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AIM-PIbd-32-Kurbanova-A-A/aimenv/Lib/site-packages/scipy/interpolate/_fitpack2.py

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lab 1 is done
2024-10-02 22:15:59 +04:00
"""
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
import numpy as np
from . import _fitpack_impl
from . import _dfitpack as 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))
Next we sample on a finer grid using interpolation (kx=ky=1 for bilinear).
>>> rbs = RectBivariateSpline(xarr, yarr, zdata, kx=1, ky=1)
>>> xarr_fine = np.linspace(-3, 3, 200)
>>> yarr_fine = np.linspace(-3, 3, 200)
>>> xgrid_fine, ygrid_fine = np.meshgrid(xarr_fine, yarr_fine, indexing="ij")
>>> zdata_interp = rbs(xgrid_fine, ygrid_fine, grid=False)
And check that the result agrees with the input by plotting both.
>>> import matplotlib.pyplot as plt
>>> fig = plt.figure()
>>> ax1 = fig.add_subplot(1, 2, 1, aspect="equal")
>>> ax2 = fig.add_subplot(1, 2, 2, aspect="equal")
>>> ax1.imshow(zdata)
>>> ax2.imshow(zdata_interp)
>>> plt.show()
"""
x = np.asarray(x)
y = np.asarray(y)
tx, ty, c = self.tck[:3]
kx, ky = self.degrees
if grid:
if x.size == 0 or y.size == 0:
return np.zeros((x.size, y.size), dtype=self.tck[2].dtype)
if (x.size >= 2) and (not np.all(np.diff(x) >= 0.0)):
raise ValueError("x must be strictly increasing when `grid` is True")
if (y.size >= 2) and (not np.all(np.diff(y) >= 0.0)):
raise ValueError("y must be strictly increasing when `grid` is True")
if dx or dy:
z, ier = dfitpack.parder(tx, ty, c, kx, ky, dx, dy, x, y)
if not ier == 0:
raise ValueError("Error code returned by parder: %s" % ier)
else:
z, ier = dfitpack.bispev(tx, ty, c, kx, ky, x, y)
if not ier == 0:
raise ValueError("Error code returned by bispev: %s" % ier)
else:
# standard Numpy broadcasting
if x.shape != y.shape:
x, y = np.broadcast_arrays(x, y)
shape = x.shape
x = x.ravel()
y = y.ravel()
if x.size == 0 or y.size == 0:
return np.zeros(shape, dtype=self.tck[2].dtype)
if dx or dy:
z, ier = dfitpack.pardeu(tx, ty, c, kx, ky, dx, dy, x, y)
if not ier == 0:
raise ValueError("Error code returned by pardeu: %s" % ier)
else:
z, ier = dfitpack.bispeu(tx, ty, c, kx, ky, x, y)
if not ier == 0:
raise ValueError("Error code returned by bispeu: %s" % ier)
z = z.reshape(shape)
return z
def partial_derivative(self, dx, dy):
"""Construct a new spline representing a partial derivative of this
spline.
Parameters
----------
dx, dy : int
Orders of the derivative in x and y respectively. They must be
non-negative integers and less than the respective degree of the
original spline (self) in that direction (``kx``, ``ky``).
Returns
-------
spline :
A new spline of degrees (``kx - dx``, ``ky - dy``) representing the
derivative of this spline.
Notes
-----
.. versionadded:: 1.9.0
"""
if dx == 0 and dy == 0:
return self
else:
kx, ky = self.degrees
if not (dx >= 0 and dy >= 0):
raise ValueError("order of derivative must be positive or"
" zero")
if not (dx < kx and dy < ky):
raise ValueError("order of derivative must be less than"
" degree of spline")
tx, ty, c = self.tck[:3]
newc, ier = dfitpack.pardtc(tx, ty, c, kx, ky, dx, dy)
if ier != 0:
# This should not happen under normal conditions.
raise ValueError("Unexpected error code returned by"
" pardtc: %d" % ier)
nx = len(tx)
ny = len(ty)
newtx = tx[dx:nx - dx]
newty = ty[dy:ny - dy]
newkx, newky = kx - dx, ky - dy
newclen = (nx - dx - kx - 1) * (ny - dy - ky - 1)
return _DerivedBivariateSpline._from_tck((newtx, newty,
newc[:newclen],
newkx, newky))
_surfit_messages = {1: """
The required storage space exceeds the available storage space: nxest
or nyest too small, or s too small.
The weighted least-squares spline corresponds to the current set of
knots.""",
2: """
A theoretically impossible result was found during the iteration
process for finding a smoothing spline with fp = s: s too small or
badly chosen eps.
Weighted sum of squared residuals does not satisfy 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.
Weighted sum of squared residuals does not satisfy abs(fp-s)/s < tol.""",
4: """
No more knots can be added because the number of b-spline coefficients
(nx-kx-1)*(ny-ky-1) already exceeds the number of data points m:
either s or m too small.
The weighted least-squares spline corresponds to the current set of
knots.""",
5: """
No more knots can be added because the additional knot would (quasi)
coincide with an old one: s too small or too large a weight to an
inaccurate data point.
The weighted least-squares spline corresponds to the current set of
knots.""",
10: """
Error on entry, no approximation returned. The following conditions
must hold:
xb<=x[i]<=xe, yb<=y[i]<=ye, w[i]>0, i=0..m-1
If iopt==-1, then
xb<tx[kx+1]<tx[kx+2]<...<tx[nx-kx-2]<xe
yb<ty[ky+1]<ty[ky+2]<...<ty[ny-ky-2]<ye""",
-3: """
The coefficients of the spline returned have been computed as the
minimal norm least-squares solution of a (numerically) rank deficient
system (deficiency=%i). If deficiency is large, the results may be
inaccurate. Deficiency may strongly depend on the value of eps."""
}
class BivariateSpline(_BivariateSplineBase):
"""
Base class for bivariate splines.
This describes a spline ``s(x, y)`` of degrees ``kx`` and ``ky`` on
the rectangle ``[xb, xe] * [yb, ye]`` calculated from a given set
of data points ``(x, y, z)``.
This class is meant to be subclassed, not instantiated directly.
To construct these splines, call either `SmoothBivariateSpline` or
`LSQBivariateSpline` or `RectBivariateSpline`.
See Also
--------
UnivariateSpline :
a smooth univariate spline to fit a given set of data points.
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.
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
"""
def ev(self, xi, yi, dx=0, dy=0):
"""
Evaluate the spline at points
Returns the interpolated value at ``(xi[i], yi[i]),
i=0,...,len(xi)-1``.
Parameters
----------
xi, yi : array_like
Input coordinates. Standard Numpy broadcasting is obeyed.
The ordering of axes is consistent with
``np.meshgrid(..., indexing="ij")`` and inconsistent with the
default ordering ``np.meshgrid(..., indexing="xy")``.
dx : int, optional
Order of x-derivative
.. versionadded:: 0.14.0
dy : int, optional
Order of y-derivative
.. 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
>>> def f(x, y):
... return np.exp(-np.sqrt((x / 2) ** 2 + y**2))
We sample the function on a coarse grid and set up the interpolator. Note that
the default ``indexing="xy"`` of meshgrid would result in an unexpected
(transposed) result after interpolation.
>>> xarr = np.linspace(-3, 3, 21)
>>> yarr = np.linspace(-3, 3, 21)
>>> xgrid, ygrid = np.meshgrid(xarr, yarr, indexing="ij")
>>> zdata = f(xgrid, ygrid)
>>> rbs = RectBivariateSpline(xarr, yarr, zdata, kx=1, ky=1)
Next we sample the function along a diagonal slice through the coordinate space
on a finer grid using interpolation.
>>> xinterp = np.linspace(-3, 3, 201)
>>> yinterp = np.linspace(3, -3, 201)
>>> zinterp = rbs.ev(xinterp, yinterp)
And check that the interpolation passes through the function evaluations as a
function of the distance from the origin along the slice.
>>> import matplotlib.pyplot as plt
>>> fig = plt.figure()
>>> ax1 = fig.add_subplot(1, 1, 1)
>>> ax1.plot(np.sqrt(xarr**2 + yarr**2), np.diag(zdata), "or")
>>> ax1.plot(np.sqrt(xinterp**2 + yinterp**2), zinterp, "-b")
>>> plt.show()
"""
return self.__call__(xi, yi, dx=dx, dy=dy, grid=False)
def integral(self, xa, xb, ya, yb):
"""
Evaluate the integral of the spline over area [xa,xb] x [ya,yb].
Parameters
----------
xa, xb : float
The end-points of the x integration interval.
ya, yb : float
The end-points of the y integration interval.
Returns
-------
integ : float
The value of the resulting integral.
"""
tx, ty, c = self.tck[:3]
kx, ky = self.degrees
return dfitpack.dblint(tx, ty, c, kx, ky, xa, xb, ya, yb)
@staticmethod
def _validate_input(x, y, z, w, kx, ky, eps):
x, y, z = np.asarray(x), np.asarray(y), np.asarray(z)
if not x.size == y.size == z.size:
raise ValueError('x, y, and z should have a same length')
if w is not None:
w = np.asarray(w)
if x.size != w.size:
raise ValueError('x, y, z, and w should have a same length')
elif not np.all(w >= 0.0):
raise ValueError('w should be positive')
if (eps is not None) and (not 0.0 < eps < 1.0):
raise ValueError('eps should be between (0, 1)')
if not x.size >= (kx + 1) * (ky + 1):
raise ValueError('The length of x, y and z should be at least'
' (kx+1) * (ky+1)')
return x, y, z, w
class _DerivedBivariateSpline(_BivariateSplineBase):
"""Bivariate spline constructed from the coefficients and knots of another
spline.
Notes
-----
The class is not meant to be instantiated directly from the data to be
interpolated or smoothed. As a result, its ``fp`` attribute and
``get_residual`` method are inherited but overridden; ``AttributeError`` is
raised when they are accessed.
The other inherited attributes can be used as usual.
"""
_invalid_why = ("is unavailable, because _DerivedBivariateSpline"
" instance is not constructed from data that are to be"
" interpolated or smoothed, but derived from the"
" underlying knots and coefficients of another spline"
" object")
@property
def fp(self):
raise AttributeError("attribute \"fp\" %s" % self._invalid_why)
def get_residual(self):
raise AttributeError("method \"get_residual\" %s" % self._invalid_why)
class SmoothBivariateSpline(BivariateSpline):
"""
Smooth bivariate spline approximation.
Parameters
----------
x, y, z : array_like
1-D sequences of data points (order is not important).
w : array_like, optional
Positive 1-D sequence of weights, of same length as `x`, `y` and `z`.
bbox : array_like, optional
Sequence of length 4 specifying the boundary of the rectangular
approximation domain. By default,
``bbox=[min(x), max(x), min(y), max(y)]``.
kx, ky : ints, optional
Degrees of the bivariate spline. Default is 3.
s : float, optional
Positive smoothing factor defined for estimation condition:
``sum((w[i]*(z[i]-s(x[i], y[i])))**2, axis=0) <= s``
Default ``s=len(w)`` which should be a good value if ``1/w[i]`` is an
estimate of the standard deviation of ``z[i]``.
eps : float, optional
A threshold for determining the effective rank of an over-determined
linear system of equations. `eps` should have a value within the open
interval ``(0, 1)``, the default is 1e-16.
See Also
--------
BivariateSpline :
a base class for bivariate splines.
UnivariateSpline :
a smooth univariate spline to fit a given set of data 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
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
Notes
-----
The length of `x`, `y` and `z` should be at least ``(kx+1) * (ky+1)``.
If the input data is such that input dimensions have incommensurate
units and differ by many orders of magnitude, the interpolant may have
numerical artifacts. Consider rescaling the data before interpolating.
This routine constructs spline knot vectors automatically via the FITPACK
algorithm. The spline knots may be placed away from the data points. For
some data sets, this routine may fail to construct an interpolating spline,
even if one is requested via ``s=0`` parameter. In such situations, it is
recommended to use `bisplrep` / `bisplev` directly instead of this routine
and, if needed, increase the values of ``nxest`` and ``nyest`` parameters
of `bisplrep`.
For linear interpolation, prefer `LinearNDInterpolator`.
See ``https://gist.github.com/ev-br/8544371b40f414b7eaf3fe6217209bff``
for discussion.
"""
def __init__(self, x, y, z, w=None, bbox=[None] * 4, kx=3, ky=3, s=None,
eps=1e-16):
x, y, z, w = self._validate_input(x, y, z, w, kx, ky, eps)
bbox = ravel(bbox)
if not bbox.shape == (4,):
raise ValueError('bbox shape should be (4,)')
if s is not None and not s >= 0.0:
raise ValueError("s should be s >= 0.0")
xb, xe, yb, ye = bbox
nx, tx, ny, ty, c, fp, wrk1, ier = dfitpack.surfit_smth(x, y, z, w,
xb, xe, yb,
ye, kx, ky,
s=s, eps=eps,
lwrk2=1)
if ier > 10: # lwrk2 was to small, re-run
nx, tx, ny, ty, c, fp, wrk1, ier = dfitpack.surfit_smth(x, y, z, w,
xb, xe, yb,
ye, kx, ky,
s=s,
eps=eps,
lwrk2=ier)
if ier in [0, -1, -2]: # normal return
pass
else:
message = _surfit_messages.get(ier, 'ier=%s' % (ier))
warnings.warn(message, stacklevel=2)
self.fp = fp
self.tck = tx[:nx], ty[:ny], c[:(nx-kx-1)*(ny-ky-1)]
self.degrees = kx, ky
class LSQBivariateSpline(BivariateSpline):
"""
Weighted least-squares bivariate spline approximation.
Parameters
----------
x, y, z : array_like
1-D sequences of data points (order is not important).
tx, ty : array_like
Strictly ordered 1-D sequences of knots coordinates.
w : array_like, optional
Positive 1-D array of weights, of the same length as `x`, `y` and `z`.
bbox : (4,) array_like, optional
Sequence of length 4 specifying the boundary of the rectangular
approximation domain. By default,
``bbox=[min(x,tx),max(x,tx), min(y,ty),max(y,ty)]``.
kx, ky : ints, optional
Degrees of the bivariate spline. Default is 3.
eps : float, optional
A threshold for determining the effective rank of an over-determined
linear system of equations. `eps` should have a value within the open
interval ``(0, 1)``, the default is 1e-16.
See Also
--------
BivariateSpline :
a base class for bivariate splines.
UnivariateSpline :
a smooth univariate spline to fit a given set of data points.
SmoothBivariateSpline :
a smoothing bivariate spline through the given points
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.
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
Notes
-----
The length of `x`, `y` and `z` should be at least ``(kx+1) * (ky+1)``.
If the input data is such that input dimensions have incommensurate
units and differ by many orders of magnitude, the interpolant may have
numerical artifacts. Consider rescaling the data before interpolating.
"""
def __init__(self, x, y, z, tx, ty, w=None, bbox=[None]*4, kx=3, ky=3,
eps=None):
x, y, z, w = self._validate_input(x, y, z, w, kx, ky, eps)
bbox = ravel(bbox)
if not bbox.shape == (4,):
raise ValueError('bbox shape should be (4,)')
nx = 2*kx+2+len(tx)
ny = 2*ky+2+len(ty)
# The Fortran subroutine "surfit" (called as dfitpack.surfit_lsq)
# requires that the knot arrays passed as input should be "real
# array(s) of dimension nmax" where "nmax" refers to the greater of nx
# and ny. We pad the tx1/ty1 arrays here so that this is satisfied, and
# slice them to the desired sizes upon return.
nmax = max(nx, ny)
tx1 = zeros((nmax,), float)
ty1 = zeros((nmax,), float)
tx1[kx+1:nx-kx-1] = tx
ty1[ky+1:ny-ky-1] = ty
xb, xe, yb, ye = bbox
tx1, ty1, c, fp, ier = dfitpack.surfit_lsq(x, y, z, nx, tx1, ny, ty1,
w, xb, xe, yb, ye,
kx, ky, eps, lwrk2=1)
if ier > 10:
tx1, ty1, c, fp, ier = dfitpack.surfit_lsq(x, y, z,
nx, tx1, ny, ty1, w,
xb, xe, yb, ye,
kx, ky, eps, lwrk2=ier)
if ier in [0, -1, -2]: # normal return
pass
else:
if ier < -2:
deficiency = (nx-kx-1)*(ny-ky-1)+ier
message = _surfit_messages.get(-3) % (deficiency)
else:
message = _surfit_messages.get(ier, 'ier=%s' % (ier))
warnings.warn(message, stacklevel=2)
self.fp = fp
self.tck = tx1[:nx], ty1[:ny], c
self.degrees = kx, ky
class RectBivariateSpline(BivariateSpline):
"""
Bivariate spline approximation over a rectangular mesh.
Can be used for both smoothing and interpolating data.
Parameters
----------
x,y : array_like
1-D arrays of coordinates in strictly ascending order.
Evaluated points outside the data range will be extrapolated.
z : array_like
2-D array of data with shape (x.size,y.size).
bbox : array_like, optional
Sequence of length 4 specifying the boundary of the rectangular
approximation domain, which means the start and end spline knots of
each dimension are set by these values. By default,
``bbox=[min(x), max(x), min(y), max(y)]``.
kx, ky : ints, optional
Degrees of the bivariate spline. Default is 3.
s : float, optional
Positive smoothing factor defined for estimation condition:
``sum((z[i]-f(x[i], y[i]))**2, axis=0) <= s`` where f is a spline
function. Default is ``s=0``, which is for interpolation.
See Also
--------
BivariateSpline :
a base class for bivariate splines.
UnivariateSpline :
a smooth univariate spline to fit a given set of data points.
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
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
Notes
-----
If the input data is such that input dimensions have incommensurate
units and differ by many orders of magnitude, the interpolant may have
numerical artifacts. Consider rescaling the data before interpolating.
"""
def __init__(self, x, y, z, bbox=[None] * 4, kx=3, ky=3, s=0):
x, y, bbox = ravel(x), ravel(y), ravel(bbox)
z = np.asarray(z)
if not np.all(diff(x) > 0.0):
raise ValueError('x must be strictly increasing')
if not np.all(diff(y) > 0.0):
raise ValueError('y must be strictly increasing')
if not x.size == z.shape[0]:
raise ValueError('x dimension of z must have same number of '
'elements as x')
if not y.size == z.shape[1]:
raise ValueError('y dimension of z must have same number of '
'elements as y')
if not bbox.shape == (4,):
raise ValueError('bbox shape should be (4,)')
if s is not None and not s >= 0.0:
raise ValueError("s should be s >= 0.0")
z = ravel(z)
xb, xe, yb, ye = bbox
nx, tx, ny, ty, c, fp, ier = dfitpack.regrid_smth(x, y, z, xb, xe, yb,
ye, kx, ky, s)
if ier not in [0, -1, -2]:
msg = _surfit_messages.get(ier, 'ier=%s' % (ier))
raise ValueError(msg)
self.fp = fp
self.tck = tx[:nx], ty[:ny], c[:(nx - kx - 1) * (ny - ky - 1)]
self.degrees = kx, ky
_spherefit_messages = _surfit_messages.copy()
_spherefit_messages[10] = """
ERROR. On entry, the input data are controlled on validity. The following
restrictions must be satisfied:
-1<=iopt<=1, m>=2, ntest>=8 ,npest >=8, 0<eps<1,
0<=teta(i)<=pi, 0<=phi(i)<=2*pi, w(i)>0, i=1,...,m
lwrk1 >= 185+52*v+10*u+14*u*v+8*(u-1)*v**2+8*m
kwrk >= m+(ntest-7)*(npest-7)
if iopt=-1: 8<=nt<=ntest , 9<=np<=npest
0<tt(5)<tt(6)<...<tt(nt-4)<pi
0<tp(5)<tp(6)<...<tp(np-4)<2*pi
if iopt>=0: s>=0
if one of these conditions is found to be violated,control
is immediately repassed to the calling program. in that
case there is no approximation returned."""
_spherefit_messages[-3] = """
WARNING. The coefficients of the spline returned have been computed as the
minimal norm least-squares solution of a (numerically) rank
deficient system (deficiency=%i, rank=%i). Especially if the rank
deficiency, which is computed by 6+(nt-8)*(np-7)+ier, is large,
the results may be inaccurate. They could also seriously depend on
the value of eps."""
class SphereBivariateSpline(_BivariateSplineBase):
"""
Bivariate spline s(x,y) of degrees 3 on a sphere, calculated from a
given set of data points (theta,phi,r).
.. versionadded:: 0.11.0
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
UnivariateSpline :
a smooth univariate spline to fit a given set of data points.
SmoothBivariateSpline :
a smoothing bivariate spline through the given points
LSQUnivariateSpline :
a univariate spline using weighted least-squares fitting
"""
def __call__(self, theta, phi, dtheta=0, dphi=0, grid=True):
"""
Evaluate the spline or its derivatives at given positions.
Parameters
----------
theta, phi : array_like
Input coordinates.
If `grid` is False, evaluate the spline at points
``(theta[i], phi[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 theta, phi. 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")``.
dtheta : int, optional
Order of theta-derivative
.. versionadded:: 0.14.0
dphi : int
Order of phi-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 use splines to interpolate a bivariate function on a
sphere. The value of the function is known on a grid of longitudes and
colatitudes.
>>> import numpy as np
>>> from scipy.interpolate import RectSphereBivariateSpline
>>> def f(theta, phi):
... return np.sin(theta) * np.cos(phi)
We evaluate the function on the grid. Note that the default indexing="xy"
of meshgrid would result in an unexpected (transposed) result after
interpolation.
>>> thetaarr = np.linspace(0, np.pi, 22)[1:-1]
>>> phiarr = np.linspace(0, 2 * np.pi, 21)[:-1]
>>> thetagrid, phigrid = np.meshgrid(thetaarr, phiarr, indexing="ij")
>>> zdata = f(thetagrid, phigrid)
We next set up the interpolator and use it to evaluate the function
on a finer grid.
>>> rsbs = RectSphereBivariateSpline(thetaarr, phiarr, zdata)
>>> thetaarr_fine = np.linspace(0, np.pi, 200)
>>> phiarr_fine = np.linspace(0, 2 * np.pi, 200)
>>> zdata_fine = rsbs(thetaarr_fine, phiarr_fine)
Finally we plot the coarsly-sampled input data alongside the
finely-sampled interpolated data to check that they agree.
>>> import matplotlib.pyplot as plt
>>> fig = plt.figure()
>>> ax1 = fig.add_subplot(1, 2, 1)
>>> ax2 = fig.add_subplot(1, 2, 2)
>>> ax1.imshow(zdata)
>>> ax2.imshow(zdata_fine)
>>> plt.show()
"""
theta = np.asarray(theta)
phi = np.asarray(phi)
if theta.size > 0 and (theta.min() < 0. or theta.max() > np.pi):
raise ValueError("requested theta out of bounds.")
return _BivariateSplineBase.__call__(self, theta, phi,
dx=dtheta, dy=dphi, grid=grid)
def ev(self, theta, phi, dtheta=0, dphi=0):
"""
Evaluate the spline at points
Returns the interpolated value at ``(theta[i], phi[i]),
i=0,...,len(theta)-1``.
