`scipy.interpolate._fitpack_py:sproot`

source: /scipy/interpolate/_fitpack_py.py :473

Signature

def sproot ( tck , mest = 10 )

Summary

Find the roots of a cubic B-spline.

Extended Summary

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.

Notes

Manipulating the tck-tuples directly is not recommended. In new code, prefer using the BSpline objects.

Array API Standard Support

sproot is not in-scope for support of Python Array API Standard compatible backends other than NumPy.

See dev-arrayapi for more information.

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)

✓

Converting to a PPoly object does find the roots at ``x=2``:

ppoly = PPoly.from_spline(tck)
ppoly.roots(extrapolate=False)

✓

Further examples are given :ref:`in the tutorial <tutorial-interpolate_splXXX>`.

Aliases

scipy.interpolate.sproot