/
_fitpack2.py
2362 lines (1965 loc) · 87.1 KB
/
_fitpack2.py
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"""
fitpack --- curve and surface fitting with splines
fitpack is based on a collection of Fortran routines DIERCKX
by P. Dierckx (see http://www.netlib.org/dierckx/) transformed
to double routines by Pearu Peterson.
"""
# Created by Pearu Peterson, June,August 2003
__all__ = [
'UnivariateSpline',
'InterpolatedUnivariateSpline',
'LSQUnivariateSpline',
'BivariateSpline',
'LSQBivariateSpline',
'SmoothBivariateSpline',
'LSQSphereBivariateSpline',
'SmoothSphereBivariateSpline',
'RectBivariateSpline',
'RectSphereBivariateSpline']
import warnings
from numpy import zeros, concatenate, ravel, diff, array, ones # noqa:F401
import numpy as np
from . import _fitpack_impl
from . import dfitpack
dfitpack_int = dfitpack.types.intvar.dtype
# ############### Univariate spline ####################
_curfit_messages = {1: """
The required storage space exceeds the available storage space, as
specified by the parameter nest: nest too small. If nest is already
large (say nest > m/2), it may also indicate that s is too small.
The approximation returned is the weighted least-squares spline
according to the knots t[0],t[1],...,t[n-1]. (n=nest) the parameter fp
gives the corresponding weighted sum of squared residuals (fp>s).
""",
2: """
A theoretically impossible result was found during the iteration
process for finding a smoothing spline with fp = s: s too small.
There is an approximation returned but the corresponding weighted sum
of squared residuals does not satisfy the condition abs(fp-s)/s < tol.""",
3: """
The maximal number of iterations maxit (set to 20 by the program)
allowed for finding a smoothing spline with fp=s has been reached: s
too small.
There is an approximation returned but the corresponding weighted sum
of squared residuals does not satisfy the condition abs(fp-s)/s < tol.""",
10: """
Error on entry, no approximation returned. The following conditions
must hold:
xb<=x[0]<x[1]<...<x[m-1]<=xe, w[i]>0, i=0..m-1
if iopt=-1:
xb<t[k+1]<t[k+2]<...<t[n-k-2]<xe"""
}
# UnivariateSpline, ext parameter can be an int or a string
_extrap_modes = {0: 0, 'extrapolate': 0,
1: 1, 'zeros': 1,
2: 2, 'raise': 2,
3: 3, 'const': 3}
class UnivariateSpline:
"""
1-D smoothing spline fit to a given set of data points.
Fits a spline y = spl(x) of degree `k` to the provided `x`, `y` data. `s`
specifies the number of knots by specifying a smoothing condition.
Parameters
----------
x : (N,) array_like
1-D array of independent input data. Must be increasing;
must be strictly increasing if `s` is 0.
y : (N,) array_like
1-D array of dependent input data, of the same length as `x`.
w : (N,) array_like, optional
Weights for spline fitting. Must be positive. If `w` is None,
weights are all 1. Default is None.
bbox : (2,) array_like, optional
2-sequence specifying the boundary of the approximation interval. If
`bbox` is None, ``bbox=[x[0], x[-1]]``. Default is None.
k : int, optional
Degree of the smoothing spline. Must be 1 <= `k` <= 5.
``k = 3`` is a cubic spline. Default is 3.
s : float or None, optional
Positive smoothing factor used to choose the number of knots. Number
of knots will be increased until the smoothing condition is satisfied::
sum((w[i] * (y[i]-spl(x[i])))**2, axis=0) <= s
However, because of numerical issues, the actual condition is::
abs(sum((w[i] * (y[i]-spl(x[i])))**2, axis=0) - s) < 0.001 * s
If `s` is None, `s` will be set as `len(w)` for a smoothing spline
that uses all data points.
If 0, spline will interpolate through all data points. This is
equivalent to `InterpolatedUnivariateSpline`.
Default is None.
The user can use the `s` to control the tradeoff between closeness
and smoothness of fit. Larger `s` means more smoothing while smaller
values of `s` indicate less smoothing.
Recommended values of `s` depend on the weights, `w`. If the weights
represent the inverse of the standard-deviation of `y`, then a good
`s` value should be found in the range (m-sqrt(2*m),m+sqrt(2*m))
where m is the number of datapoints in `x`, `y`, and `w`. This means
``s = len(w)`` should be a good value if ``1/w[i]`` is an
estimate of the standard deviation of ``y[i]``.
ext : int or str, optional
Controls the extrapolation mode for elements
not in the interval defined by the knot sequence.
