Question

我有一个0和1的矩阵：

       1 0 0 0 0
       0 0 1 1 0
  a =  0 0 1 1 0
       1 0 0 0 0
       1 1 0 0 0

我希望使用python将其分解为连接组件的总和，其中“连接”是指矩阵，其中每个“1”至少具有上/下/左/右的另一个“1”。否则必须隔离“1”：

       1 0 0 0 0   1 0 0 0 0   0 0 0 0 0   0 0 0 0 0
       0 0 1 1 0   0 0 0 0 0   0 0 1 1 0   0 0 0 0 0
  a =  0 0 1 1 0 = 0 0 0 0 0 + 0 0 1 1 0 + 0 0 0 0 0
       1 0 0 0 0   0 0 0 0 0   0 0 0 0 0   1 0 0 0 0
       1 1 0 0 0   0 0 0 0 0   0 0 0 0 0   1 1 0 0 0

可能有趣的是，在这个问题（Largest rectangle of 1's in 2d binary matrix）中，Guy Adini建议使用BFS分解来分解连接组件中的矩阵。但是我找不到它的python实现，也没有找到如何使用BFS来解决我的问题。

Answer 1

有效的算法如下：

对于算法访问的元素，保留一个大小相同的visited矩阵（或者等效地，访问元素的一组坐标）。
您逐个浏览矩阵的所有元素：
- 如果未访问元素且为1，则将其标记为已访问，并以相同方式递归浏览其所有邻居。递归函数必须返回连接的函数集（带有这些的矩阵，带坐标的集合等）。

Answer 2

这是一个自定义实现。如果你愿意，我允许你修改它以删除重复项。

import itertools

class Matrix():
    def __init__(self, matrix):
        self.matrix = matrix

    def computeConnectedComponents(self):
        rows_id = list(range(len(self.matrix)))
        cols_id = list(range(len(self.matrix[0])))

        #here are the position of every 1 in the grid. ( row number, column number) indexes start at 0
        positions = [couple for couple in self.generate_pairs(rows_id, cols_id) if self.matrix[couple[0]][couple[1]]==1]

        #here we store all the connected component finded
        allConnectedComponents = []

        #while there is position not affected to any connected component
        while positions != [] :
            #the first element is taken as start of the connected component
            to_explore = [positions.pop(0)]
            currentConnectedComponent = set()
            #while there is node connected to a node connected to the component
            while list(to_explore) != []:
                currentNode = to_explore.pop()
                currentConnectedComponent.add(currentNode)

                to_explore += [coord for coord in self.node_neighbourhood(currentNode) if (self.matrix[coord[0]][coord[1]]==1 and (coord not in to_explore) and (coord not in currentConnectedComponent))]

                allConnectedComponents.append(currentConnectedComponent)
                positions = [position for position in positions if position not in currentConnectedComponent]

        return allConnectedComponents

    #http://stackoverflow.com/questions/16135833/generate-combinations-of-elements-from-multiple-lists
    def generate_pairs(self, *args):
        for i, l in enumerate(args, 1):
            for x, y in itertools.product(l, itertools.chain(*args[i:])):
                yield (x, y)

    def node_neighbourhood(self, coordinates):
        row, column = coordinates[0], coordinates[1]
        result = []
        if (row - 1) >= 0 :
            result.append((row-1, column))
        if (row + 1) < len(self.matrix):
            result.append((row+1, column))
        if (column - 1) >= 0:
            result.append((row, column-1))
        if (column + 1) < len(self.matrix[0]):
            result.append((row, column+1))
        return result


if __name__ == "__main__":
    data = [[1,0,0,0,0],
           [0,0,1,1,0],
           [0,0,1,1,0],
           [1,0,0,0,0],
           [1,1,0,0,0]]

    matrix = Matrix(data)
    for connectedComponent in matrix.computeConnectedComponents():
        print(connectedComponent)

如何将矩阵分解为连接组件的总和？

2 个答案: