package scipy

One-dimensional spline with explicit internal knots.

Fits a spline y = spl(x) of degree `k` to the provided `x`, `y` data. `t` specifies the internal knots of the spline

Parameters ---------- x : (N,) array_like Input dimension of data points -- must be increasing y : (N,) array_like Input dimension of data points t : (M,) array_like interior knots of the spline. Must be in ascending order and::

bbox0 < t0 < ... < t-1 < bbox-1

w : (N,) array_like, optional weights for spline fitting. Must be positive. If None (default), weights are all equal. bbox : (2,) array_like, optional 2-sequence specifying the boundary of the approximation interval. If None (default), ``bbox = x[0], x[-1]``. k : int, optional Degree of the smoothing spline. Must be 1 <= `k` <= 5. Default is k=3, a cubic spline. ext : int or str, optional Controls the extrapolation mode for elements not in the interval defined by the knot sequence.

* if ext=0 or 'extrapolate', return the extrapolated value. * if ext=1 or 'zeros', return 0 * if ext=2 or 'raise', raise a ValueError * if ext=3 of 'const', return the boundary value.

The default value is 0.

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 or non-sensical results) if the inputs do contain infinities or NaNs. Default is False.

Raises ------ ValueError If the interior knots do not satisfy the Schoenberg-Whitney conditions

See Also -------- UnivariateSpline : Superclass -- knots are specified by setting a smoothing condition InterpolatedUnivariateSpline : spline passing through all points splrep : An older, non object-oriented wrapping of FITPACK splev, sproot, splint, spalde BivariateSpline : A similar class for two-dimensional spline interpolation

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 ``xj`` such that ``tj < xj < tj+k+1``, for ``j=0, 1,...,n-k-2``.

Examples -------- >>> from scipy.interpolate import LSQUnivariateSpline, UnivariateSpline >>> import matplotlib.pyplot as plt >>> x = np.linspace(-3, 3, 50) >>> y = np.exp(-x**2) + 0.1 * np.random.randn(50)

Fit a smoothing spline with a pre-defined internal knots:

>>> t = -1, 0, 1 >>> spl = LSQUnivariateSpline(x, y, t)

>>> xs = np.linspace(-3, 3, 1000) >>> plt.plot(x, y, 'ro', ms=5) >>> plt.plot(xs, spl(xs), 'g-', lw=3) >>> plt.show()

Check the knot vector:

>>> spl.get_knots() array(-3., -1., 0., 1., 3.)

Constructing lsq spline using the knots from another spline:

>>> x = np.arange(10) >>> s = UnivariateSpline(x, x, s=0) >>> s.get_knots() array( 0., 2., 3., 4., 5., 6., 7., 9.) >>> knt = s.get_knots() >>> s1 = LSQUnivariateSpline(x, x, knt1:-1) # Chop 1st and last knot >>> s1.get_knots() array( 0., 2., 3., 4., 5., 6., 7., 9.)

Construct a new spline representing the antiderivative of this spline.

Parameters ---------- n : int, optional Order of antiderivative to evaluate. Default: 1

Returns ------- spline : UnivariateSpline Spline of order k2=k+n representing the antiderivative of this spline.

Notes -----

.. versionadded:: 0.13.0

See Also -------- splantider, derivative

Examples -------- >>> from scipy.interpolate import UnivariateSpline >>> x = np.linspace(0, np.pi/2, 70) >>> y = 1 / np.sqrt(1 - 0.8*np.sin(x)**2) >>> spl = UnivariateSpline(x, y, s=0)

The derivative is the inverse operation of the antiderivative, although some floating point error accumulates:

>>> spl(1.7), spl.antiderivative().derivative()(1.7) (array(2.1565429877197317), array(2.1565429877201865))

Antiderivative can be used to evaluate definite integrals:

>>> ispl = spl.antiderivative() >>> ispl(np.pi/2) - ispl(0) 2.2572053588768486

This is indeed an approximation to the complete elliptic integral :math:`K(m) = \int_0^\pi/2 1 - m\sin^2 x^

1/2

}

dx`:

>>> from scipy.special import ellipk >>> ellipk(0.8) 2.2572053268208538

Construct a new spline representing the derivative of this spline.

Parameters ---------- n : int, optional Order of derivative to evaluate. Default: 1

Returns ------- spline : UnivariateSpline Spline of order k2=k-n representing the derivative of this spline.

See Also -------- splder, antiderivative

Notes -----

.. versionadded:: 0.13.0

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

>>> from scipy.interpolate import UnivariateSpline >>> x = np.linspace(0, 10, 70) >>> y = np.sin(x) >>> spl = UnivariateSpline(x, y, k=4, s=0)

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):

>>> spl.derivative().roots() / 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)`.

Return all derivatives of the spline at the point x.

Parameters ---------- x : float The point to evaluate the derivatives at.

Returns ------- der : ndarray, shape(k+1,) Derivatives of the orders 0 to k.

Examples -------- >>> from scipy.interpolate import UnivariateSpline >>> x = np.linspace(0, 3, 11) >>> y = x**2 >>> spl = UnivariateSpline(x, y) >>> spl.derivatives(1.5) array(2.25, 3.0, 2.0, 0)

Return spline coefficients.

Return positions of interior knots of the spline.

Internally, the knot vector contains ``2*k`` additional boundary knots.

Return weighted sum of squared residuals of the spline approximation.

This is equivalent to::

sum((wi * (yi-spl(xi)))**2, axis=0)

Return definite integral of the spline between two given points.

Parameters ---------- a : float Lower limit of integration. b : float Upper limit of integration.

Returns ------- integral : float The value of the definite integral of the spline between limits.

Examples -------- >>> from scipy.interpolate import UnivariateSpline >>> x = np.linspace(0, 3, 11) >>> y = x**2 >>> spl = UnivariateSpline(x, y) >>> spl.integral(0, 3) 9.0

which agrees with :math:`\int x^2 dx = x^3 / 3` between the limits of 0 and 3.

A caveat is that this routine assumes the spline to be zero outside of the data limits:

>>> spl.integral(-1, 4) 9.0 >>> spl.integral(-1, 0) 0.0

Return the zeros of the spline.

Restriction: only cubic splines are supported by fitpack.

Continue spline computation with the given smoothing factor s and with the knots found at the last call.

This routine modifies the spline in place.

Print the object to a human-readable representation.

Print the object to a human-readable representation.

Pretty-print the object to a formatter.