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])

    ``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]))

    r$