Parameters
----------
theta, phi : array_like
Input coordinates. Standard Numpy broadcasting is obeyed.
The ordering of axes is consistent with
np.meshgrid(..., indexing="ij") and inconsistent with the
default ordering np.meshgrid(..., indexing="xy").
dtheta : int, optional
Order of theta-derivative
.. versionadded:: 0.14.0
dphi : int, optional
Order of phi-derivative
.. versionadded:: 0.14.0
Examples
--------
Suppose that we want to use splines to interpolate a bivariate function on a
sphere. The value of the function is known on a grid of longitudes and
colatitudes.
>>> import numpy as np
>>> from scipy.interpolate import RectSphereBivariateSpline
>>> def f(theta, phi):
... return np.sin(theta) * np.cos(phi)
We evaluate the function on the grid. Note that the default indexing="xy"
of meshgrid would result in an unexpected (transposed) result after
interpolation.
>>> thetaarr = np.linspace(0, np.pi, 22)[1:-1]
>>> phiarr = np.linspace(0, 2 * np.pi, 21)[:-1]
>>> thetagrid, phigrid = np.meshgrid(thetaarr, phiarr, indexing="ij")
>>> zdata = f(thetagrid, phigrid)
We next set up the interpolator and use it to evaluate the function
at points not on the original grid.
>>> rsbs = RectSphereBivariateSpline(thetaarr, phiarr, zdata)
>>> thetainterp = np.linspace(thetaarr[0], thetaarr[-1], 200)
>>> phiinterp = np.linspace(phiarr[0], phiarr[-1], 200)
>>> zinterp = rsbs.ev(thetainterp, phiinterp)
Finally we plot the original data for a diagonal slice through the
initial grid, and the spline approximation along the same slice.
>>> import matplotlib.pyplot as plt
>>> fig = plt.figure()
>>> ax1 = fig.add_subplot(1, 1, 1)
>>> ax1.plot(np.sin(thetaarr) * np.sin(phiarr), np.diag(zdata), "or")
>>> ax1.plot(np.sin(thetainterp) * np.sin(phiinterp), zinterp, "-b")
>>> plt.show()
"""
return self.__call__(theta, phi, dtheta=dtheta, dphi=dphi, grid=False)
class SmoothSphereBivariateSpline(SphereBivariateSpline):
"""
Smooth bivariate spline approximation in spherical coordinates.
.. versionadded:: 0.11.0
Parameters
----------
theta, phi, r : array_like
1-D sequences of data points (order is not important). Coordinates
must be given in radians. Theta must lie within the interval
``[0, pi]``, and phi must lie within the interval ``[0, 2pi]``.
w : array_like, optional
Positive 1-D sequence of weights.
s : float, optional
Positive smoothing factor defined for estimation condition:
``sum((w(i)*(r(i) - s(theta(i), phi(i))))**2, axis=0) <= s``
Default ``s=len(w)`` which should be a good value if ``1/w[i]`` is an
estimate of the standard deviation of ``r[i]``.
eps : float, optional
A threshold for determining the effective rank of an over-determined
linear system of equations. `eps` should have a value within the open
interval ``(0, 1)``, the default is 1e-16.
See Also
--------
BivariateSpline :
a base class for bivariate splines.
UnivariateSpline :
a smooth univariate spline to fit a given set of data points.
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
LSQSphereBivariateSpline :
a bivariate spline in spherical coordinates using weighted
least-squares fitting
RectBivariateSpline :
a bivariate spline over a rectangular mesh.
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
Notes
-----
For more information, see the FITPACK_ site about this function.
.. _FITPACK: http://www.netlib.org/dierckx/sphere.f
Examples
--------
Suppose we have global data on a coarse grid (the input data does not
have to be on a grid):
>>> import numpy as np
>>> theta = np.linspace(0., np.pi, 7)
>>> phi = np.linspace(0., 2*np.pi, 9)
>>> data = np.empty((theta.shape[0], phi.shape[0]))
>>> data[:,0], data[0,:], data[-1,:] = 0., 0., 0.
>>> data[1:-1,1], data[1:-1,-1] = 1., 1.
>>> data[1,1:-1], data[-2,1:-1] = 1., 1.
>>> data[2:-2,2], data[2:-2,-2] = 2., 2.
>>> data[2,2:-2], data[-3,2:-2] = 2., 2.
>>> data[3,3:-2] = 3.
