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scipy.interpolate.

# RectSphereBivariateSpline

class scipy.interpolate.RectSphereBivariateSpline(u, v, r, s=0.0, pole_continuity=False, pole_values=None, pole_exact=False, pole_flat=False)

  Bivariate spline approximation over a rectangular mesh on a sphere.

  Can be used for smoothing data.

  Added in version 0.11.0.

  Parameters:
    uarray_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)`.

    varray_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`.

    rarray_like

      2-D array of data with shape `(u.size, v.size)`.

    sfloat, optional

      Positive smoothing factor defined for estimation condition (`s=0` is for interpolation).

    pole_continuitybool 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_valuesfloat 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_exactbool 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_flatbool 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.

  Methods

  `__call__`(theta, phi[, dtheta, dphi, grid]) | Evaluate the spline or its derivatives at given positions.
  `ev`(theta, phi[, dtheta, dphi]) | Evaluate the spline at points
  `get_coeffs`() | Return spline coefficients.
  `get_knots`() | Return a tuple (tx,ty) where tx,ty contain knots positions of the spline with respect to x-, y-variable, respectively.
  `get_residual`() | Return weighted sum of squared residuals of the spline approximation: sum ((w[i]*(z[i]-s(x[i],y[i])))**2,axis=0)
  `partial_derivative`(dx, dy) | Construct a new spline representing a partial derivative of this spline.

  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.

  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()