`scipy.interpolate._fitpack2:BivariateSpline.ev`

source: /scipy/interpolate/_fitpack2.py :1229

Signature

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

Summary

Evaluate the spline at points

Extended Summary

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()

✓

Aliases

scipy.interpolate.BivariateSpline.ev