我正在尝试创建
[[ 1, 1, 1, 0, 0, 0, 0, 0, 0],
[-1, 0, 0, 1, 1, 0, 0, 0, 0],
[ 0, -1, 0, -1, 0, 1, 1, 0, 0],
[ 0, 0, 0, 0, -1, -1, 0, 1, 0],
[ 0, 0, -1, 0, 0, 0, -1, 0, 1],
[ 0, 0, 0, 0, 0, 0, 0, -1, -1]]
S=[1, 2, 3, 4, 5, 6]
D=[[1, 2], [1, 3], [1, 5], [2, 3], [2, 4], [3, 4], [3, 5], [4, 6], [5, 6]]
INC = [[0]*len(D) for _ in range(len(S))]
for i in range(len(D)):
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在此之后,我在得到零矩阵的地方做错了
for j in S:
if i == j:
INC.append(1)
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我曾尝试将 D 吐到两个不同的列表中,但对我来说它开始变得复杂
my_list1 = [i[0] for i in D]
my_list2 = [i[1] for i in D]
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我对你想要什么的最好猜测......你的变量名很差。我会使用NetworkX ( networkx)并让它做所有的数学运算。
import networkx as nx
nodes = [1, 2, 3, 4, 5, 6]
edges = [[1, 2], [1, 3], [1, 5], [2, 3], [2, 4], [3, 4], [3, 5], [4, 6], [5, 6]]
G = nx.DiGraph()
G.add_nodes_from(nodes)
G.add_edges_from(edges)
incidence_matrix = -nx.incidence_matrix(G, oriented=True)
# ^ this returns a scipy sparse matrix, can convert into the full array as below
# (as long as your node count is reasonable: this'll have that squared elements)
print(incidence_matrix.toarray())
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输出:
[[ 1. 1. 1. 0. 0. 0. 0. 0. 0.]
[-1. 0. 0. 1. 1. 0. 0. 0. 0.]
[ 0. -1. 0. -1. 0. 1. 1. 0. 0.]
[ 0. 0. 0. 0. -1. -1. 0. 1. 0.]
[ 0. 0. -1. 0. 0. 0. -1. 0. 1.]
[ 0. 0. 0. 0. 0. 0. 0. -1. -1.]]
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