__all__ = ['splrep', 'splprep', 'splev', 'splint', 'sproot', 'spalde', 'bisplrep', 'bisplev', 'insert', 'splder', 'splantider'] import numpy as np # These are in the API for fitpack even if not used in fitpack.py itself. from ._fitpack_impl import bisplrep, bisplev, dblint # noqa: F401 from . import _fitpack_impl as _impl from ._bsplines import BSpline def splprep(x, w=None, u=None, ub=None, ue=None, k=3, task=0, s=None, t=None, full_output=0, nest=None, per=0, quiet=1): """ Find the B-spline representation of an N-D curve. .. legacy:: function Specifically, we recommend using `make_splprep` in new code. Given a list of N rank-1 arrays, `x`, which represent a curve in N-dimensional space parametrized by `u`, find a smooth approximating spline curve g(`u`). Uses the FORTRAN routine parcur from FITPACK. Parameters ---------- x : array_like A list of sample vector arrays representing the curve. w : array_like, optional Strictly positive rank-1 array of weights the same length as `x[0]`. The weights are used in computing the weighted least-squares spline fit. If the errors in the `x` values have standard-deviation given by the vector d, then `w` should be 1/d. Default is ``ones(len(x[0]))``. u : array_like, optional An array of parameter values. If not given, these values are calculated automatically as ``M = len(x[0])``, where v[0] = 0 v[i] = v[i-1] + distance(`x[i]`, `x[i-1]`) u[i] = v[i] / v[M-1] ub, ue : int, optional The end-points of the parameters interval. Defaults to u[0] and u[-1]. k : int, optional Degree of the spline. Cubic splines are recommended. Even values of `k` should be avoided especially with a small s-value. ``1 <= k <= 5``, default is 3. task : int, optional If task==0 (default), find t and c for a given smoothing factor, s. If task==1, find t and c for another value of the smoothing factor, s. There must have been a previous call with task=0 or task=1 for the same set of data. If task=-1 find the weighted least square spline for a given set of knots, t. s : float, optional A smoothing condition. The amount of smoothness is determined by satisfying the conditions: ``sum((w * (y - g))**2,axis=0) <= s``, where g(x) is the smoothed interpolation of (x,y). The user can use `s` to control the trade-off between closeness and smoothness of fit. Larger `s` means more smoothing while smaller values of `s` indicate less smoothing. Recommended values of `s` depend on the weights, w. If the weights represent the inverse of the standard-deviation of y, then a good `s` value should be found in the range ``(m-sqrt(2*m),m+sqrt(2*m))``, where m is the number of data points in x, y, and w. t : array, optional The knots needed for ``task=-1``. There must be at least ``2*k+2`` knots. full_output : int, optional If non-zero, then return optional outputs. nest : int, optional An over-estimate of the total number of knots of the spline to help in determining the storage space. By default nest=m/2. Always large enough is nest=m+k+1. per : int, optional If non-zero, data points are considered periodic with period ``x[m-1] - x[0]`` and a smooth periodic spline approximation is returned. Values of ``y[m-1]`` and ``w[m-1]`` are not used. quiet : int, optional Non-zero to suppress messages. Returns ------- tck : tuple A tuple, ``(t,c,k)`` containing the vector of knots, the B-spline coefficients, and the degree of the spline. u : array An array of the values of the parameter. fp : float The weighted sum of squared residuals of the spline approximation. ier : int An integer flag about splrep success. Success is indicated if ier<=0. If ier in [1,2,3] an error occurred but was not raised. Otherwise an error is raised. msg : str A message corresponding to the integer flag, ier. See Also -------- splrep, splev, sproot, spalde, splint, bisplrep, bisplev UnivariateSpline, BivariateSpline BSpline make_interp_spline Notes ----- See `splev` for evaluation of the spline and its derivatives. The number of dimensions N must be smaller than 11. The number of coefficients in the `c` array is ``k+1`` less than the number of knots, ``len(t)``. This is in contrast with `splrep`, which zero-pads the array of coefficients to have the same length as the array of knots. These