如何在网络矩阵（networkx）中找到 n 最大的边权重？

Question

我尝试使用以下代码生成网络矩阵。有了这个矩阵，我想找到不在对角线上的 20 个权重最高的边（i.e. i!=j 在矩阵中）。我还想获取由这些边组成的节点的名称（成对）。

import heapq
def mapper_network(self, _, info):
    G = nx.Graph()  #create a graph
    for i in range(len(info)):   
        edge_from = info[0]  # edge from
        edge_to = info[1]    # edge to
        weight = info[2]     # edge weight
        G.add_edge(edge_from, edge_to, weight=weight) #insert the edge to the graph
    A = nx.adjacency_matrix(G)  # create an adjacency matrix
    A_square = A * A  # find the product of the matrix
    print heapq.nlargest(20, A_square) # to print out the 20 highest weighted edges

但是，使用此代码，我未能生成 20 个权重最大的边。我得到raise ValueError("The truth value of an array with more than one " ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all().

相反，用这个

    print heapq.nlargest(20, range(len(A_square)), A_square.take)

它给了我：

    raise TypeError("sparse matrix length is ambiguous; use getnnz()"
    TypeError: sparse matrix length is ambiguous; use getnnz() or shape[0]

有

    def mapper_network(self, _, info):
    G = nx.Graph()
    for i in range(len(info)):
        edge_from = info[0]
        edge_to = info[1]
        weight = info[2]
        G.add_edge(edge_from, edge_to, weight=weight)
    A = nx.adjacency_matrix(G)
    A_square = A * A #can print (A_square.todense())
    weight = nx.get_edge_attributes(A_square, weight)
    edges = A_square.edges(data = True)
    s = sorted(G.edges(data=True), key=lambda (source, target, data): data['weight'])
    print s

我收到了

  File         "/tmp/MRQ7_trevor.vagrant.20160814.040827.770006/job_local_dir/1/mapper/0/mrjob.tar.gz/mrjob/job.py", line 433, in run
mr_job.execute()
 File "/tmp/MRQ7_trevor.vagrant.20160814.040827.770006/job_local_dir/1/mapper/0/mrjob.tar.gz/mrjob/job.py", line 442, in execute
self.run_mapper(self.options.step_num)
 File "/tmp/MRQ7_trevor.vagrant.20160814.040827.770006/job_local_dir/1/mapper/0/mrjob.tar.gz/mrjob/job.py", line 507, in run_mapper
for out_key, out_value in mapper(key, value) or ():
File "MRQ7_trevor.py", line 90, in mapper_network
weight = nx.get_edge_attributes(A_square, weight)
File "/home/vagrant/anaconda/lib/python2.7/site-packages/networkx/classes/function.py", line 428, in get_edge_attributes
if G.is_multigraph():
File "/home/vagrant/anaconda/lib/python2.7/site-packages/scipy/sparse/base.py", line 499, in __getattr__
raise AttributeError(attr + " not found")

AttributeError: is_multigraph not found

有人可以帮我解决这个问题吗？非常感谢！

Answer 1

这一行的问题：

heapq.nlargest(20, A_square)

那nlargest 不期望可迭代的可迭代对象，而是数字的可交互对象。

所以你可以这样做：

heapq.nlargest(20, itertools.chain.from_iterable(A_square))

itertools.chain.iterables 接受一个可迭代的迭代，并使用所有内部迭代的内容创建一个新的迭代。

---

但是，这并不能完全解决您最初的问题，原因有两个：

1. 为什么要取邻接矩阵的平方？这样做只会为您提供图中长度为 2 的路径的最高加权和，这与您想要的完全不同。只需使用邻接矩阵即可。

2. 在您的代码中没有任何地方可以删除对角线。你可以这样做：for n in G.number_of_nodes(): A[n][n] = 0