具有属性的不同长度的节点,加边如果> = 1的属性是相同的(Having nodes with

2019-10-29 18:39发布

在NetworkX“M努力实现以下目标:

  • 一个图形中创建的母亲节点“和“儿童节点”,其中孩子节点仅具有1属性,母节点具有几个(4)。
  • 创建母节点和子节点之间的边是否至少一个属性(键 - 值对)是相同的,
  • 只有一个母节点和孩子节点之间创建一个优势:即使两个母节点具有4的一个重叠的属性,不应该有在两者之间的边缘

到目前为止,我有第一部分工作,并在第二个michaelg非常有帮助,但仍有一个错误。

import networkx as nx
from itertools import product

# Mother-nodes
M = [('E_%d' % h, {'a': i, 'b': j, 'c': k, 'd': l})
 for h, (i, j, k, l) in enumerate(product(range(2), repeat=4), start=1)]

# children-nodes
a = [ (  'a_%d' % i, {'a' : i}) for i in range(0,2)  ]
b = [ (  'b_%d' % i, {'b' : i}) for i in range(0,2)  ]
c = [ (  'c_%d' % i, {'c' : i}) for i in range(0,2)  ]
d = [ (  'd_%d' % i, {'d' : i}) for i in range(0,2)  ]

# graph containing both
M_c = nx.Graph()
M_c.add_nodes_from(M)
ls_children = [a, b, c , d]
for ls_c in ls_children:
    M_c.add_nodes_from(ls_c)

# what it looks like so far
list(M_c.nodes(data=True))[0:20]

[('E_9', {'a': 1, 'b': 0, 'c': 0, 'd': 0}),
 ('d_0', {'d': 0}),
 ('E_10', {'a': 1, 'b': 0, 'c': 0, 'd': 1}),
 ('b_0', {'b': 0}),
 ('E_2', {'a': 0, 'b': 0, 'c': 0, 'd': 1}),
 ('E_1', {'a': 0, 'b': 0, 'c': 0, 'd': 0}),
 ('c_1', {'c': 1}),
 ...
    ] 

然后第二部分,其产生一个错误:

for start in M_c.nodes(data=True):
    for end in M_c.nodes(data=True):
        for attr in list(start[1].keys()):
            if start[1][attr]:
                if end[1][attr]:
                    if start[1][attr] == end[1][attr]:
                        M_c.add_edge(start[0], end[0] )
    # Adding an else and continue statement does not affect the error, 
    # even adding three of them, for each if statement
    #        else:
    #            continue

---------------------------------------------------------------------------
KeyError                                  Traceback (most recent call last)
<ipython-input-5-32ae2a6095e5> in <module>()
      3         for attr in list(start[1].keys()):
      4             if start[1][attr]:
----> 5                 if end[1][attr]:
      6                     if start[1][attr] == end[1][attr]:
      7                         M_c.add_edge(start[0], end[0] )

KeyError: 'a'

我也许忽视的东西 - 任何建议是极大的赞赏。

编辑-1:

正如ducminh建议我跑:

for mother_node in M:
    for child_node in chain(a, b, c, d):
        if child_node[1].items() <= mother_node[1].items():
            M_c.add_edge(child_node, mother_node)

其返回此错误:

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-6-24f1a24a49e8> in <module>()
      2     for child_node in chain(a, b, c, d):
      3         if child_node[1].items() <= mother_node[1].items():
----> 4             M_c.add_edge(child_node, mother_node)
      5 

/usr/local/lib/python3.5/dist-packages/networkx/classes/graph.py in add_edge(self, u_of_edge, v_of_edge, **attr)
    873         u, v = u_of_edge, v_of_edge
    874         # add nodes
--> 875         if u not in self._node:
    876             self._adj[u] = self.adjlist_inner_dict_factory()
    877             self._node[u] = {}

TypeError: unhashable type: 'dict'

Answer 1:

在验证端节点的属性的存在

既然你在迭代开始节点的属性,您还需要验证在终端节点它的存在:

这里是一个修改后的代码:

for start in M_c.nodes(data=True):
for end in M_c.nodes(data=True):
    for attr in list(start[1].keys()):
        # verify that the attribute is in the end node
        if attr in end[1]:
            if start[1][attr] == end[1][attr]:
                M_c.add_edge(start[0], end[0] )


Answer 2:

如果一个节点是母节点,它总是有所有4个属性a, b, c, d ,在子节点只有一个属性。 因此,它可能是在字典的情况下start[1]有钥匙attr ,而end[1]没有。

为了母亲节点和子节点之间的正确添加的边缘,我们需要遍历所有可能对(母节点,子节点),然后检查子节点的属性的字典是的,母亲节点的子字典。

from itertools import chain

for mother_node in M:
    for child_node in chain(a, b, c, d):
        if child_node[1].items() <= mother_node[1].items():
            M_c.add_edge(child_node[0], mother_node[0])


文章来源: Having nodes with differing lengths of attributes, add edges if >=1 attribute is the same