additional coefficients are ignored by evaluation routines, `splev` and `BSpline`. References ---------- .. [1] P. Dierckx, "Algorithms for smoothing data with periodic and parametric splines, Computer Graphics and Image Processing", 20 (1982) 171-184. .. [2] P. Dierckx, "Algorithms for smoothing data with periodic and parametric splines", report tw55, Dept. Computer Science, K.U.Leuven, 1981. .. [3] P. Dierckx, "Curve and surface fitting with splines", Monographs on Numerical Analysis, Oxford University Press, 1993. Examples -------- Generate a discretization of a limacon curve in the polar coordinates: >>> import numpy as np >>> phi = np.linspace(0, 2.*np.pi, 40) >>> r = 0.5 + np.cos(phi) # polar coords >>> x, y = r * np.cos(phi), r * np.sin(phi) # convert to cartesian And interpolate: >>> from scipy.interpolate import splprep, splev >>> tck, u = splprep([x, y], s=0) >>> new_points = splev(u, tck) Notice that (i) we force interpolation by using ``s=0``, (ii) the parameterization, ``u``, is generated automatically. Now plot the result: >>> import matplotlib.pyplot as plt >>> fig, ax = plt.subplots() >>> ax.plot(x, y, 'ro') >>> ax.plot(new_points[0], new_points[1], 'r-') >>> plt.show() """ res = _impl.splprep(x, w, u, ub, ue, k, task, s, t, full_output, nest, per, quiet) return res def splrep(x, y, w=None, xb=None, xe=None, k=3, task=0, s=None, t=None, full_output=0, per=0, quiet=1): """ Find the B-spline representation of a 1-D curve. .. legacy:: function Specifically, we recommend using `make_splrep` in new code. Given the set of data points ``(x[i], y[i])`` determine a smooth spline approximation of degree k on the interval ``xb <= x <= xe``. Parameters ---------- x, y : array_like The data points defining a curve ``y = f(x)``. w : array_like, optional Strictly positive rank-1 array of weights the same length as `x` and `y`. The weights are used in computing the weighted least-squares spline fit. If the errors in the `y` values have standard-deviation given by the vector ``d``, then `w` should be ``1/d``. Default is ``ones(len(x))``. xb, xe : float, optional The interval to fit. If None, these default to ``x[0]`` and ``x[-1]`` respectively. k : int, optional The degree of the spline fit. It is recommended to use cubic splines. Even values of `k` should be avoided especially with small `s` values. ``1 <= k <= 5``. task : {1, 0, -1}, optional If ``task==0``, find ``t`` and ``c`` for a given smoothing factor, `s`. If ``task==1`` find ``t`` and ``c`` for another value of the smoothing factor, `s`. There must have been a previous call with ``task=0`` or ``task=1`` for the same set of data (``t`` will be stored an used internally) If ``task=-1`` find the weighted least square spline for a given set of knots, ``t``. These should be interior knots as knots on the ends will be added automatically. s : float, optional A smoothing condition. The amount of smoothness is determined by satisfying the conditions: ``sum((w * (y - g))**2,axis=0) <= s`` where ``g(x)`` is the smoothed interpolation of ``(x,y)``. The user can use `s` to control the tradeoff between closeness and smoothness of fit. Larger `s` means more smoothing while smaller values of `s` indicate less smoothing. Recommended values of `s` depend on the weights, `w`. If the weights represent the inverse of the standard-deviation of `y`, then a good `s` value should be found in the range ``(m-sqrt(2*m),m+sqrt(2*m))`` where ``m`` is the number of datapoints in `x`, `y`, and `w`. default : ``s=m-sqrt(2*m)`` if weights are supplied. ``s = 0.0`` (interpolating) if no weights are supplied. t : array_like, optional The knots needed for ``task=-1``. If given then task is automatically set to ``-1``. full_output : bool, optional If non-zero, then return optional outputs. per : bool, optional If non-zero, data points are considered periodic with period ``x[m-1]`` - ``x[0]`` and a smooth periodic spline approximation is returned. Values of ``y[m-1]`` and ``w[m-1]`` are not used. The default is zero, corresponding to boundary condition 'not-a-knot'. quiet : bool, optional Non-zero to suppress messages. Returns ------- tck : tuple A tuple ``(t,c,k)`` containing the vector of knots, the B-spline