>>> data = np.roll(data, 4, 1)
We need to set up the interpolator object
>>> lats, lons = np.meshgrid(theta, phi)
>>> from scipy.interpolate import SmoothSphereBivariateSpline
>>> lut = SmoothSphereBivariateSpline(lats.ravel(), lons.ravel(),
... data.T.ravel(), s=3.5)
As a first test, we'll see what the algorithm returns when run on the
input coordinates
>>> data_orig = lut(theta, phi)
Finally we interpolate the data to a finer grid
>>> fine_lats = np.linspace(0., np.pi, 70)
>>> fine_lons = np.linspace(0., 2 * np.pi, 90)
>>> data_smth = lut(fine_lats, fine_lons)
>>> import matplotlib.pyplot as plt
>>> fig = plt.figure()
>>> ax1 = fig.add_subplot(131)
>>> ax1.imshow(data, interpolation='nearest')
>>> ax2 = fig.add_subplot(132)
>>> ax2.imshow(data_orig, interpolation='nearest')
>>> ax3 = fig.add_subplot(133)
>>> ax3.imshow(data_smth, interpolation='nearest')
>>> plt.show()
"""
def __init__(self, theta, phi, r, w=None, s=0., eps=1E-16):
theta, phi, r = np.asarray(theta), np.asarray(phi), np.asarray(r)
# input validation
if not ((0.0 <= theta).all() and (theta <= np.pi).all()):
raise ValueError('theta should be between [0, pi]')
if not ((0.0 <= phi).all() and (phi <= 2.0 * np.pi).all()):
raise ValueError('phi should be between [0, 2pi]')
if w is not None:
w = np.asarray(w)
if not (w >= 0.0).all():
raise ValueError('w should be positive')
if not s >= 0.0:
raise ValueError('s should be positive')
if not 0.0 < eps < 1.0:
raise ValueError('eps should be between (0, 1)')
nt_, tt_, np_, tp_, c, fp, ier = dfitpack.spherfit_smth(theta, phi,
r, w=w, s=s,
eps=eps)
if ier not in [0, -1, -2]:
message = _spherefit_messages.get(ier, 'ier=%s' % (ier))
raise ValueError(message)
self.fp = fp
self.tck = tt_[:nt_], tp_[:np_], c[:(nt_ - 4) * (np_ - 4)]
self.degrees = (3, 3)
def __call__(self, theta, phi, dtheta=0, dphi=0, grid=True):
theta = np.asarray(theta)
phi = np.asarray(phi)
if phi.size > 0 and (phi.min() < 0. or phi.max() > 2. * np.pi):
raise ValueError("requested phi out of bounds.")
return SphereBivariateSpline.__call__(self, theta, phi, dtheta=dtheta,
dphi=dphi, grid=grid)
class LSQSphereBivariateSpline(SphereBivariateSpline):
"""
Weighted least-squares bivariate spline approximation in spherical
coordinates.
Determines a smoothing bicubic spline according to a given
set of knots in the `theta` and `phi` directions.
.. versionadded:: 0.11.0
Parameters
----------
theta, phi, r : array_like
1-D sequences of data points (order is not important). Coordinates
must be given in radians. Theta must lie within the interval
``[0, pi]``, and phi must lie within the interval ``[0, 2pi]``.
tt, tp : array_like
Strictly ordered 1-D sequences of knots coordinates.
Coordinates must satisfy ``0 < tt[i] < pi``, ``0 < tp[i] < 2*pi``.
w : array_like, optional
Positive 1-D sequence of weights, of the same length as `theta`, `phi`
and `r`.
eps : float, optional
A threshold for determining the effective rank of an over-determined
linear system of equations. `eps` should have a value within the
open interval ``(0, 1)``, the default is 1e-16.
See Also
--------
BivariateSpline :
a base class for bivariate splines.
UnivariateSpline :
a smooth univariate spline to fit a given set of data points.
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
RectBivariateSpline :
a bivariate spline over a rectangular mesh.
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
Notes
-----
For more information, see the FITPACK_ site about this function.
.. _FITPACK: http://www.netlib.org/dierckx/sphere.f
Examples
--------
Suppose we have global data on a coarse grid (the input data does not
have to be on a grid):
>>> from scipy.interpolate import LSQSphereBivariateSpline
>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> theta = np.linspace(0, np.pi, num=7)
>>> phi = np.linspace(0, 2*np.pi, num=9)
>>> data = np.empty((theta.shape[0], phi.shape[0]))
>>> data[:,0], data[0,:], data[-1,:] = 0., 0., 0.
>>> data[1:-1,1], data[1:-1,-1] = 1., 1.
>>> data[1,1:-1], data[-2,1:-1] = 1., 1.
>>> data[2:-2,2], data[2:-2,-2] = 2., 2.
>>> data[2,2:-2], data[-3,2:-2] = 2., 2.
>>> data[3,3:-2] = 3.
>>> data = np.roll(data, 4, 1)
We need to set up the interpolator object. Here, we must also specify the
coordinates of the knots to use.
>>> lats, lons = np.meshgrid(theta, phi)
>>> knotst, knotsp = theta.copy(), phi.copy()
>>> knotst[0] += .0001
>>> knotst[-1] -= .0001
>>> knotsp[0] += .0001
>>> knotsp[-1] -= .0001
>>> lut = LSQSphereBivariateSpline(lats.ravel(), lons.ravel(),
... data.T.ravel(), knotst, knotsp)
As a first test, we'll see what the algorithm returns when run on the
input coordinates
>>> data_orig = lut(theta, phi)
Finally we interpolate the data to a finer grid
>>> fine_lats = np.linspace(0., np.pi, 70)
>>> fine_lons = np.linspace(0., 2*np.pi, 90)
>>> data_lsq = lut(fine_lats, fine_lons)
>>> fig = plt.figure()
>>> ax1 = fig.add_subplot(131)
>>> ax1.imshow(data, interpolation='nearest')
>>> ax2 = fig.add_subplot(132)
>>> ax2.imshow(data_orig, interpolation='nearest')
>>> ax3 = fig.add_subplot(133)
>>> ax3.imshow(data_lsq, interpolation='nearest')
>>> plt.show()
"""
def __init__(self, theta, phi, r, tt, tp, w=None, eps=1E-16):
theta, phi, r = np.asarray(theta), np.asarray(phi), np.asarray(r)
tt, tp = np.asarray(tt), np.asarray(tp)
if not ((0.0 <= theta).all() and (theta <= np.pi).all()):
raise ValueError('theta should be between [0, pi]')
if not ((0.0 <= phi).all() and (phi <= 2*np.pi).all()):
raise ValueError('phi should be between [0, 2pi]')
if not ((0.0 < tt).all() and (tt < np.pi).all()):
raise ValueError('tt should be between (0, pi)')
if not ((0.0 < tp).all() and (tp < 2*np.pi).all()):
raise ValueError('tp should be between (0, 2pi)')
if w is not None:
w = np.asarray(w)
if not (w >= 0.0).all():
raise ValueError('w should be positive')
if not 0.0 < eps < 1.0:
raise ValueError('eps should be between (0, 1)')
nt_, np_ = 8 + len(tt), 8 + len(tp)
tt_, tp_ = zeros((nt_,), float), zeros((np_,), float)
tt_[4:-4], tp_[4:-4] = tt, tp
tt_[-4:], tp_[-4:] = np.pi, 2. * np.pi
tt_, tp_, c, fp, ier = dfitpack.spherfit_lsq(theta, phi, r, tt_, tp_,
w=w, eps=eps)
if ier > 0:
message = _spherefit_messages.get(ier, 'ier=%s' % (ier))
raise ValueError(message)
self.fp = fp
self.tck = tt_, tp_, c
self.degrees = (3, 3)
def __call__(self, theta, phi, dtheta=0, dphi=0, grid=True):
theta = np.asarray(theta)
phi = np.asarray(phi)
if phi.size > 0 and (phi.min() < 0. or phi.max() > 2. * np.pi):
raise ValueError("requested phi out of bounds.")