* if ext=0 or 'extrapolate', return the extrapolated value.
* if ext=1 or 'zeros', return 0
* if ext=2 or 'raise', raise a ValueError
* if ext=3 or 'const', return the boundary value.
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 or non-sensical results) if the inputs
do contain infinities or NaNs.
Default is False.
See Also
--------
BivariateSpline :
a base class for bivariate splines.
SmoothBivariateSpline :
a smoothing bivariate spline through the given points
LSQBivariateSpline :
a bivariate spline using weighted least-squares fitting
RectSphereBivariateSpline :
a bivariate spline over a rectangular mesh on a sphere
SmoothSphereBivariateSpline :
a smoothing bivariate spline in spherical coordinates
LSQSphereBivariateSpline :
a bivariate spline in spherical coordinates using weighted
least-squares fitting
RectBivariateSpline :
a bivariate spline over a rectangular mesh
InterpolatedUnivariateSpline :
a interpolating univariate spline for a given set of data points.
bisplrep :
a function to find a bivariate B-spline representation of a surface
bisplev :
a function to evaluate a bivariate B-spline and its derivatives
splrep :
a function to find the B-spline representation of a 1-D curve
splev :
a function to evaluate a B-spline or its derivatives
sproot :
a function to find the roots of a cubic B-spline
splint :
a function to evaluate the definite integral of a B-spline between two
given points
spalde :
a function to evaluate all derivatives of a B-spline
Notes
-----
The number of data points must be larger than the spline degree `k`.
**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:
>>> import numpy as np
>>> from scipy.interpolate import UnivariateSpline
>>> x, y = np.array([1, 2, 3, 4]), np.array([1, np.nan, 3, 4])
>>> w = np.isnan(y)
>>> y[w] = 0.
>>> spl = UnivariateSpline(x, y, 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.)
References
----------
Based on algorithms described in [1]_, [2]_, [3]_, and [4]_:
.. [1] P. Dierckx, "An algorithm for smoothing, differentiation and
integration of experimental data using spline functions",
J.Comp.Appl.Maths 1 (1975) 165-184.
.. [2] P. Dierckx, "A fast algorithm for smoothing data on a rectangular
grid while using spline functions", SIAM J.Numer.Anal. 19 (1982)
1286-1304.
.. [3] P. Dierckx, "An improved algorithm for curve fitting with spline
functions", report tw54, Dept. Computer Science,K.U. Leuven, 1981.
.. [4] P. Dierckx, "Curve and surface fitting with splines", Monographs on
Numerical Analysis, Oxford University Press, 1993.
Examples
--------
>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> from scipy.interpolate import UnivariateSpline
>>> rng = np.random.default_rng()
>>> x = np.linspace(-3, 3, 50)
>>> y = np.exp(-x**2) + 0.1 * rng.standard_normal(50)
>>> plt.plot(x, y, 'ro', ms=5)
Use the default value for the smoothing parameter:
>>> spl = UnivariateSpline(x, y)
>>> xs = np.linspace(-3, 3, 1000)
>>> plt.plot(xs, spl(xs), 'g', lw=3)
Manually change the amount of smoothing:
>>> spl.set_smoothing_factor(0.5)
>>> plt.plot(xs, spl(xs), 'b', lw=3)
>>> plt.show()
"""
def __init__(self, x, y, w=None, bbox=[None]*2, k=3, s=None,
ext=0, check_finite=False):
x, y, w, bbox, self.ext = self.validate_input(x, y, w, bbox, k, s, ext,
check_finite)
# _data == x,y,w,xb,xe,k,s,n,t,c,fp,fpint,nrdata,ier
data = dfitpack.fpcurf0(x, y, k, w=w, xb=bbox[0],
xe=bbox[1], s=s)
if data[-1] == 1:
# nest too small, setting to maximum bound
data = self._reset_nest(data)
self._data = data
self._reset_class()
@staticmethod
def validate_input(x, y, w, bbox, k, s, ext, check_finite):
x, y, bbox = np.asarray(x), np.asarray(y), np.asarray(bbox)
if w is not None:
w = np.asarray(w)
if check_finite:
w_finite = np.isfinite(w).all() if w is not None else True
if (not np.isfinite(x).all() or not np.isfinite(y).all() or
not w_finite):
raise ValueError("x and y array must not contain "
"NaNs or infs.")
if s is None or s > 0:
if not np.all(diff(x) >= 0.0):
raise ValueError("x must be increasing if s > 0")
else:
if not np.all(diff(x) > 0.0):
raise ValueError("x must be strictly increasing if s = 0")
if x.size != y.size:
raise ValueError("x and y should have a same length")
elif w is not None and not x.size == y.size == w.size:
raise ValueError("x, y, and w should have a same length")
elif bbox.shape != (2,):
raise ValueError("bbox shape should be (2,)")
elif not (1 <= k <= 5):
raise ValueError("k should be 1 <= k <= 5")
elif s is not None and not s >= 0.0:
raise ValueError("s should be s >= 0.0")
try:
ext = _extrap_modes[ext]
except KeyError as e:
raise ValueError("Unknown extrapolation mode %s." % ext) from e
return x, y, w, bbox, ext
@classmethod
def _from_tck(cls, tck, ext=0):
"""Construct a spline object from given tck"""
self = cls.__new__(cls)
t, c, k = tck
self._eval_args = tck
# _data == x,y,w,xb,xe,k,s,n,t,c,fp,fpint,nrdata,ier
self._data = (None, None, None, None, None, k, None, len(t), t,
c, None, None, None, None)
self.ext = ext
return self
def _reset_class(self):
data = self._data
n, t, c, k, ier = data[7], data[8], data[9], data[5], data[-1]
self._eval_args = t[:n], c[:n], k
if ier == 0:
# the spline returned has a residual sum of squares fp
# such that abs(fp-s)/s <= tol with tol a relative
# tolerance set to 0.001 by the program
pass
elif ier == -1:
# the spline returned is an interpolating spline
self._set_class(InterpolatedUnivariateSpline)
elif ier == -2:
# the spline returned is the weighted least-squares
# polynomial of degree k. In this extreme case fp gives
# the upper bound fp0 for the smoothing factor s.
self._set_class(LSQUnivariateSpline)
else:
# error
if ier == 1:
self._set_class(LSQUnivariateSpline)
message = _curfit_messages.get(ier, 'ier=%s' % (ier))
warnings.warn(message, stacklevel=3)
def _set_class(self, cls):
self._spline_class = cls
if self.__class__ in (UnivariateSpline, InterpolatedUnivariateSpline,
LSQUnivariateSpline):
self.__class__ = cls
else:
# It's an unknown subclass -- don't change class. cf. #731
pass
def _reset_nest(self, data, nest=None):
n = data[10]
if nest is None:
k, m = data[5], len(data[0])
nest = m+k+1 # this is the maximum bound for nest
else:
if not n <= nest:
raise ValueError("`nest` can only be increased")
t, c, fpint, nrdata = (np.resize(data[j], nest) for j in
[8, 9, 11, 12])
args = data[:8] + (t, c, n, fpint, nrdata, data[13])
data = dfitpack.fpcurf1(*args)
return data
def set_smoothing_factor(self, s):
""" Continue spline computation with the given smoothing
factor s and with the knots found at the last call.
This routine modifies the spline in place.
"""
data = self._data
if data[6] == -1:
warnings.warn('smoothing factor unchanged for'
'LSQ spline with fixed knots',
stacklevel=2)
return
args = data[:6] + (s,) + data[7:]
data = dfitpack.fpcurf1(*args)
if data[-1] == 1:
# nest too small, setting to maximum bound
data = self._reset_nest(data)
self._data = data
self._reset_class()
def __call__(self, x, nu=0, ext=None):
"""
Evaluate spline (or its nu-th derivative) at positions x.
Parameters
----------
x : array_like
A 1-D array of points at which to return the value of the smoothed
spline or its derivatives. Note: `x` can be unordered but the
evaluation is more efficient if `x` is (partially) ordered.
nu : int
The order of derivative of the spline to compute.
ext : int
Controls the value returned for elements of `x` not in the
interval defined by the knot sequence.
* if ext=0 or 'extrapolate', return the extrapolated value.
* if ext=1 or 'zeros', return 0
* if ext=2 or 'raise', raise a ValueError
* if ext=3 or 'const', return the boundary value.
The default value is 0, passed from the initialization of
UnivariateSpline.