coefficients, and the degree of the spline. fp : array, optional The weighted sum of squared residuals of the spline approximation. ier : int, optional An integer flag about splrep success. Success is indicated if ``ier<=0``. If ``ier in [1,2,3]``, an error occurred but was not raised. Otherwise an error is raised. msg : str, optional A message corresponding to the integer flag, `ier`. See Also -------- UnivariateSpline, BivariateSpline splprep, splev, sproot, spalde, splint bisplrep, bisplev BSpline make_interp_spline Notes ----- See `splev` for evaluation of the spline and its derivatives. Uses the FORTRAN routine ``curfit`` from FITPACK. The user is responsible for assuring that the values of `x` are unique. Otherwise, `splrep` will not return sensible results. If provided, 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``. This routine zero-pads the coefficients array ``c`` to have the same length as the array of knots ``t`` (the trailing ``k + 1`` coefficients are ignored by the evaluation routines, `splev` and `BSpline`.) This is in contrast with `splprep`, which does not zero-pad the coefficients. The default boundary condition is 'not-a-knot', i.e. the first and second segment at a curve end are the same polynomial. More boundary conditions are available in `CubicSpline`. References ---------- Based on algorithms described in [1]_, [2]_, [3]_, and [4]_: .. [1] P. Dierckx, "An algorithm for smoothing, differentiation and integration of experimental data using spline functions", J.Comp.Appl.Maths 1 (1975) 165-184. .. [2] P. Dierckx, "A fast algorithm for smoothing data on a rectangular grid while using spline functions", SIAM J.Numer.Anal. 19 (1982) 1286-1304. .. [3] P. Dierckx, "An improved algorithm for curve fitting with spline functions", report tw54, Dept. Computer Science,K.U. Leuven, 1981. .. [4] P. Dierckx, "Curve and surface fitting with splines", Monographs on Numerical Analysis, Oxford University Press, 1993. Examples -------- You can interpolate 1-D points with a B-spline curve. Further examples are given in :ref:`in the tutorial `. >>> import numpy as np >>> import matplotlib.pyplot as plt >>> from scipy.interpolate import splev, splrep >>> x = np.linspace(0, 10, 10) >>> y = np.sin(x) >>> spl = splrep(x, y) >>> x2 = np.linspace(0, 10, 200) >>> y2 = splev(x2, spl) >>> plt.plot(x, y, 'o', x2, y2) >>> plt.show() """ res = _impl.splrep(x, y, w, xb, xe, k, task, s, t, full_output, per, quiet) return res def splev(x, tck, der=0, ext=0): """ Evaluate a B-spline or its derivatives. .. legacy:: function Specifically, we recommend constructing a `BSpline` object and using its ``__call__`` method. Given the knots and coefficients of a B-spline representation, evaluate the value of the smoothing polynomial and its derivatives. This is a wrapper around the FORTRAN routines splev and splder of FITPACK. Parameters ---------- x : array_like An array of points at which to return the value of the smoothed spline or its derivatives. If `tck` was returned from `splprep`, then the parameter values, u should be given. tck : BSpline instance or tuple If a tuple, then it should be a sequence of length 3 returned by `splrep` or `splprep` containing the knots, coefficients, and degree of the spline. (Also see Notes.) der : int, optional The order of derivative of the spline to compute (must be less than or equal to k, the degree of the spline). ext : int, optional Controls the value returned for elements of ``x`` not in the interval defined by the knot sequence. * if ext=0, return the extrapolated value. * if ext=1, return 0 * if ext=2, raise a ValueError * if ext=3, return the boundary value. The default value is 0. Returns ------- y : ndarray or list of ndarrays An array of values representing the spline function evaluated at the points in `x`. If `tck` was returned from `splprep`, then this is a list of arrays representing the curve in an N-D space. See Also -------- splprep, splrep, sproot, spalde, splint bisplrep, bisplev BSpline Notes ----- Manipulating the tck-tuples directly is not recommended. In new code, prefer using `BSpline` objects. References ---------- .. [1] C. de Boor, "On calculating with b-splines", J. Approximation Theory, 6, p.50-62, 1972. .. [2] M. G. Cox, "The numerical evaluation of b-splines", J. Inst. Maths Applics, 10, p.134-149, 1972. .. [3] P. Dierckx, "Curve and surface fitting with splines", Monographs on Numerical Analysis, Oxford University Press, 1993. Examples -------- Examples are given :ref:`in the tutorial `. A comparison between `splev`, `splder` and `spalde` to compute the derivatives of a B-spline can be found in the `spalde` examples section. """ if isinstance(tck, BSpline): if tck.c.ndim > 1: mesg = ("Calling splev() with BSpline objects with c.ndim > 1 is " "not allowed. Use BSpline.