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make_lsq_spline(x, y, t, k=3, w=None, axis=0, check_finite=True)

The result is a linear combination

$$S(x) = \sum_j c_j B_j(x; t)$$

of the B-spline basis elements, $B_j(x; t)$ , which minimizes

$$\sum_{j} \left( w_j \times (S(x_j) - y_j) \right)^2$$

Notes

The number of data points must be larger than the spline degree k.

Knots t must satisfy the Schoenberg-Whitney conditions, i.e., there must be a subset of data points x[j] such that t[j] < x[j] < t[j+k+1] , for j=0, 1,...,n-k-2 .

Parameters

x : array_like, shape (m,): Abscissas.
y : array_like, shape (m, ...): Ordinates.
t : array_like, shape (n + k + 1,).: Knots. Knots and data points must satisfy Schoenberg-Whitney conditions.
k : int, optional: B-spline degree. Default is cubic, k=3.
w : array_like, shape (n,), optional: Weights for spline fitting. Must be positive. If None , then weights are all equal. Default is None .
axis : int, optional: Interpolation axis. Default is zero.
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` with knots `t`.

Compute the (coefficients of) an LSQ B-spline.

Examples

Generate some noisy data:

>>> rng = np.random.default_rng()
... x = np.linspace(-3, 3, 50)
... y = np.exp(-x**2) + 0.1 * rng.standard_normal(50)

Now fit a smoothing cubic spline with a pre-defined 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 with nan . A workaround is to use zero weights for not-a-number 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.)

See :

Back References

The following pages refer to to this document either explicitly or contain code examples using this.

scipy.interpolate._bsplines.make_lsq_spline scipy.interpolate._bsplines.make_interp_spline

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