"""
x = np.asarray(x)
# empty input yields empty output
if x.size == 0:
return array([])
if ext is None:
ext = self.ext
else:
try:
ext = _extrap_modes[ext]
except KeyError as e:
raise ValueError("Unknown extrapolation mode %s." % ext) from e
return _fitpack_impl.splev(x, self._eval_args, der=nu, ext=ext)
def get_knots(self):
""" Return positions of interior knots of the spline.
Internally, the knot vector contains ``2*k`` additional boundary knots.
"""
data = self._data
k, n = data[5], data[7]
return data[8][k:n-k]
def get_coeffs(self):
"""Return spline coefficients."""
data = self._data
k, n = data[5], data[7]
return data[9][:n-k-1]
def get_residual(self):
"""Return weighted sum of squared residuals of the spline approximation.
This is equivalent to::
sum((w[i] * (y[i]-spl(x[i])))**2, axis=0)
"""
return self._data[10]
def integral(self, a, b):
""" Return definite integral of the spline between two given points.
Parameters
----------
a : float
Lower limit of integration.
b : float
Upper limit of integration.
Returns
-------
integral : float
The value of the definite integral of the spline between limits.
Examples
--------
>>> import numpy as np
>>> from scipy.interpolate import UnivariateSpline
>>> x = np.linspace(0, 3, 11)
>>> y = x**2
>>> spl = UnivariateSpline(x, y)
>>> spl.integral(0, 3)
9.0
which agrees with :math:`\\int x^2 dx = x^3 / 3` between the limits
of 0 and 3.
A caveat is that this routine assumes the spline to be zero outside of
the data limits:
>>> spl.integral(-1, 4)
9.0
>>> spl.integral(-1, 0)
0.0
"""
return _fitpack_impl.splint(a, b, self._eval_args)
def derivatives(self, x):
""" Return all derivatives of the spline at the point x.
Parameters
----------
x : float
The point to evaluate the derivatives at.
Returns
-------
der : ndarray, shape(k+1,)
Derivatives of the orders 0 to k.
Examples
--------
>>> import numpy as np
>>> from scipy.interpolate import UnivariateSpline
>>> x = np.linspace(0, 3, 11)
>>> y = x**2
>>> spl = UnivariateSpline(x, y)
>>> spl.derivatives(1.5)
array([2.25, 3.0, 2.0, 0])
"""
return _fitpack_impl.spalde(x, self._eval_args)
def roots(self):
""" Return the zeros of the spline.
Notes
-----
Restriction: only cubic splines are supported by FITPACK. For non-cubic
splines, use `PPoly.root` (see below for an example).
Examples
--------
For some data, this method may miss a root. This happens when one of
the spline knots (which FITPACK places automatically) happens to
coincide with the true root. A workaround is to convert to `PPoly`,
which uses a different root-finding algorithm.
For example,
>>> x = [1.96, 1.97, 1.98, 1.99, 2.00, 2.01, 2.02, 2.03, 2.04, 2.05]
>>> y = [-6.365470e-03, -4.790580e-03, -3.204320e-03, -1.607270e-03,
... 4.440892e-16, 1.616930e-03, 3.243000e-03, 4.877670e-03,
... 6.520430e-03, 8.170770e-03]
>>> from scipy.interpolate import UnivariateSpline
>>> spl = UnivariateSpline(x, y, s=0)
>>> spl.roots()
array([], dtype=float64)
Converting to a PPoly object does find the roots at `x=2`:
>>> from scipy.interpolate import splrep, PPoly
>>> tck = splrep(x, y, s=0)
>>> ppoly = PPoly.from_spline(tck)
>>> ppoly.roots(extrapolate=False)
array([2.])
See Also
--------
sproot
PPoly.roots
"""
k = self._data[5]
if k == 3:
t = self._eval_args[0]
mest = 3 * (len(t) - 7)
return _fitpack_impl.sproot(self._eval_args, mest=mest)
raise NotImplementedError('finding roots unsupported for '
'non-cubic splines')
def derivative(self, n=1):
"""
Construct a new spline representing the derivative of this spline.
Parameters
----------
n : int, optional
Order of derivative to evaluate. Default: 1
Returns
-------
spline : UnivariateSpline
Spline of order k2=k-n representing the derivative of this
spline.
