python - Numpy:如何找到矩阵 A 中子矩阵的唯一局部最小值？

Question

给定一个 MxN (4x4) 维度的矩阵 A，如何找到每个 2x2 子矩阵的次优最小值？

A = array([[ 32673.    ,  15108.2   ,  26767.2   ,   9420.   ],
           [ 32944.2   ,  14604.01  ,  26757.01  ,   9127.2  ],
           [ 26551.2   ,  9257.01   ,  26595.01  ,   9309.2  ],
           [ 26624.    ,   8935.2   ,  26673.2   ,   8982.   ]])

一组子矩阵的次优最小值，是该子矩阵中与其他最小值的局部位置不冲突的最小值:

示例算法:

1. Find the minimum in A: 8935.2 global coords[3,1], local coords [1,1]
2. No other matrix has been evaluated so no conflict yet.
3. Find the next submatrix min: 8982. gc [3,3], lc [1,1]
4. Conflict exists, find next min in same submatrix: 9309.2 gc [2,3], lc [0,1]
5. Find next submatrix min: 9420 gc [0,3] lc[0,1]
6. Conflict exists, find next min: 26757.01 gc [1,2] lc [1,0]
7. Find next submatrix min: 14604 -- conflict with lc[1,1]
8. Find next submatrix min: 15108.2 -- conflict with lc [0,1]
9. Find next submatrix min: 32673. gc [0,0], lc [0,0]

我想尝试的一种方法是遵循上面的算法，但我没有再次详尽地搜索每个子矩阵，而是用“高”值 (>> max(A)) 全局更新每个子矩阵的局部位置，这是每次成功找到最小值时递增。

预期的输出将是一个列表:

[((0, 0), (0, 0), 32673), ((0, 1), (1, 0), 26757.01), ((1, 0), (1, 1), 8935.2), ((1, 1), (0, 1), 9309.2)]

形式为 [((t1), (t2), value) ... ]，其中 t1 是 A 中子矩阵的坐标，t2 是子矩阵中所选最小值的坐标。

编辑:子矩阵定义为 ZxZ，其中 MxN 模 ZxZ == 0，并且从 (0,0) 开始不重叠，并平铺以匹配 MxN 的维度。

编辑:下面是我构建的解决方案，但速度很慢。我怀疑如果我在每次迭代中从矩阵中删除子矩阵，那么性能可能会提高，但我不确定该怎么做。

    def get_mins(self, result):
    # result is the 2d array
    dim = 2  # 2x2 submatrix
    mins = []
    count = 0
    while count < dim**2:
        a, b = result.shape
        M4D = result.reshape(a//dim, dim, b//dim, dim)
        lidx = M4D.transpose(0, 2, 1, 3).reshape(-1, b//dim, dim**2).argmin(-1)
        r, c = numpy.unravel_index(lidx, [dim, dim])

        yy = M4D.min(axis=(1, 3))
        ww = numpy.dstack((r, c))

        super_min = numpy.unravel_index(numpy.argmin(yy), (dim, dim))

        rows = super_min[0]
        cols = super_min[1]

        # ww[rows,cols] g_ves us 2x2 position
        offset_r, offset_c = ww[rows, cols]
        # super_min gives us submatrix position

        mins.append((tuple(super_min), (offset_r, offset_c), yy.min()))

        if dim > 1:
            # update all other positions with inf >> max(result)
            result[numpy.ix_([offset_r + (d * dim) for d in range(dim)], [offset_c + (d * dim) for d in range(dim)])] = numpy.inf
            # update the submatrix to all == numpy.inf
            result[rows*dim:((rows*dim)+dim), cols*dim:((cols*dim)+dim)] = numpy.inf
        count += 1
    return mins

Answer 1

考虑到在选择全局最小值时迭代之间的依赖性，这是一种单循环方法 -

def unq_localmin(A, dim):
    m, n = A.shape
    M4D = A.reshape(m//dim, dim, n//dim, dim)
    M2Dr = M4D.swapaxes(1,2).reshape(-1,dim**2)
    a = M2Dr.copy()

