# scipy.interpolate.make_interp_spline

scipy.interpolate.make_interp_spline(x, y, k=3, t=None, bc_type=None, axis=0, check_finite=True)

  Compute the (coefficients of) interpolating B-spline.

  Parameters
    x array_like, shape (n,)

      Abscissas.

    y array_like, shape (n, …)

      Ordinates.

    k int, optional

      B-spline degree. Default is cubic, k=3.

    t array_like, shape (nt + k + 1,), optional.

      Knots. The number of knots needs to agree with the number of datapoints and the number of derivatives at the edges. Specifically, `nt - n` must equal `len(deriv_l) + len(deriv_r)`.

    bc_type 2-tuple or None

      Boundary conditions. Default is None, which means choosing the boundary conditions automatically. Otherwise, it must be a length-two tuple where the first element sets the boundary conditions at `x[0]` and the second element sets the boundary conditions at `x[-1]`. Each of these must be an iterable of pairs `(order, value)` which gives the values of derivatives of specified orders at the given edge of the interpolation interval. Alternatively, the following string aliases are recognized:

      - `"clamped"`: The first derivatives at the ends are zero. This is

          equivalent to `bc_type=([(1, 0.0)], [(1, 0.0)])`.

      - `"natural"`: The second derivatives at ends are zero. This is equivalent to `bc_type=([(2, 0.0)], [(2, 0.0)])`.

      - `"not-a-knot"` (default): The first and second segments are the same polynomial. This is equivalent to having `bc_type=None`.

      - `"periodic"`: The values and the first `k-1` derivatives at the ends are equivalent.

    axis int, optional

      Interpolation axis. Default 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) if the inputs do contain infinities or NaNs. Default is True.

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

  See also

  `BSpline`

    base class representing the B-spline objects

  `CubicSpline`

    a cubic spline in the polynomial basis

  `make_lsq_spline`

    a similar factory function for spline fitting

  `UnivariateSpline`

    a wrapper over FITPACK spline fitting routines

  `splrep`

    a wrapper over FITPACK spline fitting routines

  Examples

  Use cubic interpolation on Chebyshev nodes:

      >>> def cheb_nodes(N):
      ...     jj = 2.*np.arange(N) + 1
      ...     x = np.cos(np.pi * jj / 2 / N)[::-1]
      ...     return x

      >>> x = cheb_nodes(20)
      >>> y = np.sqrt(1 - x**2)

      >>> from scipy.interpolate import BSpline, make_interp_spline
      >>> b = make_interp_spline(x, y)
      >>> np.allclose(b(x), y)
      True

  Note that the default is a cubic spline with a not-a-knot boundary condition

      >>> b.k
      3

  Here we use a ‘natural’ spline, with zero 2nd derivatives at edges:

      >>> l, r = [(2, 0.0)], [(2, 0.0)]
      >>> b_n = make_interp_spline(x, y, bc_type=(l, r))  # or, bc_type="natural"
      >>> np.allclose(b_n(x), y)
      True
      >>> x0, x1 = x[0], x[-1]
      >>> np.allclose([b_n(x0, 2), b_n(x1, 2)], [0, 0])
      True

  Interpolation of parametric curves is also supported. As an example, we compute a discretization of a snail curve in polar coordinates

      >>> phi = np.linspace(0, 2.*np.pi, 40)
      >>> r = 0.3 + np.cos(phi)
      >>> x, y = r*np.cos(phi), r*np.sin(phi)  # convert to Cartesian coordinates

  Build an interpolating curve, parameterizing it by the angle

      >>> from scipy.interpolate import make_interp_spline
      >>> spl = make_interp_spline(phi, np.c_[x, y])

  Evaluate the interpolant on a finer grid (note that we transpose the result to unpack it into a pair of x- and y-arrays)

      >>> phi_new = np.linspace(0, 2.*np.pi, 100)
      >>> x_new, y_new = spl(phi_new).T

  Plot the result

      >>> import matplotlib.pyplot as plt
      >>> plt.plot(x, y, 'o')
      >>> plt.plot(x_new, y_new, '-')
      >>> plt.show()

  Build a B-spline curve with 2 dimensional y

      >>> x = np.linspace(0, 2*np.pi, 10)
      >>> y = np.array([np.sin(x), np.cos(x)])

  Periodic condition is satisfied because y coordinates of points on the ends are equivalent

      >>> ax = plt.axes(projection='3d')
      >>> xx = np.linspace(0, 2*np.pi, 100)
      >>> bspl = make_interp_spline(x, y, k=5, bc_type='periodic', axis=1)
      >>> ax.plot3D(xx, *bspl(xx))
      >>> ax.scatter3D(x, *y, color='red')
      >>> plt.show()

scipy.interpolate.BSpline.construct_fast scipy.interpolate.make_lsq_spline