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bundles / scipy latest / scipy / interpolate / _bsplines / BSpline / design_matrix

classmethod

scipy.interpolate._bsplines:BSpline.design_matrix

source: /scipy/interpolate/_bsplines.py :374

Summary

Returns a design matrix as a CSR format sparse array.

Parameters

x : array_like, shape (n,)

Points to evaluate the spline at.

t : array_like, shape (nt,)

Sorted 1D array of knots.

k : int

B-spline degree.

extrapolate : bool or 'periodic', optional

Whether to extrapolate based on the first and last intervals or raise an error. If 'periodic', periodic extrapolation is used. Default is False.

Returns

design_matrix : `csr_array` object

Sparse matrix in CSR format where each row contains all the basis elements of the input row (first row = basis elements of x[0], ..., last row = basis elements x[-1]).

Notes

In each row of the design matrix all the basis elements are evaluated at the certain point (first row - x[0], ..., last row - x[-1]).

nt is a length of the vector of knots: as far as there are nt - k - 1 basis elements, nt should be not less than 2 * k + 2 to have at least k + 1 basis element.

Out of bounds x raises a ValueError.

Examples

Construct a design matrix for a B-spline 
from scipy.interpolate import make_interp_spline, BSpline
import numpy as np
x = np.linspace(0, np.pi * 2, 4)
y = np.sin(x)
k = 3
bspl = make_interp_spline(x, y, k=k)
design_matrix = bspl.design_matrix(x, bspl.t, k)
✓
design_matrix.toarray()
✗
Construct a design matrix for some vector of knots 
k = 2
t = [-1, 0, 1, 2, 3, 4, 5, 6]
x = [1, 2, 3, 4]
design_matrix = BSpline.design_matrix(x, t, k).toarray()
✓
design_matrix
✗
This result is equivalent to the one created in the sparse format 
c = np.eye(len(t) - k - 1)
design_matrix_gh = BSpline(t, c, k)(x)
np.allclose(design_matrix, design_matrix_gh, atol=1e-14)
✓

Aliases

  • scipy.interpolate.BSpline.design_matrix

Referenced by

This package