scipy.interpolate.make_lsq_spline¶

scipy.interpolate.
make_lsq_spline
(x, y, t, k=3, w=None, axis=0, check_finite=True)[source]¶ Compute the (coefficients of) an LSQ Bspline.
The result is a linear combination
\[S(x) = \sum_j c_j B_j(x; t)\]of the Bspline basis elements, \(B_j(x; t)\), which minimizes
\[\sum_{j} \left( w_j \times (S(x_j)  y_j) \right)^2\] Parameters
 xarray_like, shape (m,)
Abscissas.
 yarray_like, shape (m, …)
Ordinates.
 tarray_like, shape (n + k + 1,).
Knots. Knots and data points must satisfy SchoenbergWhitney conditions.
 kint, optional
Bspline degree. Default is cubic, k=3.
 warray_like, shape (n,), optional
Weights for spline fitting. Must be positive. If
None
, then weights are all equal. Default isNone
. axisint, optional
Interpolation axis. Default is zero.
 check_finitebool, optional
Whether to check that the input arrays contain only finite numbers. Disabling may give a performance gain, but may result in problems (crashes, nontermination) if the inputs do contain infinities or NaNs. Default is True.
 Returns
 ba BSpline object of the degree k with knots t.
See also
BSpline
base class representing the Bspline objects
make_interp_spline
a similar factory function for interpolating splines
LSQUnivariateSpline
a FITPACKbased spline fitting routine
splrep
a FITPACKbased fitting routine
Notes
The number of data points must be larger than the spline degree k.
Knots t must satisfy the SchoenbergWhitney conditions, i.e., there must be a subset of data points
x[j]
such thatt[j] < x[j] < t[j+k+1]
, forj=0, 1,...,nk2
.Examples
Generate some noisy data:
>>> x = np.linspace(3, 3, 50) >>> y = np.exp(x**2) + 0.1 * np.random.randn(50)
Now fit a smoothing cubic spline with a predefined internal knots. Here we make the knot vector (k+1)regular by adding boundary knots:
>>> from scipy.interpolate import make_lsq_spline, BSpline >>> t = [1, 0, 1] >>> k = 3 >>> t = np.r_[(x[0],)*(k+1), ... t, ... (x[1],)*(k+1)] >>> spl = make_lsq_spline(x, y, t, k)
For comparison, we also construct an interpolating spline for the same set of data:
>>> from scipy.interpolate import make_interp_spline >>> spl_i = make_interp_spline(x, y)
Plot both:
>>> import matplotlib.pyplot as plt >>> xs = np.linspace(3, 3, 100) >>> plt.plot(x, y, 'ro', ms=5) >>> plt.plot(xs, spl(xs), 'g', lw=3, label='LSQ spline') >>> plt.plot(xs, spl_i(xs), 'b', lw=3, alpha=0.7, label='interp spline') >>> plt.legend(loc='best') >>> plt.show()
NaN handling: If the input arrays contain
nan
values, the result is not useful since the underlying spline fitting routines cannot deal withnan
. A workaround is to use zero weights for notanumber data points:>>> y[8] = np.nan >>> w = np.isnan(y) >>> y[w] = 0. >>> tck = make_lsq_spline(x, y, t, w=~w)
Notice the need to replace a
nan
by a numerical value (precise value does not matter as long as the corresponding weight is zero.)