return SphereBivariateSpline.__call__(self, theta, phi, dtheta=dtheta,
dphi=dphi, grid=grid)
_spfit_messages = _surfit_messages.copy()
_spfit_messages[10] = """
ERROR: on entry, the input data are controlled on validity
the following restrictions must be satisfied.
-1<=iopt(1)<=1, 0<=iopt(2)<=1, 0<=iopt(3)<=1,
-1<=ider(1)<=1, 0<=ider(2)<=1, ider(2)=0 if iopt(2)=0.
-1<=ider(3)<=1, 0<=ider(4)<=1, ider(4)=0 if iopt(3)=0.
mu >= mumin (see above), mv >= 4, nuest >=8, nvest >= 8,
kwrk>=5+mu+mv+nuest+nvest,
lwrk >= 12+nuest*(mv+nvest+3)+nvest*24+4*mu+8*mv+max(nuest,mv+nvest)
0< u(i-1)<u(i)< pi,i=2,..,mu,
-pi<=v(1)< pi, v(1)<v(i-1)<v(i)<v(1)+2*pi, i=3,...,mv
if iopt(1)=-1: 8<=nu<=min(nuest,mu+6+iopt(2)+iopt(3))
0<tu(5)<tu(6)<...<tu(nu-4)< pi
8<=nv<=min(nvest,mv+7)
v(1)<tv(5)<tv(6)<...<tv(nv-4)<v(1)+2*pi
the schoenberg-whitney conditions, i.e. there must be
subset of grid coordinates uu(p) and vv(q) such that
tu(p) < uu(p) < tu(p+4) ,p=1,...,nu-4
(iopt(2)=1 and iopt(3)=1 also count for a uu-value
tv(q) < vv(q) < tv(q+4) ,q=1,...,nv-4
(vv(q) is either a value v(j) or v(j)+2*pi)
if iopt(1)>=0: s>=0
if s=0: nuest>=mu+6+iopt(2)+iopt(3), nvest>=mv+7
if one of these conditions is found to be violated,control is
immediately repassed to the calling program. in that case there is no
approximation returned."""
class RectSphereBivariateSpline(SphereBivariateSpline):
"""
Bivariate spline approximation over a rectangular mesh on a sphere.
Can be used for smoothing data.
.. versionadded:: 0.11.0
Parameters
----------
u : array_like
1-D array of colatitude coordinates in strictly ascending order.
Coordinates must be given in radians and lie within the open interval
``(0, pi)``.
v : array_like
1-D array of longitude coordinates in strictly ascending order.
Coordinates must be given in radians. First element (``v[0]``) must lie
within the interval ``[-pi, pi)``. Last element (``v[-1]``) must satisfy
``v[-1] <= v[0] + 2*pi``.
r : array_like
2-D array of data with shape ``(u.size, v.size)``.
s : float, optional
Positive smoothing factor defined for estimation condition
(``s=0`` is for interpolation).
pole_continuity : bool or (bool, bool), optional
Order of continuity at the poles ``u=0`` (``pole_continuity[0]``) and
``u=pi`` (``pole_continuity[1]``). The order of continuity at the pole
will be 1 or 0 when this is True or False, respectively.
Defaults to False.
pole_values : float or (float, float), optional
Data values at the poles ``u=0`` and ``u=pi``. Either the whole
parameter or each individual element can be None. Defaults to None.
pole_exact : bool or (bool, bool), optional
Data value exactness at the poles ``u=0`` and ``u=pi``. If True, the
value is considered to be the right function value, and it will be
fitted exactly. If False, the value will be considered to be a data
value just like the other data values. Defaults to False.
pole_flat : bool or (bool, bool), optional
For the poles at ``u=0`` and ``u=pi``, specify whether or not the
approximation has vanishing derivatives. Defaults to False.
See Also
--------
BivariateSpline :
a base class for bivariate splines.
UnivariateSpline :
a smooth univariate spline to fit a given set of data points.
SmoothBivariateSpline :
a smoothing bivariate spline through the given points
LSQBivariateSpline :
a bivariate spline using weighted least-squares fitting
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.
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
Notes
-----
Currently, only the smoothing spline approximation (``iopt[0] = 0`` and
``iopt[0] = 1`` in the FITPACK routine) is supported. The exact
least-squares spline approximation is not implemented yet.
When actually performing the interpolation, the requested `v` values must
lie within the same length 2pi interval that the original `v` values were
chosen from.
For more information, see the FITPACK_ site about this function.
.. _FITPACK: http://www.netlib.org/dierckx/spgrid.f
Examples
--------
Suppose we have global data on a coarse grid
>>> import numpy as np
>>> lats = np.linspace(10, 170, 9) * np.pi / 180.
>>> lons = np.linspace(0, 350, 18) * np.pi / 180.
