scipy.interpolate.UnivariateSpline.roots#
- UnivariateSpline.roots()[source]#
Return the zeros of the spline.
See also
Notes
Restriction: only cubic splines are supported by FITPACK. For non-cubic splines, use PPoly.root (see below for an example).
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 UnivariateSpline >>> spl = UnivariateSpline(x, y, s=0) >>> spl.roots() array([], dtype=float64)
Converting to a PPoly object does find the roots at x=2:
>>> from scipy.interpolate import splrep, PPoly >>> tck = splrep(x, y, s=0) >>> ppoly = PPoly.from_spline(tck) >>> ppoly.roots(extrapolate=False) array([2.])