Module « scipy.linalg »

Fonction lstsq - module scipy.linalg

Signature de la fonction lstsq

def lstsq(a, b, cond=None, overwrite_a=False, overwrite_b=False, check_finite=True, lapack_driver=None) 

Description

lstsq.__doc__

    Compute least-squares solution to equation Ax = b.

    Compute a vector x such that the 2-norm ``|b - A x|`` is minimized.

    Parameters
    ----------
    a : (M, N) array_like
        Left-hand side array
    b : (M,) or (M, K) array_like
        Right hand side array
    cond : float, optional
        Cutoff for 'small' singular values; used to determine effective
        rank of a. Singular values smaller than
        ``rcond * largest_singular_value`` are considered zero.
    overwrite_a : bool, optional
        Discard data in `a` (may enhance performance). Default is False.
    overwrite_b : bool, optional
        Discard data in `b` (may enhance performance). Default is False.
    check_finite : bool, optional
        Whether to check that the input matrices 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.
    lapack_driver : str, optional
        Which LAPACK driver is used to solve the least-squares problem.
        Options are ``'gelsd'``, ``'gelsy'``, ``'gelss'``. Default
        (``'gelsd'``) is a good choice.  However, ``'gelsy'`` can be slightly
        faster on many problems.  ``'gelss'`` was used historically.  It is
        generally slow but uses less memory.

        .. versionadded:: 0.17.0

    Returns
    -------
    x : (N,) or (N, K) ndarray
        Least-squares solution.  Return shape matches shape of `b`.
    residues : (K,) ndarray or float
        Square of the 2-norm for each column in ``b - a x``, if ``M > N`` and
        ``ndim(A) == n`` (returns a scalar if b is 1-D). Otherwise a
        (0,)-shaped array is returned.
    rank : int
        Effective rank of `a`.
    s : (min(M, N),) ndarray or None
        Singular values of `a`. The condition number of a is
        ``abs(s[0] / s[-1])``.

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

    ValueError
        When parameters are not compatible.

    See Also
    --------
    scipy.optimize.nnls : linear least squares with non-negativity constraint

    Notes
    -----
    When ``'gelsy'`` is used as a driver, `residues` is set to a (0,)-shaped
    array and `s` is always ``None``.

    Examples
    --------
    >>> from scipy.linalg import lstsq
    >>> import matplotlib.pyplot as plt

    Suppose we have the following data:

    >>> x = np.array([1, 2.5, 3.5, 4, 5, 7, 8.5])
    >>> y = np.array([0.3, 1.1, 1.5, 2.0, 3.2, 6.6, 8.6])

    We want to fit a quadratic polynomial of the form ``y = a + b*x**2``
    to this data.  We first form the "design matrix" M, with a constant
    column of 1s and a column containing ``x**2``:

    >>> M = x[:, np.newaxis]**[0, 2]
    >>> M
    array([[  1.  ,   1.  ],
           [  1.  ,   6.25],
           [  1.  ,  12.25],
           [  1.  ,  16.  ],
           [  1.  ,  25.  ],
           [  1.  ,  49.  ],
           [  1.  ,  72.25]])

    We want to find the least-squares solution to ``M.dot(p) = y``,
    where ``p`` is a vector with length 2 that holds the parameters
    ``a`` and ``b``.

    >>> p, res, rnk, s = lstsq(M, y)
    >>> p
    array([ 0.20925829,  0.12013861])

    Plot the data and the fitted curve.

    >>> plt.plot(x, y, 'o', label='data')
    >>> xx = np.linspace(0, 9, 101)
    >>> yy = p[0] + p[1]*xx**2
    >>> plt.plot(xx, yy, label='least squares fit, $y = a + bx^2$')
    >>> plt.xlabel('x')
    >>> plt.ylabel('y')
    >>> plt.legend(framealpha=1, shadow=True)
    >>> plt.grid(alpha=0.25)
    >>> plt.show()