__call__(x) instead.") raise ValueError(mesg) # remap the out-of-bounds behavior try: extrapolate = {0: True, }[ext] except KeyError as e: raise ValueError(f"Extrapolation mode {ext} is not supported " "by BSpline.") from e return tck(x, der, extrapolate=extrapolate) else: return _impl.splev(x, tck, der, ext) def splint(a, b, tck, full_output=0): """ Evaluate the definite integral of a B-spline between two given points. .. legacy:: function Specifically, we recommend constructing a `BSpline` object and using its ``integrate`` method. Parameters ---------- a, b : float The end-points of the integration interval. tck : tuple or a BSpline instance If a tuple, then it should be a sequence of length 3, containing the vector of knots, the B-spline coefficients, and the degree of the spline (see `splev`). full_output : int, optional Non-zero to return optional output. Returns ------- integral : float The resulting integral. wrk : ndarray An array containing the integrals of the normalized B-splines defined on the set of knots. (Only returned if `full_output` is non-zero) See Also -------- splprep, splrep, sproot, spalde, splev bisplrep, bisplev BSpline Notes ----- `splint` silently assumes that the spline function is zero outside the data interval (`a`, `b`). Manipulating the tck-tuples directly is not recommended. In new code, prefer using the `BSpline` objects. References ---------- .. [1] P.W. Gaffney, The calculation of indefinite integrals of b-splines", J. Inst. Maths Applics, 17, p.37-41, 1976. .. [2] P. Dierckx, "Curve and surface fitting with splines", Monographs on Numerical Analysis, Oxford University Press, 1993. Examples -------- Examples are given :ref:`in the tutorial `. """ if isinstance(tck, BSpline): if tck.c.ndim > 1: mesg = ("Calling splint() with BSpline objects with c.ndim > 1 is " "not allowed. Use BSpline.integrate() instead.") raise ValueError(mesg) if full_output != 0: mesg = (f"full_output = {full_output} is not supported. Proceeding as if " "full_output = 0") return tck.integrate(a, b, extrapolate=False) else: return _impl.splint(a, b, tck, full_output) def sproot(tck, mest=10): """ Find the roots of a cubic B-spline. .. legacy:: function Specifically, we recommend constructing a `BSpline` object and using the following pattern: `PPoly.from_spline(spl).roots()`. Given the knots (>=8) and coefficients of a cubic B-spline return the roots of the spline. Parameters ---------- tck : tuple or a BSpline object If a tuple, then it should be a sequence of length 3, containing the vector of knots, the B-spline coefficients, and the degree of the spline. The number of knots must be >= 8, and the degree must be 3. The knots must be a montonically increasing sequence. mest : int, optional An estimate of the number of zeros (Default is 10). Returns ------- zeros : ndarray An array giving the roots of the spline. See Also -------- splprep, splrep, splint, spalde, splev bisplrep, bisplev BSpline Notes ----- Manipulating the tck-tuples directly is not recommended. In new code, prefer using the `BSpline` objects. References ---------- .. [1] C. de Boor, "On calculating with b-splines", J. Approximation Theory, 6, p.50-62, 1972. .. [2] M. G. Cox, "The numerical evaluation of b-splines", J. Inst. Maths Applics, 10, p.134-149, 1972. .. [3] P. Dierckx, "Curve and surface fitting with splines", Monographs on Numerical Analysis, Oxford University Press, 1993. Examples -------- For some data, this method may miss a root. This happens when one of the spline knots (which FITPACK places automatically) happens to coincide with the true root. A workaround is to convert to `PPoly`, which uses a different root-finding algorithm. For example, >>> x = [1.96, 1.97, 1.98, 1.99, 2.00, 2.01, 2.02, 2.03, 2.04, 2.05] >>> y = [-6.365470e-03, -4.790580e-03, -3.204320e-03, -1.607270e-03, ... 