See Also
--------
splder, antiderivative
Notes
-----
.. versionadded:: 0.13.0
Examples
--------
This can be used for finding maxima of a curve:
>>> import numpy as np
>>> from scipy.interpolate import UnivariateSpline
>>> x = np.linspace(0, 10, 70)
>>> y = np.sin(x)
>>> spl = UnivariateSpline(x, y, k=4, s=0)
Now, differentiate the spline and find the zeros of the
derivative. (NB: `sproot` only works for order 3 splines, so we
fit an order 4 spline):
>>> spl.derivative().roots() / np.pi
array([ 0.50000001, 1.5 , 2.49999998])
This agrees well with roots :math:`\\pi/2 + n\\pi` of
:math:`\\cos(x) = \\sin'(x)`.
"""
tck = _fitpack_impl.splder(self._eval_args, n)
# if self.ext is 'const', derivative.ext will be 'zeros'
ext = 1 if self.ext == 3 else self.ext
return UnivariateSpline._from_tck(tck, ext=ext)
def antiderivative(self, n=1):
"""
Construct a new spline representing the antiderivative of this spline.
Parameters
----------
n : int, optional
Order of antiderivative to evaluate. Default: 1
Returns
-------
spline : UnivariateSpline
Spline of order k2=k+n representing the antiderivative of this
spline.
Notes
-----
.. versionadded:: 0.13.0
See Also
--------
splantider, derivative
Examples
--------
>>> import numpy as np
>>> from scipy.interpolate import UnivariateSpline
>>> x = np.linspace(0, np.pi/2, 70)
>>> y = 1 / np.sqrt(1 - 0.8*np.sin(x)**2)
>>> spl = UnivariateSpline(x, y, s=0)
The derivative is the inverse operation of the antiderivative,
although some floating point error accumulates:
>>> spl(1.7), spl.antiderivative().derivative()(1.7)
(array(2.1565429877197317), array(2.1565429877201865))
Antiderivative can be used to evaluate definite integrals:
>>> ispl = spl.antiderivative()
>>> ispl(np.pi/2) - ispl(0)
2.2572053588768486
This is indeed an approximation to the complete elliptic integral
:math:`K(m) = \\int_0^{\\pi/2} [1 - m\\sin^2 x]^{-1/2} dx`:
>>> from scipy.special import ellipk
>>> ellipk(0.8)
2.2572053268208538
"""
tck = _fitpack_impl.splantider(self._eval_args, n)
return UnivariateSpline._from_tck(tck, self.ext)
class InterpolatedUnivariateSpline(UnivariateSpline):
"""
1-D interpolating spline for a given set of data points.
Fits a spline y = spl(x) of degree `k` to the provided `x`, `y` data.
Spline function passes through all provided points. Equivalent to
`UnivariateSpline` with `s` = 0.
Parameters
----------
x : (N,) array_like
Input dimension of data points -- must be strictly increasing
y : (N,) array_like
input dimension of data points
w : (N,) array_like, optional
Weights for spline fitting. Must be positive. If None (default),
weights are all 1.
bbox : (2,) array_like, optional
2-sequence specifying the boundary of the approximation interval. If
None (default), ``bbox=[x[0], x[-1]]``.
k : int, optional
Degree of the smoothing spline. Must be ``1 <= k <= 5``. Default is
``k = 3``, a cubic spline.
ext : int or str, optional
Controls the extrapolation mode for elements
not in the interval defined by the knot sequence.
* if ext=0 or 'extrapolate', return the extrapolated value.
* if ext=1 or 'zeros', return 0
* if ext=2 or 'raise', raise a ValueError
* if ext=3 of 'const', return the boundary value.
The default value 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 or non-sensical results) if the inputs
do contain infinities or NaNs.
Default is False.
See Also
--------
UnivariateSpline :
a smooth univariate spline to fit a given set of data points.
LSQUnivariateSpline :
a spline for which knots are user-selected
SmoothBivariateSpline :
a smoothing bivariate spline through the given points
LSQBivariateSpline :
a bivariate spline using weighted least-squares fitting
splrep :
a function to find the B-spline representation of a 1-D curve
splev :
a function to evaluate a B-spline or its derivatives
sproot :
a function to find the roots of a cubic B-spline
splint :
a function to evaluate the definite integral of a B-spline between two
given points
spalde :
a function to evaluate all derivatives of a B-spline
Notes
-----
The number of data points must be larger than the spline degree `k`.
