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# scipy.interpolate.make_lsq_spline

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

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

  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

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

  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.

  See also

  `BSpline`
    base class representing the B-spline objects
  `make_interp_spline`
    a similar factory function for interpolating splines
  `LSQUnivariateSpline`
    a FITPACK-based spline fitting routine
  `splrep`
    a FITPACK-based fitting routine

  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`.

  Examples

  Generate some noisy data:

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

  (Source code)

  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.)