4.440892e-16, 1.616930e-03, 3.243000e-03, 4.877670e-03, ... 6.520430e-03, 8.170770e-03] >>> from scipy.interpolate import splrep, sproot, PPoly >>> tck = splrep(x, y, s=0) >>> sproot(tck) array([], dtype=float64) Converting to a PPoly object does find the roots at ``x=2``: >>> ppoly = PPoly.from_spline(tck) >>> ppoly.roots(extrapolate=False) array([2.]) Further examples are given :ref:`in the tutorial `. """ if isinstance(tck, BSpline): if tck.c.ndim > 1: mesg = ("Calling sproot() with BSpline objects with c.ndim > 1 is " "not allowed.") raise ValueError(mesg) t, c, k = tck.tck # _impl.sproot expects the interpolation axis to be last, so roll it. # NB: This transpose is a no-op if c is 1D. sh = tuple(range(c.ndim)) c = c.transpose(sh[1:] + (0,)) return _impl.sproot((t, c, k), mest) else: return _impl.sproot(tck, mest) def spalde(x, tck): """ Evaluate a B-spline and all its derivatives at one point (or set of points) up to order k (the degree of the spline), being 0 the spline itself. .. legacy:: function Specifically, we recommend constructing a `BSpline` object and evaluate its derivative in a loop or a list comprehension. Parameters ---------- x : array_like A point or a set of points at which to evaluate the derivatives. Note that ``t(k) <= x <= t(n-k+1)`` must hold for each `x`. tck : tuple A tuple (t,c,k) containing the vector of knots, the B-spline coefficients, and the degree of the spline whose derivatives to compute. Returns ------- results : {ndarray, list of ndarrays} An array (or a list of arrays) containing all derivatives up to order k inclusive for each point `x`, being the first element the spline itself. See Also -------- splprep, splrep, splint, sproot, splev, bisplrep, bisplev, UnivariateSpline, BivariateSpline References ---------- .. [1] de Boor C : On calculating with b-splines, J. Approximation Theory 6 (1972) 50-62. .. [2] Cox M.G. : The numerical evaluation of b-splines, J. Inst. Maths applics 10 (1972) 134-149. .. [3] Dierckx P. : Curve and surface fitting with splines, Monographs on Numerical Analysis, Oxford University Press, 1993. Examples -------- To calculate the derivatives of a B-spline there are several aproaches. In this example, we will demonstrate that `spalde` is equivalent to calling `splev` and `splder`. >>> import numpy as np >>> import matplotlib.pyplot as plt >>> from scipy.interpolate import BSpline, spalde, splder, splev >>> # Store characteristic parameters of a B-spline >>> tck = ((-2, -2, -2, -2, -1, 0, 1, 2, 2, 2, 2), # knots ... (0, 0, 0, 6, 0, 0, 0), # coefficients ... 3) # degree (cubic) >>> # Instance a B-spline object >>> # `BSpline` objects are preferred, except for spalde() >>> bspl = BSpline(tck[0], tck[1], tck[2]) >>> # Generate extra points to get a smooth curve >>> x = np.linspace(min(tck[0]), max(tck[0]), 100) Evaluate the curve and all derivatives >>> # The order of derivative must be less or equal to k, the degree of the spline >>> # Method 1: spalde() >>> f1_y_bsplin = [spalde(i, tck)[0] for i in x ] # The B-spline itself >>> f1_y_deriv1 = [spalde(i, tck)[1] for i in x ] # 1st derivative >>> f1_y_deriv2 = [spalde(i, tck)[2] for i in x ] # 2nd derivative >>> f1_y_deriv3 = [spalde(i, tck)[3] for i in x ] # 3rd derivative >>> # You can reach the same result by using `splev`and `splder` >>> f2_y_deriv3 = splev(x, bspl, der=3) >>> f3_y_deriv3 = splder(bspl, n=3)(x) >>> # Generate a figure with three axes for graphic comparison >>> fig, (ax1, ax2, ax3) = plt.subplots(1, 3, figsize=(16, 5)) >>> suptitle = fig.suptitle(f'Evaluate a B-spline and all derivatives') >>> # Plot B-spline and all derivatives using the three methods >>> orders = range(4) >>> linetypes = ['-', '--', '-.', ':'] >>> labels = ['B-Spline', '1st deriv.', '2nd deriv.', '3rd deriv.'] >>> functions = ['splev()', 'splder()', 'spalde()'] >>> for order, linetype, label in zip(orders, linetypes, labels): ... ax1.plot(x, splev(x, bspl, der=order), linetype, label=label) ... ax2.plot(x, splder(bspl, n=order)(x), linetype, label=label) ... ax3.plot(x, [spalde(i, tck)[order] for