python - 通过networkx中的规则找到不合适的节点？

Question

全部，

考虑下面的示例“树状”图。

纵向上它是一个基于节点“0”的层次结构。横向上它是一个从级别 1 开始的基于组的结构，组意味着从一个根节点继承的节点

    '''                         
                              +---+       
                              | 0 |                                    Level 0
                              +---+       
                                |
                 +--------------+---------------+         
                 |              |               |           
               +---+          +---+           +---+      
               | 1 |          | 2 |           | 3 |                    Level 1      
               +---+          +---+           +---+      
     +-----+----+        +-----+-----+        +|---+-----+            
     |     |     |        |     |     |        |    |     |           
   +---+ +---+ +---+    +---+ +---+ +---+   +---+ +---+ +---+  
   |11 | |12 | |13 |    |21 | |22 | |23 |   |31 | |32 | |33 |          Level 2 
   +---+ +---+ +---+    +---+ +---+ +---+   +---+ +---+ +---+  
     |     |   / |    /   |      |            |                 
     |     | /   | /      |      |            |  
     |   +---+ +---+    +---+ +---+           |
     |   |121|-|131|    |211| |221|           |                        Level 3                                 
     |   +---+ +---+    +---+ +---+           |
     |           |--------|------|            |
     |-----------|----------------------------|     


 |     Group 0       |       group 1     |      group 2       |
'''

在 Networkx 中创建：

# create it in networkx
import networkx as nx
G = nx.DiGraph()
G.add_edges_from([('0', '1'), ('0', '2'), ('0', '3')])
G.add_edges_from([('1', '11'), ('1', '12'), ('1', '13')])
G.add_edges_from([('2', '21'), ('2', '22'), ('2', '23')])
G.add_edges_from([('3', '31'), ('3', '32'), ('3', '33')])
#
G.add_edges_from([('12', '121'), ('13', '131')])
G.add_edges_from([('12', '121'), ('13', '131')])
G.add_edges_from([('21', '211'), ('22', '221')])
#
G.add_edges_from([('13', '121')])  
G.add_edges_from([('21', '131')])  
G.add_edges_from([('131', '211')])  
G.add_edges_from([('131', '221')])

#
G.add_edges_from([('121', '13')])            # node may not with "in_degree" link only
G.add_edges_from([('131', '21')])            # ditto

#
G.add_edges_from([('131', '31')]) 
G.add_edges_from([('131', '11')])
G.add_edges_from([('11', '131')])

#
G.add_edges_from([('121', '131')])

问题：

如何找出图中的节点和边，使用下面的节点作为示例：

“121”，在同一组中有多个链接到更高级别？（节点边缘类型“不确定”，可能是 in_degree 或 out_degree 或两者，在以下问题中相同）
“131”，与其他组的更高级别节点有多个链接？
“131”，链接到同一组中的同一级别节点
“131”，链接到同级节点但在其他组中
“21”，链接到不同组中的较低级别节点

“Graph”的新手，并尝试获取示例代码以了解如何使用 networkx 深入挖掘。

非常感谢。

score 2 · Accepted Answer

这样的事情可能会奏效。它使用节点的长度作为级别（必须在代码中注释掉您的级别 0 节点才能使其工作）和节点字符串的第一个元素作为组。我认为这就是您对数据结构的意图。

# create it in networkx
import networkx as nx
G = nx.DiGraph()
#G.add_edges_from([('0', '1'), ('0', '2'), ('0', '3')])
G.add_edges_from([('1', '11'), ('1', '12'), ('1', '13')])
G.add_edges_from([('2', '21'), ('2', '22'), ('2', '23')])
G.add_edges_from([('3', '31'), ('3', '32'), ('3', '33')])
#
G.add_edges_from([('12', '121'), ('13', '131')])
G.add_edges_from([('12', '121'), ('13', '131')])
G.add_edges_from([('21', '211'), ('22', '221')])
#
G.add_edges_from([('13', '121')])
G.add_edges_from([('21', '131')])
G.add_edges_from([('131', '211')])
G.add_edges_from([('131', '221')])

#
G.add_edges_from([('121', '13')])            # node may not with "in_degree" link only
G.add_edges_from([('131', '21')])            # ditto

