python numpy nonzero 函数(方法)介绍及使用

python numpy nonzero 函数(方法)介绍及使用

    Return the indices of the elements that are non-zero.

    Returns a tuple of arrays, one for each dimension of `a`,
    containing the indices of the non-zero elements in that
    dimension. The values in `a` are always tested and returned in
    row-major, C-style order.

    To group the indices by element, rather than dimension, use `argwhere`,
    which returns a row for each non-zero element.

    .. note::

       When called on a zero-d array or scalar, `</code>nonzero(a) is treated
       as `</code>nonzero(atleast_1d(a)).

       .. deprecated:: 1.17.0

          Use `atleast_1d` explicitly if this behavior is deliberate.

    Parameters
    ----------
    a : array_like
        Input array.

    Returns
    -------
    tuple_of_arrays : tuple
        Indices of elements that are non-zero.

    See Also
    --------
    flatnonzero :
        Return indices that are non-zero in the flattened version of the input
        array.
    ndarray.nonzero :
        Equivalent ndarray method.
    count_nonzero :
        Counts the number of non-zero elements in the input array.

    Notes
    -----
    While the nonzero values can be obtained with `</code>a[nonzero(a)], it is
    recommended to use `</code>x[x.astype(bool)] or `</code>x[x != 0] instead, which
    will correctly handle 0-d arrays.

    Examples
    --------
    >>> x = np.array([[3, 0, 0], [0, 4, 0], [5, 6, 0]])
    >>> x
    array([[3, 0, 0],
           [0, 4, 0],
           [5, 6, 0]])
    >>> np.nonzero(x)
    (array([0, 1, 2, 2]), array([0, 1, 0, 1]))

    >>> x[np.nonzero(x)]
    array([3, 4, 5, 6])
    >>> np.transpose(np.nonzero(x))
    array([[0, 0],
           [1, 1],
           [2, 0],
           [2, 1]])

    A common use for `</code>nonzero is to find the indices of an array, where
    a condition is True.  Given an array `a`, the condition `a` > 3 is a
    boolean array and since False is interpreted as 0, np.nonzero(a > 3)
    yields the indices of the `a` where the condition is true.

    >>> a = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
    >>> a > 3
    array([[False, False, False],
           [ True,  True,  True],
           [ True,  True,  True]])
    >>> np.nonzero(a > 3)
    (array([1, 1, 1, 2, 2, 2]), array([0, 1, 2, 0, 1, 2]))

    Using this result to index `a` is equivalent to using the mask directly:

    >>> a[np.nonzero(a > 3)]
    array([4, 5, 6, 7, 8, 9])
    >>> a[a > 3]  # prefer this spelling
    array([4, 5, 6, 7, 8, 9])

    `</code>nonzero can also be called as a method of the array.

    >>> (a > 3).nonzero()
    (array([1, 1, 1, 2, 2, 2]), array([0, 1, 2, 0, 1, 2]))

    ```````