Question

如何在python中使用给定的邻接矩阵打印图的所有边缘？例如，如果0与3和8相邻，则应打印： 0 3 0 8 没有重复我一直在使用Bfs，但是我不知道如何更新队列和当前元素。

到目前为止，这是我的代码

A =  [[0, 1, 0, 0, 0, 1], 
      [1, 0, 0, 0, 0, 1], 
      [0, 0, 0, 1, 1, 0], 
      [0, 0, 0, 0, 1, 0],
      [0, 0, 0, 0, 0, 0],
      [1, 0, 0, 0, 0, 0]]

def edges(A):
    visited = [False] * len(A)
    queue = []

    s = [0][0]
    queue.append(s)
    visited[s] = True

    while len(queue) > 0:
        s = queue.pop(0)
        print(s)
        for i in range(len(A)):
            print(i)

            for j in range(len(A[0])):
                if A[i][j] == 1 and visited[s]== False:

                    queue.append([i][j])

                    visited[s] = True

print(edges(A))

Answer 1

如果我理解正确，并且假设示例矩阵A是不对称的，则可以执行以下操作：

A =  [[0, 1, 0, 0, 0, 1],
      [1, 0, 0, 0, 0, 1],
      [0, 0, 0, 1, 1, 0],
      [0, 0, 0, 0, 1, 0],
      [0, 0, 0, 0, 0, 0],
      [1, 0, 0, 0, 0, 0]]

def edges(adj):

    for i, neighbors in enumerate(adj):
        for j, v in enumerate(neighbors):
            if v:
                yield (i, j)


for edge in edges(A):
    print(edge)

输出

(0, 1)
(0, 5)
(1, 0)
(1, 5)
(2, 3)
(2, 4)
(3, 4)
(5, 0)

Answer 2

不确定为什么要定义visited，就像double for循环一样，您要遍历所有元素一次。仍然与您正在执行的操作类似的简单方法是：

def edges(A):
    edges = []
    for i in range(len(A)):
        for j in range(len(A[0])):
            if (A[i][j] == 1):
                edges.append((i,j))
    return edges

print(edges(A))
[(0, 1), (0, 5), (1, 0), (1, 5), (2, 3), (2, 4), (3, 4), (5, 0)]

但是，您可以使用networkx轻松地做到这一点。您可以使用from_numpy_matrix从邻接矩阵创建图形，并使用edges打印带有边缘的列表：

A =  np.array([[0, 1, 0, 0, 0, 1], 
      [1, 0, 0, 0, 0, 1], 
      [0, 0, 0, 1, 1, 0], 
      [0, 0, 0, 0, 1, 0],
      [0, 0, 0, 0, 0, 0],
      [1, 0, 0, 0, 0, 0]])

import networkx as nx
g = nx.from_numpy_matrix(A)
print(g.edges)

[(0, 1), (0, 5), (1, 5), (2, 3), (2, 4), (3, 4)]

Answer 3

您可以将矩阵转换为邻接表，然后打印出节点和连接边：

A = [
    [0, 1, 0, 0, 0, 1],
    [1, 0, 0, 0, 0, 1],
    [0, 0, 0, 1, 1, 0],
    [0, 0, 0, 0, 1, 0],
    [0, 0, 0, 0, 0, 0],
    [1, 0, 0, 0, 0, 0],
]


def matrix_to_list(matrix):
    """Convert adjacency matrix to adjacency list"""
    graph = {}
    for i, node in enumerate(matrix):
        adj = []
        for j, connected in enumerate(node):
            if connected:
                adj.append(j)
        graph[i] = adj
    return graph


adjacency_list = matrix_to_list(A)
print(adjacency_list)
# {0: [1, 5], 1: [0, 5], 2: [3, 4], 3: [4], 4: [], 5: [0]}


connected_edges = [
    (node, edge) for node, edges in adjacency_list.items() for edge in edges
]
print(connected_edges)
# [(0, 1), (0, 5), (1, 0), (1, 5), (2, 3), (2, 4), (3, 4), (5, 0)]

在python的邻接矩阵中打印图的所有边缘

3 个答案: