求解scipy.ndimage.distance_transform_edt¶中返回ind最近距离的索引如何理解？索引值如何对应edt

 from scipy import ndimage
 a = np.array(([0,1,1,1,1],
                         [0,0,1,1,1],
                         [0,1,1,1,1],
                          [0,1,1,1,0],
                           [0,1,1,0,0]))
>>> ndimage.distance_transform_edt(a)
array([[ 0.    ,  1.    ,  1.4142,  2.2361,  3.    ],
             [ 0.    ,  0.    ,  1.    ,  2.    ,  2.    ],
             [ 0.    ,  1.    ,  1.4142,  1.4142,  1.    ],
             [ 0.    ,  1.    ,  1.4142,  1.    ,  0.    ],
             [ 0.    ,  1.    ,  1.    ,  0.    ,  0.    ]])

>>> edt, inds = ndimage.distance_transform_edt(a, return_indices=True)
>>> inds
array([[[0, 0, 1, 1, 3],
             [1, 1, 1, 1, 3],
            [2, 2, 1, 3, 3],
           [3, 3, 4, 4, 3],
           [4, 4, 4, 4, 4]],
         [[0, 0, 1, 1, 4],
        [0, 1, 1, 1, 4],
        [0, 0, 1, 4, 4],
        [0, 0, 3, 3, 4],
        [0, 0, 3, 3, 4]]])

参考链接：https://docs.scipy.org/doc/scipy-0.18.1/reference/generated/scipy.ndimage.distance_transform_edt.html