Parameters

t : ndarray, shape (nt + k + 1,)

sorted 1D array of knots

k : int

spline order

xval: float :

argument at which to evaluate the B-splines

m : int

index of the left edge of the evaluation interval, t[m] <= x < t[m+1]

nu : int, optional

Evaluate derivatives order :None:None:`nu`. Default is zero.

Returns

ndarray, shape (k+1,)

The values of B-splines $[B_{m-k}(xval), ..., B_{m}(xval)]$ if :None:None:`nu` is zero, otherwise the derivatives of order :None:None:`nu`.

Evaluate the k+1 B-splines which are non-zero on interval m .

Examples

A textbook use of this sort of routine is plotting the k+1 polynomial pieces which make up a B-spline of order k.

Consider a cubic spline

This example is valid syntax, but we were not able to check execution

>>> k = 3
... t = [0., 2., 2., 3., 4.]   # internal knots
... a, b = t[0], t[-1]    # base interval is [a, b)
... t = [a]*k + t + [b]*k  # add boundary knots

This example is valid syntax, but we were not able to check execution

>>> import matplotlib.pyplot as plt
... xx = np.linspace(a, b, 100)
... plt.plot(xx, BSpline.basis_element(t[k:-k])(xx),
...          'r-', lw=5, alpha=0.5)
... c = ['b', 'g', 'c', 'k']

Now we use slide an interval t[m]..t[m+1] along the base interval a..b and use evaluate_all_bspl to compute the restriction of the B-spline of interest to this interval:

This example is valid syntax, but we were not able to check execution

>>> for i in range(k+1):
...    x1, x2 = t[2*k - i], t[2*k - i + 1]
...    xx = np.linspace(x1 - 0.5, x2 + 0.5)
...    yy = [evaluate_all_bspl(t, k, x, 2*k - i)[i] for x in xx]
...    plt.plot(xx, yy, c[i] + '--', lw=3, label=str(i))
...

This example is valid syntax, but we were not able to check execution

>>> plt.grid(True)
... plt.legend()
... plt.show()

See :

Back References

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

scipy.interpolate._bspl.evaluate_all_bspl

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