Question

我正在尝试通过一个玩具示例，从列表中构建邻接矩阵，但现在我还不太清楚。我在考虑.loc（）的问题，但不确定如何正确编制索引。

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我开始使用以下方法构建矩阵：

{'nodes':['A', 'B', 'C', 'D', 'E'],
 'edges':[('A', 'B'), ('A', 'D'), ('B', 'C'), ('B', 'E'), ('C', 'D'), 
                      ('D', 'E'), ('E', 'A'),('E', 'B'), ('E', 'C')]}

但是现在我不确定如何填写它。我认为这里面很简单，也许有列表理解能力？

预期输出：

n = len(graph['nodes'])
adj_matr = pd.DataFrame(0, columns = graph['nodes'], index = graph['edges'])

Answer 1

获取邻接矩阵的一种简单方法是使用NetworkX

d = {'nodes':['A', 'B', 'C', 'D', 'E'],
     'edges':[('A', 'B'), ('A', 'D'), ('B', 'C'), ('B', 'E'), ('C', 'D'), 
                      ('D', 'E'), ('E', 'A'),('E', 'B'), ('E', 'C')]}

从您的邻接矩阵看来，该图是有向的。您可以创建如下所示的有向图，并使用以下命令从字典中定义其节点和边：

import networkx as nx
g = nx.DiGraph()
g.add_nodes_from(d['nodes'])
g.add_edges_from(d['edges'])

然后您可以使用nx.adjacency_matrix从网络中获得邻接矩阵：

m = nx.adjacency_matrix(g)
m.todense()

matrix([[0, 1, 0, 1, 0],
        [0, 0, 1, 0, 1],
        [0, 0, 0, 1, 0],
        [0, 0, 0, 0, 1],
        [1, 1, 1, 0, 0]], dtype=int64)

对于以相应节点为列的数据框，您可以执行以下操作：

pd.DataFrame(m.todense(), columns=nx.nodes(g))

   A  B  C  D  E
0  0  1  0  1  0
1  0  0  1  0  1
2  0  0  0  1  0
3  0  0  0  0  1
4  1  1  1  0  0

Answer 2

用于无向图

graph = {'nodes': ['A', 'B', 'C', 'D', 'E'],
         'edges': [('A', 'B'), ('A', 'D'), ('B', 'C'), ('B', 'E'), ('C', 'D'),
                   ('D', 'E'), ('E', 'A'), ('E', 'B'), ('E', 'C')]}
n = len(graph['nodes'])
adj_matr = pd.DataFrame(0, columns=graph['nodes'], index=graph['nodes'])
for i in graph['edges']:
    adj_matr.at[i[0], i[1]] = 1
    adj_matr.at[i[1], i[0]] = 1


print(adj_matr)

   A  B  C  D  E
A  0  1  0  1  1
B  1  0  1  0  1
C  0  1  0  1  1
D  1  0  1  0  1
E  1  1  1  1  0

有向图：

graph = {'nodes': ['A', 'B', 'C', 'D', 'E'],
         'edges': [('A', 'B'), ('A', 'D'), ('B', 'C'), ('B', 'E'), ('C', 'D'),
                   ('D', 'E'), ('E', 'A'), ('E', 'B'), ('E', 'C')]}
n = len(graph['nodes'])
adj_matr = pd.DataFrame(0, columns=graph['nodes'], index=graph['nodes'])
print(adj_matr)
for i in graph['edges']:
    adj_matr.at[i[0], i[1]] = 1
    # adj_matr.at[i[1], i[0]] = 1

print(adj_matr)

   A  B  C  D  E
A  0  1  0  1  0
B  0  0  1  0  1
C  0  0  0  1  0
D  0  0  0  0  1
E  1  1  1  0  0

Answer 3

对于有向图，您可以使用：

df = pd.DataFrame(graph['edges'], columns=['From', 'To'])
df['Edge'] = 1
adj = df.pivot(index='From', columns='To', values='Edge').fillna(0)

熊猫邻接表

3 个答案: