scipy.interpolate.BSpline.design_matrix#
- classmethod BSpline.design_matrix(x, t, k)[source]#
Returns a design matrix as a CSR format sparse array.
- Parameters
- xarray_like, shape (n,)
Points to evaluate the spline at.
- tarray_like, shape (nt,)
Sorted 1D array of knots.
- kint
B-spline degree.
- Returns
- design_matrixcsr_array object
Sparse matrix in CSR format where in each row all the basis elements are evaluated at the certain point (first row - x[0], …, last row - x[-1]).
Notes
New in version 1.8.0.
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 lenght 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 >>> 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() [[1. , 0. , 0. , 0. ], [0.2962963 , 0.44444444, 0.22222222, 0.03703704], [0.03703704, 0.22222222, 0.44444444, 0.2962963 ], [0. , 0. , 0. , 1. ]]
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 [[0.5, 0.5, 0. , 0. , 0. ], [0. , 0.5, 0.5, 0. , 0. ], [0. , 0. , 0.5, 0.5, 0. ], [0. , 0. , 0. , 0.5, 0.5]]
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) True