(x, t, c, k):
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... n = len(t) - k - 1
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... assert (n >= k+1) and (len(c) >= n)
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... return sum(c[i] * B(x, k, i, t) for i in range(n))
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Note that this is an inefficient (if straightforward) way to
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evaluate B-splines --- this spline class does it in an equivalent,
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but much more efficient way.
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Here we construct a quadratic spline function on the base interval
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``2 <= x <= 4`` and compare with the naive way of evaluating the spline:
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>>> from scipy.interpolate import BSpline
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>>> k = 2
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>>> t = [0, 1, 2, 3, 4, 5, 6]
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>>> c = [-1, 2, 0, -1]
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>>> spl = BSpline(t, c, k)
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>>> spl(2.5)
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array(1.375)
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>>> bspline(2.5, t, c, k)
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1.375
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Note that outside of the base interval results differ. This is because
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`BSpline` extrapolates the first and last polynomial pieces of B-spline
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functions active on the base interval.
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>>> import matplotlib.pyplot as plt
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>>> fig, ax = plt.subplots()
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>>> xx = np.linspace(1.5, 4.5, 50)
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>>> ax.plot(xx, [bspline(x, t, c ,k) for x in xx], 'r-', lw=3, label='naive')
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>>> ax.plot(xx, spl(xx), 'b-', lw=4, alpha=0.7, label='BSpline')
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>>> ax.grid(True)
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>>> ax.legend(loc='best')
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>>> plt.show()
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References
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----------
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.. [1] Tom Lyche and Knut Morken, Spline methods,
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http://www.uio.no/studier/emner/matnat/ifi/INF-MAT5340/v05/undervisningsmateriale/
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.. [2] Carl de Boor, A practical guide to splines, Springer, 2001.
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"""
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def __init__(self, t, c, k, extrapolate=True, axis=0):
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super(BSpline, self).__init__()
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self.k = operator.index(k)
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self.c = np.asarray(c)
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self.t = np.ascontiguousarray(t, dtype=np.float64)
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if extrapolate == 'periodic':
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self.extrapolate = extrapolate
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else:
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self.extrapolate = bool(extrapolate)
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n = self.t.shape[0] - self.k - 1
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axis = normalize_axis_index(axis, self.c.ndim)
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# Note that the normalized axis is stored in the object.
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self.axis = axis
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if axis != 0:
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# roll the interpolation axis to be the first one in self.c
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# More specifically, the target shape for self.c is (n, ...),
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# and axis !=0 means that we have c.shape (..., n, ...)
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# ^
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# axis
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self.c = np.rollaxis(self.c, axis)
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if k < 0:
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raise ValueError("Spline order cannot be negative.")
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if self.t.ndim != 1:
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raise ValueError("Knot vector must be one-dimensional.")
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if n < self.k + 1:
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raise ValueError("Need at least %d knots for degree %d" %
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(2*k + 2, k))
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if (np.diff(self.t) < 0).any():
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raise ValueError("Knots must be in a non-decreasing order.")
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if len(np.unique(self.t[k:n+1])) < 2:
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raise ValueError("Need at least two internal knots.")
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if not np.isfinite(self.t).all():
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raise ValueError("Knots should not have nans or infs.")
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if self.c.ndim < 1:
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raise ValueError("Coefficients must be at least 1-dimensional.")
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if self.c.shape[0] < n:
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raise ValueError("Knots, coefficients and degree are inconsistent.")
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dt = _get_dtype(self.c.dtype)
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self.c = np.ascontiguousarray(self.c, dtype=dt)
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@classmethod
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def construct_fast(cls, t, c, k, extrapolate=True, axis=0):
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"""Construct a spline without making checks.
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Accepts same parameters as the regular constructor. Input arrays
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`t` and `c` must of correct shape and dtype.
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"""
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self = object.__new__(cls)
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self.t, self.c, self.k = t, c, k
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self.extrapolate = extrapolate
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self.axis = axis
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return self
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@property
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def tck(self):
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"""Equivalent to ``(self.t, self.c, self.k)`` (read-only).
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"""
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return self.t, self.c, self.k
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@classmethod
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def basis_element(cls, t, extrapolate=True):
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"""Return a B-spline basis element ``B(x | t[0], ..., t[k+1])``.
