scipy.interpolate.BSpline.

derivative#

BSpline.derivative(nu=1)[source]#

Return a B-spline representing the derivative.

Parameters:

nuint, optional: Derivative order. Default is 1.

Returns:

bBSpline object: A new instance representing the derivative.

See also

splder, splantider

Notes

The coefficient array of the returned instance will be padded with b.k + 1 trailing zeros, so that len(b.c) == len(b.t). The trailing coefficients never enter evaluations, only the first len(b.t) - b.k - 1 coefficients are used.

Examples

>>> import numpy as np
>>> from scipy.interpolate import BSpline, make_interp_spline
>>> x = np.arange(11)
>>> y = x**4 + np.sin(x)
>>> spl = make_interp_spline(x, y, k=3)

A spline of degree k is built from polynomial pieces of degree k, joined at the knots t. The number of knots, the number of coefficients, and the degree are related by len(t) - len(c) == k + 1. Padded coefficients instead satisfy len(c) == len(t):

>>> len(spl.t), len(spl.c), spl.k    # knots, coefficients, degree of spl
(15, 11, 3)
>>> b = spl.derivative()             # b is a quadratic spline
>>> len(b.t), len(b.c), b.k          # knots, coefficients, degree of b
(13, 13, 2)
>>> b.c[len(b.t) - b.k - 1:]         # zero padding, unused in evaluations
array([0., 0., 0.])

Slice off the padding to get exactly len(b.t) - b.k - 1 coefficients:

>>> c = b.c[:len(b.t) - b.k - 1]
>>> len(c)
10

The trimmed coefficients define the same spline, since the padding never enters evaluations. Evaluating the spline before (b) and after (BSpline(b.t, c, b.k)) trimming on a grid of points spanning the data range gives the same values:

>>> xx = np.linspace(0, 10, 51)
>>> np.allclose(BSpline(b.t, c, b.k)(xx), b(xx))
True