#
G.add_edges_from([('131', '31')])
G.add_edges_from([('131', '11')])
G.add_edges_from([('11', '131')])
G.add_edges_from([('121', '131')])

# "121", with more than one link to higher level in same group? (node edge type "unsure", may in_degree or out_degree or both, same in following question)

print "more than one link to higher level in same group"
for node in G:
    l = len(node)-1 # higher level
    others = [n for n in G.successors(node)+G.predecessors(node)
              if len(n)==l and n[0]==node[0]]
    if len(others) > 1:
        print node

# "131", with more than one link to higner level nodes to other group?
print "more than one link to higher level in same group"
for node in G:
    l = len(node)-1 # higher level
    others = [n for n in G.successors(node)+G.predecessors(node)
              if len(n)==l and n[0]!=node[0]]
    if len(others) > 1:
        print node


# "131", with links to same level nodes in same group
print "same level, same group"
for u,v in G.edges():
    if len(u) == len(v):
        if u[0] == v[0]:
            print u

# "131", with links to same level nodes but in other group
print "same level, other group"
for u,v in G.edges():
    if len(u) == len(v):
        if u[0] != v[0]:
            print u


# "21", with links to lower level nodes in different group
print "same level, other group"
for u,v in G.edges():
    if len(u) == len(v)-1:
        if u[0] != v[0]:
            print u

score 2 · Accepted Answer

首先，您应该以某种方式定义节点所在的组或级别，因为由于破坏树结构的附加边，它没有在图 istelf 中定义。我只是按照您的命名模式编写了这些辅助函数来将给定名称转换为级别/组：

def get_group(node):
    if node == '0':
        return 1
    return int(node[0])-1

def get_level(node):
    if node == '0':
        return 0
    return len(node)

def equal_group(a,b):
    return get_group(a) == get_group(b)

def lower_level(a,b):
    return get_level(a) < get_level(b)

def equal_level(a,b):
    return get_level(a) == get_level(b)

然后你可以根据你的规范去过滤节点：

def filter_q1(node):
    k = len([predecessor for predecessor in G.predecessors_iter(node) if equal_group(node, predecessor) and lower_level(predecessor, node)] )
    return k > 1
q1_result = filter(filter_q1, G)
print 'Q1:', q1_result


def filter_q2(node):
    k = len([predecessor for predecessor in G.predecessors_iter(node) if lower_level(predecessor, node)])
    return k > 1
q2_result = filter(filter_q2, G)
print 'Q2:', q2_result


def filter_q3(node):
    k = len([neighbour for neighbour in G.neighbors_iter(node) if equal_level(node, neighbour) and equal_group(node, neighbour)])
    return k > 0
q3_result = filter(filter_q3, G)
print 'Q3:', q3_result


def filter_q4(node):
    k = len([neighbour for neighbour in G.neighbors_iter(node) if equal_level(node, neighbour)])
    return k > 0
q4_result = filter(filter_q4, G)
print 'Q4:', q4_result


def filter_q5(node):
    k = len([neighbour for neighbour in G.neighbors_iter(node) if not equal_group(node, neighbour)])
    return k > 0
q5_result = filter(filter_q5, G)
print 'Q5:', q5_result

结果如下所示：

>>>Q1: ['131', '121']
>>>Q2: ['131', '121']
>>>Q3: ['121']
>>>Q4: ['131', '121']
>>>Q5: ['21', '131', '0']

通常，您可以像这样找到不需要的边缘：

def is_bad_edge(edge):
    a, b = edge
    if not equal_group(a, b):
        return True
    if not lower_level(a, b):
        return True
    return False
bad_edges = filter(is_bad_edge, G.edges_iter())
print 'Bad edges:', bad_edges

结果是：

>>>Bad edges: [('21', '131'), ('131', '11'), ('131', '31'), ('131', '21'), ('131', '221'), ('131', '211'), ('121', '13'), ('121', '131'), ('0', '1'), ('0', '3')]

正如您所看到的，您的边来自0那里，因为它被归类为第 1 组，但节点1和节点3不是。根据您想要对节点进行分类的方式0，您可以调整函数以包含或排除根节点。

python - 通过networkx中的规则找到不合适的节点？

2 回答 2

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