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Parameters
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----------
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t : ndarray, shape (k+1,)
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internal knots
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extrapolate : bool or 'periodic', optional
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whether to extrapolate beyond the base interval, ``t[0] .. t[k+1]``,
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or to return nans.
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If 'periodic', periodic extrapolation is used.
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Default is True.
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Returns
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-------
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basis_element : callable
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A callable representing a B-spline basis element for the knot
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vector `t`.
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Notes
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-----
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The degree of the B-spline, `k`, is inferred from the length of `t` as
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``len(t)-2``. The knot vector is constructed by appending and prepending
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``k+1`` elements to internal knots `t`.
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Examples
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--------
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Construct a cubic B-spline:
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>>> from scipy.interpolate import BSpline
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>>> b = BSpline.basis_element([0, 1, 2, 3, 4])
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>>> k = b.k
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>>> b.t[k:-k]
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array([ 0., 1., 2., 3., 4.])
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>>> k
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3
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Construct a quadratic B-spline on ``[0, 1, 1, 2]``, and compare
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to its explicit form:
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>>> t = [-1, 0, 1, 1, 2]
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>>> b = BSpline.basis_element(t[1:])
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>>> def f(x):
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... return np.where(x < 1, x*x, (2. - x)**2)
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>>> import matplotlib.pyplot as plt
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>>> fig, ax = plt.subplots()
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>>> x = np.linspace(0, 2, 51)
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>>> ax.plot(x, b(x), 'g', lw=3)
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>>> ax.plot(x, f(x), 'r', lw=8, alpha=0.4)
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>>> ax.grid(True)
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>>> plt.show()
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"""
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k = len(t) - 2
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t = _as_float_array(t)
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t = np.r_[(t[0]-1,) * k, t, (t[-1]+1,) * k]
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c = np.zeros_like(t)
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c[k] = 1.
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return cls.construct_fast(t, c, k, extrapolate)
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def __call__(self, x, nu=0, extrapolate=None):
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"""
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Evaluate a spline function.
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Parameters
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----------
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x : array_like
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points to evaluate the spline at.
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nu: int, optional
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derivative to evaluate (default is 0).
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extrapolate : bool or 'periodic', optional
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whether to extrapolate based on the first and last intervals
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or return nans. If 'periodic', periodic extrapolation is used.
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Default is `self.extrapolate`.
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Returns
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-------
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y : array_like
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Shape is determined by replacing the interpolation axis
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in the coefficient array with the shape of `x`.
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"""
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if extrapolate is None:
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extrapolate = self.extrapolate
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x = np.asarray(x)
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x_shape, x_ndim = x.shape, x.ndim
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x = np.ascontiguousarray(x.ravel(), dtype=np.float_)
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# With periodic extrapolation we map x to the segment
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# [self.t[k], self.t[n]].
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if extrapolate == 'periodic':
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n = self.t.size - self.k - 1
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x = self.t[self.k] + (x - self.t[self.k]) % (self.t[n] -
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self.t[self.k])
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extrapolate = False
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out = np.empty((len(x), prod(self.c.shape[1:])), dtype=self.c.dtype)
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self._ensure_c_contiguous()
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self._evaluate(x, nu, extrapolate, out)
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out = out.reshape(x_shape + self.c.shape[1:])
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if self.axis != 0:
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# transpose to move the calculated values to the interpolation axis
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l = list(range(out.ndim))
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l = l[x_ndim:x_ndim+self.axis] + l[:x_ndim] + l[x_ndim+self.axis:]
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out = out.transpose(l)
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return out
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def _evaluate(self, xp, nu, extrapolate, out):
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_bspl.evaluate_spline(self.t, self.c.reshape(self.c.shape[0], -1),
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self.k, xp, nu, extrapolate, out)
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def _ensure_c_contiguous(self):
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"""
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c and t may be modified by the user. The Cython code expects
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that they are C contiguous.
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"""
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if not self.t.flags.c_contiguous:
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self.t = self.t.copy()
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if not self.c.flags.c_contiguous:
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self.c = self.c.copy()
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def derivative(self, nu=1):
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"""Return a B-spline representing the derivative.
