Can be used for smoothing data.
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 :None:None:`FITPACK_` site about this function.
<Unimplemented 'target' '.. _FITPACK: http://www.netlib.org/dierckx/spgrid.f'>
1-D array of colatitude coordinates in strictly ascending order. Coordinates must be given in radians and lie within the open interval (0, pi)
.
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
.
2-D array of data with shape (u.size, v.size)
.
Positive smoothing factor defined for estimation condition ( s=0
is for interpolation).
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.
Data values at the poles u=0
and u=pi
. Either the whole parameter or each individual element can be None. Defaults to None.
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.
For the poles at u=0
and u=pi
, specify whether or not the approximation has vanishing derivatives. Defaults to False.
Bivariate spline approximation over a rectangular mesh on a sphere.
BivariateSpline
a base class for bivariate splines.
LSQBivariateSpline
a bivariate spline using weighted least-squares fitting
LSQSphereBivariateSpline
a bivariate spline in spherical coordinates using weighted least-squares fitting
RectBivariateSpline
a bivariate spline over a rectangular mesh.
SmoothBivariateSpline
a smoothing bivariate spline through the given points
SmoothSphereBivariateSpline
a smoothing bivariate spline in spherical coordinates
UnivariateSpline
a smooth univariate spline to fit a given set of data points.
bisplev
a function to evaluate a bivariate B-spline and its derivatives
bisplrep
a function to find a bivariate B-spline representation of a surface
Suppose we have global data on a coarse grid
>>> 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()
The following pages refer to to this document either explicitly or contain code examples using this.
scipy.interpolate._fitpack2.RectSphereBivariateSpline
scipy.interpolate._fitpack2.SmoothBivariateSpline
scipy.interpolate._fitpack2.BivariateSpline
scipy.interpolate._fitpack2.LSQSphereBivariateSpline
scipy.interpolate._fitpack2.RectBivariateSpline
scipy.interpolate._fitpack2.LSQBivariateSpline
scipy.interpolate._fitpack2.SmoothSphereBivariateSpline
scipy.interpolate._fitpack2.UnivariateSpline
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