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 latitude coordinates in strictly ascending order. Coordinates must be given in radians and lie within the interval (0, pi). v : array_like 1-D array of longitude coordinates in strictly ascending order. Coordinates must be given in radians. First element (v0) 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
--------
RectBivariateSpline : bivariate spline approximation over a rectangular
mesh
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
>>> 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 ii in range(len(s)):
... lut = RectSphereBivariateSpline(lats, lons, data, s=s[ii])
... data_interp = lut.ev(new_lats.ravel(),
... new_lons.ravel()).reshape((360, 180)).T
... ax = fig2.add_subplot(2, 2, ii+1)
... ax.imshow(data_interp, interpolation='nearest')
... ax.set_title('s = %g' % s[ii])
>>> plt.show()