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scipy.interpolate.

# CubicSpline

class scipy.interpolate.CubicSpline(x, y, axis=0, bc_type='not-a-knot', extrapolate=None)

  Piecewise cubic interpolator to fit values (C2 smooth).

  Interpolate data with a piecewise cubic polynomial which is twice continuously differentiable [1]. The result is represented as a `PPoly` instance with breakpoints matching the given data.

  Parameters:
    x array_like, shape (n,)

      1-D array containing values of the independent variable. Values must be real, finite and in strictly increasing order.

    y array_like

      Array containing values of the dependent variable. It can have arbitrary number of dimensions, but the length along `axis` (see below) must match the length of `x`. Values must be finite.

    axis int, optional

      Axis along which y is assumed to be varying. Meaning that for `x[i]` the corresponding values are `np.take(y, i, axis=axis)`. Default is 0.

    bc_type string or 2-tuple, optional

      Boundary condition type. Two additional equations, given by the boundary conditions, are required to determine all coefficients of polynomials on each segment [2].

      If bc_type is a string, then the specified condition will be applied at both ends of a spline. Available conditions are:

      - ‘not-a-knot’ (default): The first and second segment at a curve end are the same polynomial. It is a good default when there is no information on boundary conditions.

      - ‘periodic’: The interpolated functions is assumed to be periodic of period `x[-1] - x[0]`. The first and last value of y must be identical: `y[0] == y[-1]`. This boundary condition will result in `y'[0] == y'[-1]` and `y''[0] == y''[-1]`.

      - ‘clamped’: The first derivative at curves ends are zero. Assuming a 1D y, `bc_type=((1, 0.0), (1, 0.0))` is the same condition.

      - ‘natural’: The second derivative at curve ends are zero. Assuming a 1D y, `bc_type=((2, 0.0), (2, 0.0))` is the same condition.

      If bc_type is a 2-tuple, the first and the second value will be applied at the curve start and end respectively. The tuple values can be one of the previously mentioned strings (except ‘periodic’) or a tuple `(order, deriv_values)` allowing to specify arbitrary derivatives at curve ends:

      - order: the derivative order, 1 or 2.

      - deriv_value: array_like containing derivative values, shape must be the same as y, excluding `axis` dimension. For example, if y is 1-D, then deriv_value must be a scalar. If y is 3-D with the shape (n0, n1, n2) and axis=2, then deriv_value must be 2-D and have the shape (n0, n1).

    extrapolate{bool, ‘periodic’, None}, optional

      If bool, determines whether to extrapolate to out-of-bounds points based on first and last intervals, or to return NaNs. If ‘periodic’, periodic extrapolation is used. If None (default), `extrapolate` is set to ‘periodic’ for `bc_type='periodic'` and to True otherwise.

  Attributes:
    x ndarray, shape (n,)

      Breakpoints. The same `x` which was passed to the constructor.

    c ndarray, shape (4, n-1, …)

      Coefficients of the polynomials on each segment. The trailing dimensions match the dimensions of y, excluding `axis`. For example, if y is 1-d, then `c[k, i]` is a coefficient for `(x-x[i])**(3-k)` on the segment between `x[i]` and `x[i+1]`.

    axis int

      Interpolation axis. The same axis which was passed to the constructor.

  Methods

  `__call__`(x[, nu, extrapolate]) | Evaluate the piecewise polynomial or its derivative.
  `derivative`([nu]) | Construct a new piecewise polynomial representing the derivative.
  `antiderivative`([nu]) | Construct a new piecewise polynomial representing the antiderivative.
  `integrate`(a, b[, extrapolate]) | Compute a definite integral over a piecewise polynomial.
  `solve`([y, discontinuity, extrapolate]) | Find real solutions of the equation `pp(x) == y`.
  `roots`([discontinuity, extrapolate]) | Find real roots of the piecewise polynomial.

  See also

  `Akima1DInterpolator`

    Akima 1D interpolator.

  `PchipInterpolator`

    PCHIP 1-D monotonic cubic interpolator.

  `PPoly`

    Piecewise polynomial in terms of coefficients and breakpoints.

  Notes

  Parameters bc_type and `extrapolate` work independently, i.e. the former controls only construction of a spline, and the latter only evaluation.

  When a boundary condition is ‘not-a-knot’ and n = 2, it is replaced by a condition that the first derivative is equal to the linear interpolant slope. When both boundary conditions are ‘not-a-knot’ and n = 3, the solution is sought as a parabola passing through given points.

  When ‘not-a-knot’ boundary conditions is applied to both ends, the resulting spline will be the same as returned by `splrep` (with `s=0`) and `InterpolatedUnivariateSpline`, but these two methods use a representation in B-spline basis.

  Added in version 0.18.0.

  References

  [1]

  Cubic Spline Interpolation on Wikiversity.

  [2]

  Carl de Boor, “A Practical Guide to Splines”, Springer-Verlag, 1978.

  Examples

  In this example the cubic spline is used to interpolate a sampled sinusoid. You can see that the spline continuity property holds for the first and second derivatives and violates only for the third derivative.

      >>> import numpy as np
      >>> from scipy.interpolate import CubicSpline
      >>> import matplotlib.pyplot as plt
      >>> x = np.arange(10)
      >>> y = np.sin(x)
      >>> cs = CubicSpline(x, y)
      >>> xs = np.arange(-0.5, 9.6, 0.1)
      >>> fig, ax = plt.subplots(figsize=(6.5, 4))
      >>> ax.plot(x, y, 'o', label='data')
      >>> ax.plot(xs, np.sin(xs), label='true')
      >>> ax.plot(xs, cs(xs), label="S")
      >>> ax.plot(xs, cs(xs, 1), label="S'")
      >>> ax.plot(xs, cs(xs, 2), label="S''")
      >>> ax.plot(xs, cs(xs, 3), label="S'''")
      >>> ax.set_xlim(-0.5, 9.5)
      >>> ax.legend(loc='lower left', ncol=2)
      >>> plt.show()

  In the second example, the unit circle is interpolated with a spline. A periodic boundary condition is used. You can see that the first derivative values, ds/dx=0, ds/dy=1 at the periodic point (1, 0) are correctly computed. Note that a circle cannot be exactly represented by a cubic spline. To increase precision, more breakpoints would be required.

      >>> theta = 2 * np.pi * np.linspace(0, 1, 5)
      >>> y = np.c_[np.cos(theta), np.sin(theta)]
      >>> cs = CubicSpline(theta, y, bc_type='periodic')
      >>> print("ds/dx={:.1f} ds/dy={:.1f}".format(cs(0, 1)[0], cs(0, 1)[1]))
      ds/dx=0.0 ds/dy=1.0
      >>> xs = 2 * np.pi * np.linspace(0, 1, 100)
      >>> fig, ax = plt.subplots(figsize=(6.5, 4))
      >>> ax.plot(y[:, 0], y[:, 1], 'o', label='data')
      >>> ax.plot(np.cos(xs), np.sin(xs), label='true')
      >>> ax.plot(cs(xs)[:, 0], cs(xs)[:, 1], label='spline')
      >>> ax.axes.set_aspect('equal')
      >>> ax.legend(loc='center')
      >>> plt.show()

  The third example is the interpolation of a polynomial y = x**3 on the interval 0 <= x<= 1. A cubic spline can represent this function exactly. To achieve that we need to specify values and first derivatives at endpoints of the interval. Note that y’ = 3 * x**2 and thus y’(0) = 0 and y’(1) = 3.

      >>> cs = CubicSpline([0, 1], [0, 1], bc_type=((1, 0), (1, 3)))
      >>> x = np.linspace(0, 1)
      >>> np.allclose(x**3, cs(x))
      True