    N = M2Dr.shape[0]
    R = np.empty(N,dtype=int)
    C = np.empty(N,dtype=int)
    shp = M2Dr.shape
    for i in range(N):
        r,c = np.unravel_index(np.argmin(a),shp)
        a[r] = np.inf
        a[:,c] = np.inf
        R[i], C[i] = r, c
    out = M2Dr[R,C]
    idr = np.column_stack(np.unravel_index(R,(dim,dim)))
    idc = np.column_stack(np.unravel_index(C,(dim,dim)))
    return zip(map(tuple,idr),map(tuple,idc),out)

让我们使用更大的随机 9x9 数组和 3x3 子矩阵/子数组来验证结果，以测试针对 OP 实现的多样性 get_mins -

In [66]: A   # Input data array
Out[66]: 
array([[ 927.,  852.,   18.,  949.,  933.,  558.,  519.,  118.,   82.],
       [ 939.,  782.,  178.,  987.,  534.,  981.,  879.,  895.,  407.],
       [ 968.,  187.,  539.,  986.,  506.,  499.,  529.,  978.,  567.],
       [ 767.,  272.,  881.,  858.,  621.,  301.,  675.,  151.,  670.],
       [ 874.,  221.,   72.,  210.,  273.,  823.,  784.,  289.,  425.],
       [ 621.,  510.,  303.,  935.,   88.,  970.,  278.,  125.,  669.],
       [ 702.,  722.,  620.,   51.,  845.,  414.,  154.,  154.,  635.],
       [ 600.,  928.,  540.,  462.,  772.,  487.,  196.,  499.,  208.],
       [ 654.,  335.,  258.,  297.,  649.,  712.,  292.,  767.,  819.]])

In [67]: unq_localmin(A, dim = 3) # Using proposed approach
Out[67]: 
[((0, 0), (0, 2), 18.0),
 ((2, 1), (0, 0), 51.0),
 ((1, 0), (1, 2), 72.0),
 ((1, 1), (2, 1), 88.0),
 ((0, 2), (0, 1), 118.0),
 ((2, 2), (1, 0), 196.0),
 ((2, 0), (2, 2), 258.0),
 ((1, 2), (2, 0), 278.0),
 ((0, 1), (1, 1), 534.0)]

In [68]: out = np.empty((9,9))

In [69]: get_mins(out,A) # Using OP's soln with dim = 3 edited
Out[69]: 
[((0, 0), (0, 2), 18.0),
 ((2, 1), (0, 0), 51.0),
 ((1, 0), (1, 2), 72.0),
 ((1, 1), (2, 1), 88.0),
 ((0, 2), (0, 1), 118.0),
 ((2, 2), (1, 0), 196.0),
 ((2, 0), (2, 2), 258.0),
 ((1, 2), (2, 0), 278.0),
 ((0, 1), (1, 1), 534.0)]

简化

上面的解决方案为我们提供了可用于构造索引元组的行和列索引，如使用 get_mins 打印的那样。如果您不需要这些，我们可以稍微简化提议的方法，就像这样 -

def unq_localmin_v2(A, dim):
    m, n = A.shape
    M4D = A.reshape(m//dim, dim, n//dim, dim)
    M2Dr = M4D.swapaxes(1,2).reshape(-1,dim**2)    
    N = M2Dr.shape[0]
    out = np.empty(N)
    shp = M2Dr.shape
    for i in range(N):
        r,c = np.unravel_index(np.argmin(M2Dr),shp)
        out[i] = M2Dr[r,c]
        M2Dr[r] = np.inf
        M2Dr[:,c] = np.inf        
    return out

运行时测试 -

In [52]: A = np.random.randint(11,999,(9,9)).astype(float)

In [53]: %timeit unq_localmin_v2(A, dim=3)
10000 loops, best of 3: 93.1 µs per loop

In [54]: out = np.empty((9,9))

In [55]: %timeit get_mins(out,A)
1000 loops, best of 3: 907 µs per loop