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Module « scipy.interpolate »

Fonction splder - module scipy.interpolate

Signature de la fonction splder

def splder(tck, n=1) 

Description

help(scipy.interpolate.splder)

Compute the spline representation of the derivative of a given spline

.. legacy:: function

    Specifically, we recommend constructing a `BSpline` object and using its
    ``derivative`` method.

Parameters
----------
tck : BSpline instance or tuple
    BSpline instance or a tuple (t,c,k) containing the vector of knots,
    the B-spline coefficients, and the degree of the spline whose 
    derivative to compute
n : int, optional
    Order of derivative to evaluate. Default: 1

Returns
-------
`BSpline` instance or tuple
    Spline of order k2=k-n representing the derivative
    of the input spline.
    A tuple is returned if the input argument `tck` is a tuple, otherwise
    a BSpline object is constructed and returned.

See Also
--------
splantider, splev, spalde
BSpline

Notes
-----

.. versionadded:: 0.13.0

Examples
--------
This can be used for finding maxima of a curve:

>>> from scipy.interpolate import splrep, splder, sproot
>>> import numpy as np
>>> x = np.linspace(0, 10, 70)
>>> y = np.sin(x)
>>> spl = splrep(x, y, k=4)

Now, differentiate the spline and find the zeros of the
derivative. (NB: `sproot` only works for order 3 splines, so we
fit an order 4 spline):

>>> dspl = splder(spl)
>>> sproot(dspl) / np.pi
array([ 0.50000001,  1.5       ,  2.49999998])

This agrees well with roots :math:`\pi/2 + n\pi` of
:math:`\cos(x) = \sin'(x)`.

A comparison between `splev`, `splder` and `spalde` to compute the derivatives of a 
B-spline can be found in the `spalde` examples section.

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