Пример #1
0
def pinv(a, rcond=1e-15):
    """
    Compute the (Moore-Penrose) pseudo-inverse of a matrix.

    Calculate the generalized inverse of a matrix using its
    singular-value decomposition (SVD) and including all
    `large` singular values.

    Parameters
    ----------
    a : array_like (M, N)
      Matrix to be pseudo-inverted.
    rcond : float
      Cutoff for `small` singular values.
      Singular values smaller than rcond*largest_singular_value are
      considered zero.

    Returns
    -------
    B : ndarray (N, M)
      The pseudo-inverse of `a`. If `a` is an np.matrix instance, then so
      is `B`.


    Raises
    ------
    LinAlgError
      In case SVD computation does not converge.

    Examples
    --------
    >>> a = np.random.randn(9, 6)
    >>> B = np.linalg.pinv(a)
    >>> np.allclose(a, np.dot(a, np.dot(B, a)))
    True
    >>> np.allclose(B, np.dot(B, np.dot(a, B)))
    True

    """
    a, wrap = _makearray(a)
    _assertNonEmpty(a)
    a = a.conjugate()
    u, s, vt = svd(a, 0)
    m = u.shape[0]
    n = vt.shape[1]
    cutoff = rcond * maximum.reduce(s)
    for i in range(min(n, m)):
        if s[i] > cutoff:
            s[i] = 1. / s[i]
        else:
            s[i] = 0.
    res = dot(transpose(vt), multiply(s[:, newaxis], transpose(u)))
    return wrap(res)
Пример #2
0
def pinv(a, rcond=1e-15 ):
    """
    Compute the (Moore-Penrose) pseudo-inverse of a matrix.

    Calculate the generalized inverse of a matrix using its
    singular-value decomposition (SVD) and including all
    `large` singular values.

    Parameters
    ----------
    a : array_like (M, N)
      Matrix to be pseudo-inverted.
    rcond : float
      Cutoff for `small` singular values.
      Singular values smaller than rcond*largest_singular_value are
      considered zero.

    Returns
    -------
    B : ndarray (N, M)
      The pseudo-inverse of `a`. If `a` is an np.matrix instance, then so
      is `B`.


    Raises
    ------
    LinAlgError
      In case SVD computation does not converge.

    Examples
    --------
    >>> a = np.random.randn(9, 6)
    >>> B = np.linalg.pinv(a)
    >>> np.allclose(a, np.dot(a, np.dot(B, a)))
    True
    >>> np.allclose(B, np.dot(B, np.dot(a, B)))
    True

    """
    a, wrap = _makearray(a)
    _assertNonEmpty(a)
    a = a.conjugate()
    u, s, vt = svd(a, 0)
    m = u.shape[0]
    n = vt.shape[1]
    cutoff = rcond*maximum.reduce(s)
    for i in range(min(n, m)):
        if s[i] > cutoff:
            s[i] = 1./s[i]
        else:
            s[i] = 0.;
    res = dot(transpose(vt), multiply(s[:, newaxis],transpose(u)))
    return wrap(res)
Пример #3
0
def pinv(a, rcond=1e-15 ):
    """Compute the (Moore-Penrose) pseudo-inverse of a matrix.

    Calculate a generalized inverse of a matrix using its
    singular-value decomposition and including all 'large' singular
    values.

    Parameters
    ----------
    a : array-like, shape (M, N)
        Matrix to be pseudo-inverted
    rcond : float
        Cutoff for 'small' singular values.
        Singular values smaller than rcond*largest_singular_value are
        considered zero.

    Returns
    -------
    B : array, shape (N, M)
        If a is a matrix, then so is B.

    Raises LinAlgError if SVD computation does not converge

    Examples
    --------
    >>> from numpy import *
    >>> a = random.randn(9, 6)
    >>> B = linalg.pinv(a)
    >>> allclose(a, dot(a, dot(B, a)))
    True
    >>> allclose(B, dot(B, dot(a, B)))
    True

    """
    a, wrap = _makearray(a)
    _assertNonEmpty(a)
    a = a.conjugate()
    u, s, vt = svd(a, 0)
    m = u.shape[0]
    n = vt.shape[1]
    cutoff = rcond*maximum.reduce(s)
    for i in range(min(n, m)):
        if s[i] > cutoff:
            s[i] = 1./s[i]
        else:
            s[i] = 0.;
    res = dot(transpose(vt), multiply(s[:, newaxis],transpose(u)))
    return wrap(res)
Пример #4
0
def pinv(a, rcond=1e-15):
    """Compute the (Moore-Penrose) pseudo-inverse of a matrix.

    Calculate a generalized inverse of a matrix using its
    singular-value decomposition and including all 'large' singular
    values.

    Parameters
    ----------
    a : array-like, shape (M, N)
        Matrix to be pseudo-inverted
    rcond : float
        Cutoff for 'small' singular values.
        Singular values smaller than rcond*largest_singular_value are
        considered zero.

    Returns
    -------
    B : array, shape (N, M)
        If a is a matrix, then so is B.

    Raises LinAlgError if SVD computation does not converge

    Examples
    --------
    >>> from numpy import *
    >>> a = random.randn(9, 6)
    >>> B = linalg.pinv(a)
    >>> allclose(a, dot(a, dot(B, a)))
    True
    >>> allclose(B, dot(B, dot(a, B)))
    True

    """
    a, wrap = _makearray(a)
    _assertNonEmpty(a)
    a = a.conjugate()
    u, s, vt = svd(a, 0)
    m = u.shape[0]
    n = vt.shape[1]
    cutoff = rcond * maximum.reduce(s)
    for i in range(min(n, m)):
        if s[i] > cutoff:
            s[i] = 1. / s[i]
        else:
            s[i] = 0.
    res = dot(transpose(vt), multiply(s[:, newaxis], transpose(u)))
    return wrap(res)
Пример #5
0
def pinv(a, rcond=1e-15):
    """Return the (Moore-Penrose) pseudo-inverse of a 2-d array

    This method computes the generalized inverse using the
    singular-value decomposition and all singular values larger than
    rcond of the largest.
    """
    a, wrap = _makearray(a)
    _assertNonEmpty(a)
    a = a.conjugate()
    u, s, vt = svd(a, 0)
    m = u.shape[0]
    n = vt.shape[1]
    cutoff = rcond * maximum.reduce(s)
    for i in range(min(n, m)):
        if s[i] > cutoff:
            s[i] = 1. / s[i]
        else:
            s[i] = 0.
    return wrap(dot(transpose(vt), multiply(s[:, newaxis], transpose(u))))
Пример #6
0
def pinv(a, rcond=1e-15 ):
    """Return the (Moore-Penrose) pseudo-inverse of a 2-d array

    This method computes the generalized inverse using the
    singular-value decomposition and all singular values larger than
    rcond of the largest.
    """
    a, wrap = _makearray(a)
    a = a.conjugate()
    u, s, vt = svd(a, 0)
    m = u.shape[0]
    n = vt.shape[1]
    cutoff = rcond*maximum.reduce(s)
    for i in range(min(n, m)):
        if s[i] > cutoff:
            s[i] = 1./s[i]
        else:
            s[i] = 0.;
    return wrap(dot(transpose(vt),
                       multiply(s[:, newaxis],transpose(u))))
Пример #7
0
def _rc(s, tolerance):
    cutoff = tolerance * maximum.reduce(s)
    return cutoff