计算图中每个节点的聚类

Question

谁能帮助我如何在不使用 python 库的情况下计算图中每个节点的聚类？ 通式为2.0 * E / (V *(V - 1)) 。 代码（无法正常工作）：

def clustering():
        clust = []
        print(vertexDegree)
        E = len(list1)
   
    
    for i in vertexDegree:
        if i <= 1:
              clust.append(0)
        else:
              clust.append(2.0 * E / (i *(i - 1)))
      
    vertex = 1
    for i in clust:
        print("Vertex ", vertex, "have clustering: ", i)
        vertex += 1
        print(clust)

list1是连接节点的列表 - [[1, 2], [3, 5], [2, 4]]

E是所有连接（边）的数量

V是邻居（节点）之间可能的连接（边）的数量。

该图由字典表示 - {1: [2], 2: [1, 4], 3: [5], 4: [2], 5: [3]} ，计算vertexDegree并保存在列表 - [1, 2, 1, 1, 1]

Answer 1

将 networkx 的代码转换为使用基于字典的图形

Networkx代码： 如何使用Networkx计算Python中图中每个节点的聚类系数

代码

def clustering(data, undirected = True):
  """
    Computes clustering coefficient for each
    node in graph g
  """
  def has_edge(n1, n2):
    """ Helper function
        True if n1 has edge to n2 or n2 has edge to n1
    """
    # n1 neighbors
    neighbours = g.get(n1, [])
    if n2 in neighbours:
      return True

    # n2 neighbors
    neighbours = g.get(n2, [])
    if n2 in neighbours:
      return True

    return False

  def edges_to_dic(edges, undirected = True):
    " Generates graph dictionary from edges "
    res = {}
    for v1, v2 in edges:
      res.setdefault(v1, set())
      res[v1].add(v2)
      if undirected:
        res.setdefault(v2, set())
        res[v2].add(v1)
    return res

  if isinstance(data, list):
    # list of edges
    g = edges_to_dic(data, undirected)
  else:
    g = data

  result = {}
  for node in g:
    # Iterate over nodes of g
    neighbours = g[node]
    n_neighbors = len(neighbours)
    n_links = 0
    
    if n_neighbors > 1:
      for node1 in neighbours:
        for node2 in neighbours:
          if has_edge(node1,node2):
            n_links += 1

      n_links /= 2 #because n_links is calculated twice
      result[node] = 2*n_links/(n_neighbors*(n_neighbors-1))
    else:
      result[node] = 0

  return result

用法

与字典一起使用

g = {1: [2], 2: [1, 4], 3: [5], 4: [2], 5: [3]}

print(clustering(g))
# Output: {1: 0, 2: 0.0, 3: 0, 4: 0, 5: 0}

适用于边缘

edges = [[1, 2], [3, 5], [2, 4]]
print(clustering(edges))
# Output: {1: 0, 2: 0.0, 3: 0, 4: 0, 5: 0}

解释

所有系数为零的原因是：

1. 节点 1、3、4、5 只有一个邻居，因此它们的系数为零
2. 节点 2 有邻居 [1, 4] 但这些邻居没有连接，因此系数为零。

测试源

def show_results(edges, text):
  " Show results for set of test cases "
  print(text)
  print('\tUndirected: ', clustering(edges))
  print('\tDirected', clustering(edges, undirected = False))


tests = [
  [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]], # C = 1
  [[1, 2], [1, 3], [1, 4], [3, 4]],                 # C = 1/3
  [[1, 2], [1, 3], [1, 4]]                          # C = 0
]

texts = [
  "Case C = 1",
  "Case C = 1/3",
  "Case C = 0"
]
for edges, text in zip(tests, texts):
  show_results(edges, text)

Output

Case C = 1
    Undirected:  {1: 1.0, 2: 1.0, 3: 1.0, 4: 1.0}
    Directed {1: 0.5, 2: 0.5, 3: 0}
Case C = 1/3
    Undirected:  {1: 0.3333333333333333, 2: 0, 3: 1.0, 4: 1.0}
    Directed {1: 0.16666666666666666, 3: 0}
Case C = 0
    Undirected:  {1: 0.0, 2: 0, 3: 0, 4: 0}
    Directed {1: 0.0}

测试图