用python在邻接矩阵中打印图形的所有边
如何在python中打印具有给定邻接矩阵的图的所有边?例如,如果0与3和8相邻,则应打印: 0 3 0 8 不重复 我一直在使用Bfs,但我不知道如何更新队列和当前元素 这是到目前为止我的代码用python在邻接矩阵中打印图形的所有边,python,networkx,graph-theory,breadth-first-search,Python,Networkx,Graph Theory,Breadth First Search,如何在python中打印具有给定邻接矩阵的图的所有边?例如,如果0与3和8相邻,则应打印: 0 3 0 8 不重复 我一直在使用Bfs,但我不知道如何更新队列和当前元素 这是到目前为止我的代码 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
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))
如果我理解正确,并且假设您的示例矩阵
A
是不对称的,您可以:
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)
一种简单的方法是迭代邻接矩阵,并在存在连接的地方构建一个具有索引的元组列表:
[(i,j) for i,l in enumerate(A) for j,v in enumerate(l) if v]
# [(0, 1), (0, 5), (1, 0), (1, 5), (2, 3), (2, 4), (3, 4), (5, 0)]
但是,您可以使用networkx
轻松实现这一点。您可以使用从邻接矩阵创建图形,并使用以下命令打印带有边的列表:
可以将矩阵转换为邻接列表,然后打印节点和连接边:
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)]
你看过这个库了吗?是的,它是正确的,但是有没有办法用普通的bfs技术打印相同的输出。我是说使用队列和数组?感谢BTW感谢您的回复,但是如果不使用networkx,您又如何做到这一点呢
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)]