i in x], linetype, label=label) >>> for ax, function in zip((ax1, ax2, ax3), functions): ... ax.set_title(function) ... ax.legend() >>> plt.tight_layout() >>> plt.show() """ if isinstance(tck, BSpline): raise TypeError("spalde does not accept BSpline instances.") else: return _impl.spalde(x, tck) def insert(x, tck, m=1, per=0): """ Insert knots into a B-spline. .. legacy:: function Specifically, we recommend constructing a `BSpline` object and using its ``insert_knot`` method. Given the knots and coefficients of a B-spline representation, create a new B-spline with a knot inserted `m` times at point `x`. This is a wrapper around the FORTRAN routine insert of FITPACK. Parameters ---------- x (u) : float A knot value at which to insert a new knot. If `tck` was returned from ``splprep``, then the parameter values, u should be given. tck : a `BSpline` instance or a tuple If tuple, then it is expected to be a tuple (t,c,k) containing the vector of knots, the B-spline coefficients, and the degree of the spline. m : int, optional The number of times to insert the given knot (its multiplicity). Default is 1. per : int, optional If non-zero, the input spline is considered periodic. Returns ------- BSpline instance or a tuple A new B-spline with knots t, coefficients c, and degree k. ``t(k+1) <= x <= t(n-k)``, where k is the degree of the spline. In case of a periodic spline (``per != 0``) there must be either at least k interior knots t(j) satisfying ``t(k+1)>> from scipy.interpolate import splrep, insert >>> import numpy as np >>> x = np.linspace(0, 10, 5) >>> y = np.sin(x) >>> tck = splrep(x, y) >>> tck[0] array([ 0., 0., 0., 0., 5., 10., 10., 10., 10.]) A knot is inserted: >>> tck_inserted = insert(3, tck) >>> tck_inserted[0] array([ 0., 0., 0., 0., 3., 5., 10., 10., 10., 10.]) Some knots are inserted: >>> tck_inserted2 = insert(8, tck, m=3) >>> tck_inserted2[0] array([ 0., 0., 0., 0., 5., 8., 8., 8., 10., 10., 10., 10.]) """ if isinstance(tck, BSpline): t, c, k = tck.tck # FITPACK expects the interpolation axis to be last, so roll it over # NB: if c array is 1D, transposes are no-ops sh = tuple(range(c.ndim)) c = c.transpose(sh[1:] + (0,)) t_, c_, k_ = _impl.insert(x, (t, c, k), m, per) # and roll the last axis back c_ = np.asarray(c_) c_ = c_.transpose((sh[-1],) + sh[:-1]) return BSpline(t_, c_, k_) else: return _impl.insert(x, tck, m, per) def splder(tck, n=1): """ 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. """ if isinstance(tck, BSpline): return tck.derivative(n) else: return _impl.splder(tck, n) def splantider(tck, n=1): """ Compute the spline for the antiderivative (integral) of a given spline. .. legacy:: function Specifically, we recommend constructing a `BSpline` object and using its ``antiderivative`` method. Parameters ---------- tck : BSpline instance or a tuple of (t, c, k) Spline whose antiderivative to compute n : int, optional Order of antiderivative to evaluate. Default: 1 Returns ------- BSpline instance or a tuple of (t2, c2, k2) Spline of order k2=k+n representing the antiderivative of the input spline. A tuple is returned iff the input argument `tck` is a tuple, otherwise a BSpline object is constructed and returned. See Also -------- splder, splev, spalde BSpline Notes ----- The `splder` function is the inverse operation of this function. Namely, ``splder(splantider(tck))`` is identical to `tck`, modulo rounding error. .. versionadded:: 0.13.0 Examples -------- >>> from scipy.interpolate import splrep, splder, splantider, splev >>> import numpy as np >>> x = np.linspace(0, np.pi/2, 70) >>> y = 1 / np.sqrt(1 - 0.8*np.sin(x)**2) >>> spl = splrep(x, y) The derivative is the inverse operation of the antiderivative, although some floating point error accumulates: >>> splev(1.7, spl), splev(1.7, splder(splantider(spl))) (array(2.1565429877197317), array(2.1565429877201865)) Antiderivative can be used to evaluate definite integrals: >>> ispl = splantider(spl) >>> splev(np.pi/2, ispl) - splev(0, ispl) 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 """ if isinstance(tck, BSpline): return tck.antiderivative(n) else: return _impl.splantider(tck, n)