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Parameters
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----------
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nu : int, optional
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Derivative order.
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Default is 1.
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Returns
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-------
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b : BSpline object
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A new instance representing the derivative.
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See Also
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--------
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splder, splantider
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"""
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c = self.c
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# pad the c array if needed
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ct = len(self.t) - len(c)
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if ct > 0:
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c = np.r_[c, np.zeros((ct,) + c.shape[1:])]
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tck = _fitpack_impl.splder((self.t, c, self.k), nu)
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return self.construct_fast(*tck, extrapolate=self.extrapolate,
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axis=self.axis)
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def antiderivative(self, nu=1):
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"""Return a B-spline representing the antiderivative.
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Parameters
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----------
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nu : int, optional
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Antiderivative order. Default is 1.
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Returns
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-------
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b : BSpline object
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A new instance representing the antiderivative.
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Notes
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-----
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If antiderivative is computed and ``self.extrapolate='periodic'``,
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it will be set to False for the returned instance. This is done because
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the antiderivative is no longer periodic and its correct evaluation
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outside of the initially given x interval is difficult.
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See Also
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--------
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splder, splantider
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"""
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c = self.c
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# pad the c array if needed
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ct = len(self.t) - len(c)
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if ct > 0:
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c = np.r_[c, np.zeros((ct,) + c.shape[1:])]
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tck = _fitpack_impl.splantider((self.t, c, self.k), nu)
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if self.extrapolate == 'periodic':
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extrapolate = False
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else:
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extrapolate = self.extrapolate
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return self.construct_fast(*tck, extrapolate=extrapolate,
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axis=self.axis)
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def integrate(self, a, b, extrapolate=None):
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"""Compute a definite integral of the spline.
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Parameters
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----------
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a : float
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Lower limit of integration.
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b : float
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Upper limit of integration.
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extrapolate : bool or 'periodic', optional
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whether to extrapolate beyond the base interval,
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``t[k] .. t[-k-1]``, or take the spline to be zero outside of the
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base interval. If 'periodic', periodic extrapolation is used.
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If None (default), use `self.extrapolate`.
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Returns
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-------
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I : array_like
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Definite integral of the spline over the interval ``[a, b]``.
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Examples
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--------
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Construct the linear spline ``x if x < 1 else 2 - x`` on the base
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interval :math:`[0, 2]`, and integrate it
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>>> from scipy.interpolate import BSpline
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>>> b = BSpline.basis_element([0, 1, 2])
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>>> b.integrate(0, 1)
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array(0.5)
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If the integration limits are outside of the base interval, the result
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is controlled by the `extrapolate` parameter
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>>> b.integrate(-1, 1)
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array(0.0)
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>>> b.integrate(-1, 1, extrapolate=False)
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array(0.5)
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>>> import matplotlib.pyplot as plt
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>>> fig, ax = plt.subplots()
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>>> ax.grid(True)
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>>> ax.axvline(0, c='r', lw=5, alpha=0.5) # base interval
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>>> ax.axvline(2, c='r', lw=5, alpha=0.5)
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>>> xx = [-1, 1, 2]
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>>> ax.plot(xx, b(xx))
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>>> plt.show()
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"""
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if extrapolate is None:
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extrapolate = self.extrapolate
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# Prepare self.t and self.c.
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self._ensure_c_contiguous()
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# Swap integration bounds if needed.
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sign = 1
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if b < a:
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a, b = b, a
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sign = -1
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n = self.t.size - self.k - 1
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if extrapolate != "periodic" and not extrapolate:
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# Shrink the integration interval, if needed.
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a = max(a, self.t[self.k])
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b = min(b, self.t[n])
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if self.c.ndim == 1:
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# Fast path: use FITPACK's routine
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# (cf _fitpack_impl.splint).
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t, c, k = self.tck
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integral, wrk = _dierckx._splint(t, c, k, a, b)
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return integral * sign
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out = np.empty((2, prod(self.c.shape[1:])), dtype=self.c.dtype)
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# Compute the antiderivative.
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c = self.c
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ct = len(self.t) - len(c)
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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)
|