`.
    rcond : float, optional
        Cut-off ratio for small singular values of `a`.
        For the purposes of rank determination, singular values are treated
        as zero if they are smaller than `rcond` times the largest singular
        value of `a`.

        .. versionchanged:: 1.14.0
           If not set, a FutureWarning is given. The previous default
           of ``-1`` will use the machine precision as `rcond` parameter,
           the new default will use the machine precision times `max(M, N)`.
           To silence the warning and use the new default, use ``rcond=None``,
           to keep using the old behavior, use ``rcond=-1``.

    Returns
    -------
    x : {(N,), (N, K)} ndarray
        Least-squares solution. If `b` is two-dimensional,
        the solutions are in the `K` columns of `x`.
    residuals : {(1,), (K,), (0,)} ndarray
        Sums of squared residuals: Squared Euclidean 2-norm for each column in
        ``b - a @ x``.
        If the rank of `a` is < N or M <= N, this is an empty array.
        If `b` is 1-dimensional, this is a (1,) shape array.
        Otherwise the shape is (K,).
    rank : int
        Rank of matrix `a`.
    s : (min(M, N),) ndarray
        Singular values of `a`.

    Raises
    ------
    LinAlgError
        If computation does not converge.

    See Also
    --------
    scipy.linalg.lstsq : Similar function in SciPy.

    Notes
    -----
    If `b` is a matrix, then all array results are returned as matrices.

    Examples
    --------
    Fit a line, ``y = mx + c``, through some noisy data-points:

    >>> x = np.array([0, 1, 2, 3])
    >>> y = np.array([-1, 0.2, 0.9, 2.1])

    By examining the coefficients, we see that the line should have a
    gradient of roughly 1 and cut the y-axis at, more or less, -1.

    We can rewrite the line equation as ``y = Ap``, where ``A = [[x 1]]``
    and ``p = [[m], [c]]``.  Now use `lstsq` to solve for `p`:

    >>> A = np.vstack([x, np.ones(len(x))]).T
    >>> A
    array([[ 0.,  1.],
           [ 1.,  1.],
           [ 2.,  1.],
           [ 3.,  1.]])

    >>> m, c = np.linalg.lstsq(A, y, rcond=None)[0]
    >>> m, c
    (1.0 -0.95) # may vary

    Plot the data along with the fitted line:

    >>> import matplotlib.pyplot as plt
    >>> _ = plt.plot(x, y, 'o', label='Original data', markersize=10)
    >>> _ = plt.plot(x, m*x + c, 'r', label='Fitted line')
    >>> _ = plt.legend()
    >>> plt.show()

    rp