# scipy.interpolate.BSpline.design_matrix

classmethod BSpline.design_matrix(x, t, k, extrapolate=False)

  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.

      New in version 1.10.0.

  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

  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 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()
      [[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