Examples
--------
>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> from scipy.interpolate import InterpolatedUnivariateSpline
>>> rng = np.random.default_rng()
>>> x = np.linspace(-3, 3, 50)
>>> y = np.exp(-x**2) + 0.1 * rng.standard_normal(50)
>>> spl = InterpolatedUnivariateSpline(x, y)
>>> plt.plot(x, y, 'ro', ms=5)
>>> xs = np.linspace(-3, 3, 1000)
>>> plt.plot(xs, spl(xs), 'g', lw=3, alpha=0.7)
>>> plt.show()
Notice that the ``spl(x)`` interpolates `y`:
>>> spl.get_residual()
0.0
"""
def __init__(self, x, y, w=None, bbox=[None]*2, k=3,
ext=0, check_finite=False):
x, y, w, bbox, self.ext = self.validate_input(x, y, w, bbox, k, None,
ext, check_finite)
if not np.all(diff(x) > 0.0):
raise ValueError('x must be strictly increasing')
# _data == x,y,w,xb,xe,k,s,n,t,c,fp,fpint,nrdata,ier
self._data = dfitpack.fpcurf0(x, y, k, w=w, xb=bbox[0],
xe=bbox[1], s=0)
self._reset_class()
_fpchec_error_string = """The input parameters have been rejected by fpchec. \
This means that at least one of the following conditions is violated:
1) k+1 <= n-k-1 <= m
2) t(1) <= t(2) <= ... <= t(k+1)
t(n-k) <= t(n-k+1) <= ... <= t(n)
3) t(k+1) < t(k+2) < ... < t(n-k)
4) t(k+1) <= x(i) <= t(n-k)
5) The conditions specified by Schoenberg and Whitney must hold
for at least one subset of data points, i.e., there must be a
subset of data points y(j) such that
t(j) < y(j) < t(j+k+1), j=1,2,...,n-k-1
"""
class LSQUnivariateSpline(UnivariateSpline):
"""
1-D spline with explicit internal knots.
Fits a spline y = spl(x) of degree `k` to the provided `x`, `y` data. `t`
specifies the internal knots of the spline
Parameters
----------
x : (N,) array_like
Input dimension of data points -- must be increasing
y : (N,) array_like
Input dimension of data points
t : (M,) array_like
interior knots of the spline. Must be in ascending order and::
bbox[0] < t[0] < ... < t[-1] < bbox[-1]
w : (N,) array_like, optional
weights for spline fitting. Must be positive. If None (default),
weights are all 1.
bbox : (2,) array_like, optional
2-sequence specifying the boundary of the approximation interval. If
None (default), ``bbox = [x[0], x[-1]]``.
k : int, optional
Degree of the smoothing spline. Must be 1 <= `k` <= 5.
Default is `k` = 3, a cubic spline.
ext : int or str, optional
Controls the extrapolation mode for elements
not in the interval defined by the knot sequence.
* if ext=0 or 'extrapolate', return the extrapolated value.
* if ext=1 or 'zeros', return 0
* if ext=2 or 'raise', raise a ValueError
* if ext=3 of 'const', return the boundary value.
The default value 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 or non-sensical results) if the inputs
do contain infinities or NaNs.
Default is False.
Raises
------
ValueError
If the interior knots do not satisfy the Schoenberg-Whitney conditions
See Also
--------
UnivariateSpline :
a smooth univariate spline to fit a given set of data points.
InterpolatedUnivariateSpline :
a interpolating univariate spline for a given set of data points.
splrep :
a function to find the B-spline representation of a 1-D curve
splev :
a function to evaluate a B-spline or its derivatives
sproot :
a function to find the roots of a cubic B-spline
splint :
a function to evaluate the definite integral of a B-spline between two
given points
spalde :
a function to evaluate all derivatives of a B-spline
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
--------
>>> import numpy as np
>>> from scipy.interpolate import LSQUnivariateSpline, UnivariateSpline
>>> 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)
Fit a smoothing spline with a pre-defined internal knots:
>>> t = [-1, 0, 1]
>>> spl = LSQUnivariateSpline(x, y, t)
>>> xs = np.linspace(-3, 3, 1000)
>>> plt.plot(x, y, 'ro', ms=5)
>>> plt.plot(xs, spl(xs), 'g-', lw=3)
>>> plt.show()
Check the knot vector:
>>> spl.get_knots()
array([-3., -1., 0., 1., 3.])
Constructing lsq spline using the knots from another spline:
>>> x = np.arange(10)
>>> s = UnivariateSpline(x, x, s=0)
>>> s.get_knots()
array([ 0., 2., 3., 4., 5., 6., 7., 9.])
