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Classe « BivariateSpline »

Méthode scipy.interpolate.BivariateSpline.ev

Signature de la méthode ev

def ev(self, xi, yi, dx=0, dy=0) 

Description

help(BivariateSpline.ev)

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()

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