>>> data = np.dot(np.atleast_2d(90. - np.linspace(-80., 80., 18)).T,
... np.atleast_2d(180. - np.abs(np.linspace(0., 350., 9)))).T
We want to interpolate it to a global one-degree grid
>>> new_lats = np.linspace(1, 180, 180) * np.pi / 180
>>> new_lons = np.linspace(1, 360, 360) * np.pi / 180
>>> new_lats, new_lons = np.meshgrid(new_lats, new_lons)
We need to set up the interpolator object
>>> from scipy.interpolate import RectSphereBivariateSpline
>>> lut = RectSphereBivariateSpline(lats, lons, data)
Finally we interpolate the data. The `RectSphereBivariateSpline` object
only takes 1-D arrays as input, therefore we need to do some reshaping.
>>> data_interp = lut.ev(new_lats.ravel(),
... new_lons.ravel()).reshape((360, 180)).T
Looking at the original and the interpolated data, one can see that the
interpolant reproduces the original data very well:
>>> import matplotlib.pyplot as plt
>>> fig = plt.figure()
>>> ax1 = fig.add_subplot(211)
>>> ax1.imshow(data, interpolation='nearest')
>>> ax2 = fig.add_subplot(212)
>>> ax2.imshow(data_interp, interpolation='nearest')
>>> plt.show()
Choosing the optimal value of ``s`` can be a delicate task. Recommended
values for ``s`` depend on the accuracy of the data values. If the user
has an idea of the statistical errors on the data, she can also find a
proper estimate for ``s``. By assuming that, if she specifies the
right ``s``, the interpolator will use a spline ``f(u,v)`` which exactly
reproduces the function underlying the data, she can evaluate
``sum((r(i,j)-s(u(i),v(j)))**2)`` to find a good estimate for this ``s``.
For example, if she knows that the statistical errors on her
``r(i,j)``-values are not greater than 0.1, she may expect that a good
``s`` should have a value not larger than ``u.size * v.size * (0.1)**2``.
If nothing is known about the statistical error in ``r(i,j)``, ``s`` must
be determined by trial and error. The best is then to start with a very
large value of ``s`` (to determine the least-squares polynomial and the
corresponding upper bound ``fp0`` for ``s``) and then to progressively
decrease the value of ``s`` (say by a factor 10 in the beginning, i.e.
``s = fp0 / 10, fp0 / 100, ...`` and more carefully as the approximation
shows more detail) to obtain closer fits.
The interpolation results for different values of ``s`` give some insight
into this process:
>>> fig2 = plt.figure()
>>> s = [3e9, 2e9, 1e9, 1e8]
>>> for idx, sval in enumerate(s, 1):
... lut = RectSphereBivariateSpline(lats, lons, data, s=sval)
... data_interp = lut.ev(new_lats.ravel(),
... new_lons.ravel()).reshape((360, 180)).T
... ax = fig2.add_subplot(2, 2, idx)
... ax.imshow(data_interp, interpolation='nearest')
... ax.set_title(f"s = {sval:g}")
>>> plt.show()
"""
def __init__(self, u, v, r, s=0., pole_continuity=False, pole_values=None,
pole_exact=False, pole_flat=False):
iopt = np.array([0, 0, 0], dtype=dfitpack_int)
ider = np.array([-1, 0, -1, 0], dtype=dfitpack_int)
if pole_values is None:
pole_values = (None, None)
elif isinstance(pole_values, (float, np.float32, np.float64)):
pole_values = (pole_values, pole_values)
if isinstance(pole_continuity, bool):
pole_continuity = (pole_continuity, pole_continuity)
if isinstance(pole_exact, bool):
pole_exact = (pole_exact, pole_exact)
if isinstance(pole_flat, bool):
pole_flat = (pole_flat, pole_flat)
r0, r1 = pole_values
iopt[1:] = pole_continuity
if r0 is None:
ider[0] = -1
else:
ider[0] = pole_exact[0]
if r1 is None:
ider[2] = -1
else:
ider[2] = pole_exact[1]
ider[1], ider[3] = pole_flat
u, v = np.ravel(u), np.ravel(v)
r = np.asarray(r)
if not (0.0 < u[0] and u[-1] < np.pi):
raise ValueError('u should be between (0, pi)')
if not -np.pi <= v[0] < np.pi:
raise ValueError('v[0] should be between [-pi, pi)')
if not v[-1] <= v[0] + 2*np.pi:
raise ValueError('v[-1] should be v[0] + 2pi or less ')
if not np.all(np.diff(u) > 0.0):
raise ValueError('u must be strictly increasing')
if not np.all(np.diff(v) > 0.0):
raise ValueError('v must be strictly increasing')
if not u.size == r.shape[0]:
raise ValueError('u dimension of r must have same number of '
'elements as u')
if not v.size == r.shape[1]:
raise ValueError('v dimension of r must have same number of '
'elements as v')
if pole_continuity[1] is False and pole_flat[1] is True:
raise ValueError('if pole_continuity is False, so must be '
'pole_flat')
if pole_continuity[0] is False and pole_flat[0] is True:
raise ValueError('if pole_continuity is False, so must be '
'pole_flat')
if not s >= 0.0:
raise ValueError('s should be positive')
r = np.ravel(r)
nu, tu, nv, tv, c, fp, ier = dfitpack.regrid_smth_spher(iopt, ider,
u.copy(),
v.copy(),
r.copy(),
r0, r1, s)
if ier not in [0, -1, -2]:
msg = _spfit_messages.get(ier, 'ier=%s' % (ier))
raise ValueError(msg)
self.fp = fp
self.tck = tu[:nu], tv[:nv], c[:(nu - 4) * (nv-4)]
self.degrees = (3, 3)
self.v0 = v[0]
def __call__(self, theta, phi, dtheta=0, dphi=0, grid=True):
theta = np.asarray(theta)
phi = np.asarray(phi)
return SphereBivariateSpline.__call__(self, theta, phi, dtheta=dtheta,
dphi=dphi, grid=grid)
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