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使用NXN矩阵计算最近的邻居

发布于 2025-01-30 19:09:28 字数 118 浏览 1 评论 0原文

假设我有nxn t随机矩阵，我如何找到矩阵中最近的邻居元素？

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木緿 2025-02-06 19:09:28

根据您的图像描述，我通常的方法是生成索引矩阵，计算每个元素和给定位置之间的欧几里得距离，然后判断距离是否为1：

>>> m = np.random.rand(5, 5)
>>> m
array([[0.15993681, 0.34212   , 0.27103587, 0.47880503, 0.92797347],
       [0.56698489, 0.16124083, 0.63037649, 0.79914445, 0.55247301],
       [0.05826773, 0.26793143, 0.62384919, 0.8248818 , 0.75127237],
       [0.38831836, 0.51175059, 0.70764241, 0.08915257, 0.63465847],
       [0.17459285, 0.74014131, 0.42850117, 0.54282051, 0.61795445]])
>>> ii, jj = np.indices(m.shape)
>>> i, j = 2, 3
>>> mask = np.hypot(ii - i, jj - j) == 1
>>> mask
array([[False, False, False, False, False],
       [False, False, False,  True, False],
       [False, False,  True, False,  True],
       [False, False, False,  True, False],
       [False, False, False, False, False]])
>>> m[mask]
array([0.79914445, 0.62384919, 0.75127237, 0.08915257])

此方法的优势是：

通常不慢，除非您的矩阵很大，但是仍然有解决方法。
您不必考虑某些边界条件。例如，目标位置位于矩阵的边缘。
它具有强大的扩展性。例如，如果您将邻居重新定义为距离小于或等于2的元素，或者在自身周围的一个圆圈中的一个元素，则只需进行一点修改即可达到期望。

对于大型矩阵，考虑到边界条件，您可以通过这样的功能实现它：

>>> def neighbor_indices(arr, pos, dist_func=np.hypot, dist=1):
...     start = [max(p - dist, 0) for p in pos]
...     shape = [s - p for s, p in zip(arr.shape, start)]
...     indices = np.indices(shape)
...     ii, jj = [i + s for i, s in zip(indices, start)]
...     i, j = pos
...     mask = dist_func(ii - i, jj - j) == dist
...     return ii[mask], jj[mask]
...
>>> neighbor_indices(m, (2, 3))
(array([1, 2, 2, 3]), array([3, 2, 4, 3]))
>>> neighbor_indices(m, (3, 4))
(array([2, 3, 4]), array([4, 3, 4]))
>>> neighbor_indices(m, (-1, -1))
(array([], dtype=int32), array([], dtype=int32))
>>> neighbor_indices(m, (5, 5))
(array([], dtype=int32), array([], dtype=int32))
>>> neighbor_indices(m, (2, 3), dist=2)
(array([0, 2, 4]), array([3, 1, 3]))
>>> neighbor_indices(m, (2, 3), lambda i, j: np.maximum(np.abs(i), np.abs(j)))
(array([1, 1, 1, 2, 2, 3, 3, 3]), array([2, 3, 4, 2, 4, 2, 3, 4]))

According to your image description, my usual way is to generate an index matrix, calculate the Euclidean distance between each element and a given position, and then judge whether the distance is 1:

>>> m = np.random.rand(5, 5)
>>> m
array([[0.15993681, 0.34212   , 0.27103587, 0.47880503, 0.92797347],
       [0.56698489, 0.16124083, 0.63037649, 0.79914445, 0.55247301],
       [0.05826773, 0.26793143, 0.62384919, 0.8248818 , 0.75127237],
       [0.38831836, 0.51175059, 0.70764241, 0.08915257, 0.63465847],
       [0.17459285, 0.74014131, 0.42850117, 0.54282051, 0.61795445]])
>>> ii, jj = np.indices(m.shape)
>>> i, j = 2, 3
>>> mask = np.hypot(ii - i, jj - j) == 1
>>> mask
array([[False, False, False, False, False],
       [False, False, False,  True, False],
       [False, False,  True, False,  True],
       [False, False, False,  True, False],
       [False, False, False, False, False]])
>>> m[mask]
array([0.79914445, 0.62384919, 0.75127237, 0.08915257])

The advantage of this method are:

it is usually not slow, unless your matrix is very large, but there are still workarounds.
you needn't to consider some boundary conditions. For example, the target position is at the edge of the matrix.
it has strong expansibility. For example, if you redefine the neighbor as an element with a distance of less than or equal to 2, or an element in a circle around itself, you can achieve your expectation with only a little modification.

For a large matrix, considering the boundary conditions, you can implement it through such a function:

>>> def neighbor_indices(arr, pos, dist_func=np.hypot, dist=1):
...     start = [max(p - dist, 0) for p in pos]
...     shape = [s - p for s, p in zip(arr.shape, start)]
...     indices = np.indices(shape)
...     ii, jj = [i + s for i, s in zip(indices, start)]
...     i, j = pos
...     mask = dist_func(ii - i, jj - j) == dist
...     return ii[mask], jj[mask]
...
>>> neighbor_indices(m, (2, 3))
(array([1, 2, 2, 3]), array([3, 2, 4, 3]))
>>> neighbor_indices(m, (3, 4))
(array([2, 3, 4]), array([4, 3, 4]))
>>> neighbor_indices(m, (-1, -1))
(array([], dtype=int32), array([], dtype=int32))
>>> neighbor_indices(m, (5, 5))
(array([], dtype=int32), array([], dtype=int32))
>>> neighbor_indices(m, (2, 3), dist=2)
(array([0, 2, 4]), array([3, 1, 3]))
>>> neighbor_indices(m, (2, 3), lambda i, j: np.maximum(np.abs(i), np.abs(j)))
(array([1, 1, 1, 2, 2, 3, 3, 3]), array([2, 3, 4, 2, 4, 2, 3, 4]))

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