>>> knt = s.get_knots()
>>> s1 = LSQUnivariateSpline(x, x, knt[1:-1]) # Chop 1st and last knot
>>> s1.get_knots()
array([ 0., 2., 3., 4., 5., 6., 7., 9.])
"""
def __init__(self, x, y, t, w=None, bbox=[None]*2, k=3,
ext=0, check_finite=False):
x, y, w, bbox, self.ext = self.validate_input(x, y, w, bbox, k, None,
ext, check_finite)
if not np.all(diff(x) >= 0.0):
raise ValueError('x must be increasing')
# _data == x,y,w,xb,xe,k,s,n,t,c,fp,fpint,nrdata,ier
xb = bbox[0]
xe = bbox[1]
if xb is None:
xb = x[0]
if xe is None:
xe = x[-1]
t = concatenate(([xb]*(k+1), t, [xe]*(k+1)))
n = len(t)
if not np.all(t[k+1:n-k]-t[k:n-k-1] > 0, axis=0):
raise ValueError('Interior knots t must satisfy '
'Schoenberg-Whitney conditions')
if not dfitpack.fpchec(x, t, k) == 0:
raise ValueError(_fpchec_error_string)
data = dfitpack.fpcurfm1(x, y, k, t, w=w, xb=xb, xe=xe)
self._data = data[:-3] + (None, None, data[-1])
self._reset_class()
# ############### Bivariate spline ####################
class _BivariateSplineBase:
""" Base class for Bivariate spline s(x,y) interpolation on the rectangle
[xb,xe] x [yb, ye] calculated from a given set of data points
(x,y,z).
See Also
--------
bisplrep :
a function to find a bivariate B-spline representation of a surface
bisplev :
a function to evaluate a bivariate B-spline and its derivatives
BivariateSpline :
a base class for bivariate splines.
SphereBivariateSpline :
a bivariate spline on a spherical grid
"""
@classmethod
def _from_tck(cls, tck):
"""Construct a spline object from given tck and degree"""
self = cls.__new__(cls)
if len(tck) != 5:
raise ValueError("tck should be a 5 element tuple of tx,"
" ty, c, kx, ky")
self.tck = tck[:3]
self.degrees = tck[3:]
return self
def get_residual(self):
""" Return weighted sum of squared residuals of the spline
approximation: sum ((w[i]*(z[i]-s(x[i],y[i])))**2,axis=0)
"""
return self.fp
def get_knots(self):
""" Return a tuple (tx,ty) where tx,ty contain knots positions
of the spline with respect to x-, y-variable, respectively.
The position of interior and additional knots are given as
t[k+1:-k-1] and t[:k+1]=b, t[-k-1:]=e, respectively.
"""
return self.tck[:2]
def get_coeffs(self):
""" Return spline coefficients."""
return self.tck[2]
def __call__(self, x, y, dx=0, dy=0, grid=True):
"""
Evaluate the spline or its derivatives at given positions.
Parameters
----------
x, y : array_like
Input coordinates.
If `grid` is False, evaluate the spline at points ``(x[i],
y[i]), i=0, ..., len(x)-1``. Standard Numpy broadcasting
is obeyed.
If `grid` is True: evaluate spline at the grid points
defined by the coordinate arrays x, y. The arrays must be
sorted to increasing order.
The ordering of axes is consistent with
``np.meshgrid(..., indexing="ij")`` and inconsistent with the
default ordering ``np.meshgrid(..., indexing="xy")``.
dx : int
Order of x-derivative
.. versionadded:: 0.14.0
dy : int
Order of y-derivative
.. versionadded:: 0.14.0
grid : bool
Whether to evaluate the results on a grid spanned by the
input arrays, or at points specified by the input arrays.
.. versionadded:: 0.14.0
Examples
--------
Suppose that we want to bilinearly interpolate an exponentially decaying
function in 2 dimensions.
>>> import numpy as np
>>> from scipy.interpolate import RectBivariateSpline
We sample the function on a coarse grid. Note that the default indexing="xy"
of meshgrid would result in an unexpected (transposed) result after
interpolation.
>>> xarr = np.linspace(-3, 3, 100)
>>> yarr = np.linspace(-3, 3, 100)
>>> xgrid, ygrid = np.meshgrid(xarr, yarr, indexing="ij")
The function to interpolate decays faster along one axis than the other.
>>> zdata = np.exp(-np.sqrt((xgrid / 2) ** 2 + ygrid**2))