Module « scipy.interpolate »

Fonction make_lsq_spline - module scipy.interpolate

Signature de la fonction make_lsq_spline

def make_lsq_spline(x, y, t, k=3, w=None, axis=0, check_finite=True) 

Description

make_lsq_spline.__doc__

Compute the (coefficients of) an LSQ B-spline.

    The result is a linear combination

    .. math::

            S(x) = \sum_j c_j B_j(x; t)

    of the B-spline basis elements, :math:`B_j(x; t)`, which minimizes

    .. math::

        \sum_{j} \left( w_j \times (S(x_j) - y_j) \right)^2

    Parameters
    ----------
    x : array_like, shape (m,)
        Abscissas.
    y : array_like, shape (m, ...)
        Ordinates.
    t : array_like, shape (n + k + 1,).
        Knots.
        Knots and data points must satisfy Schoenberg-Whitney conditions.
    k : int, optional
        B-spline degree. Default is cubic, k=3.
    w : array_like, shape (n,), optional
        Weights for spline fitting. Must be positive. If ``None``,
        then weights are all equal.
        Default is ``None``.
    axis : int, optional
        Interpolation axis. Default is zero.
    check_finite : bool, optional
        Whether to check that the input arrays contain only finite numbers.
        Disabling may give a performance gain, but may result in problems
        (crashes, non-termination) if the inputs do contain infinities or NaNs.
        Default is True.

    Returns
    -------
    b : a BSpline object of the degree `k` with knots `t`.

    Notes
    -----

    The number of data points must be larger than the spline degree `k`.

    Knots `t` must satisfy the Schoenberg-Whitney conditions,
    i.e., there must be a subset of data points ``x[j]`` such that
    ``t[j] < x[j] < t[j+k+1]``, for ``j=0, 1,...,n-k-2``.

    Examples
    --------
    Generate some noisy data:

    >>> rng = np.random.default_rng()
    >>> x = np.linspace(-3, 3, 50)
    >>> y = np.exp(-x**2) + 0.1 * rng.standard_normal(50)

    Now fit a smoothing cubic spline with a pre-defined internal knots.
    Here we make the knot vector (k+1)-regular by adding boundary knots:

    >>> from scipy.interpolate import make_lsq_spline, BSpline
    >>> t = [-1, 0, 1]
    >>> k = 3
    >>> t = np.r_[(x[0],)*(k+1),
    ...           t,
    ...           (x[-1],)*(k+1)]
    >>> spl = make_lsq_spline(x, y, t, k)

    For comparison, we also construct an interpolating spline for the same
    set of data:

    >>> from scipy.interpolate import make_interp_spline
    >>> spl_i = make_interp_spline(x, y)

    Plot both:

    >>> import matplotlib.pyplot as plt
    >>> xs = np.linspace(-3, 3, 100)
    >>> plt.plot(x, y, 'ro', ms=5)
    >>> plt.plot(xs, spl(xs), 'g-', lw=3, label='LSQ spline')
    >>> plt.plot(xs, spl_i(xs), 'b-', lw=3, alpha=0.7, label='interp spline')
    >>> plt.legend(loc='best')
    >>> plt.show()

    **NaN handling**: If the input arrays contain ``nan`` values, the result is
    not useful since the underlying spline fitting routines cannot deal with
    ``nan``. A workaround is to use zero weights for not-a-number data points:

    >>> y[8] = np.nan
    >>> w = np.isnan(y)
    >>> y[w] = 0.
    >>> tck = make_lsq_spline(x, y, t, w=~w)

    Notice the need to replace a ``nan`` by a numerical value (precise value
    does not matter as long as the corresponding weight is zero.)

    See Also
    --------
    BSpline : base class representing the B-spline objects
    make_interp_spline : a similar factory function for interpolating splines
    LSQUnivariateSpline : a FITPACK-based spline fitting routine
    splrep : a FITPACK-based fitting routine