有没有更快、更好的构建 Networkx 树的方法。目前，我的代码是
for numb in range(0,len(previous)):
nodos = list(chunks(current,3))
for i in range(0,3):
G.add_edge(previous[numb],nodos[numb][i])
其工作方式如下: 1. 树有 3 个分支(或边)。我有两个数组:
previous = [x,y,z] #These nodes have already been added to the graph
current = [xx,xy,xz, xy,yy,yz, xz,yz,zz] #This nodes need to be added.
理想情况下，我应该执行以下操作:
1. Start with x in previous:
1.1 Pick the first 3 nodes in current (i.e. xx,xy,xz)
1.1.1 Add the nodes-edges: x->xx, x->xy, x->xz
到目前为止我的代码是这样的:
1. Start with x in previous
2. Partition current into chunks of 3 items: [[xx,xy,xz], [xy,yy,yz], [xz,yz,zz]]
3. Loop through all the nodes in these chunks:
4. Add x->xx, loop again, add x->xy... etc.
我的实现效率极低。你将如何有效地做到这一点？谢谢
请您参考如下方法:
您可以使用 https://github.com/networkx/networkx/blob/master/networkx/generators/classic.py#L50 中的辅助函数
def _tree_edges(n,r):
# helper function for trees
# yields edges in rooted tree at 0 with n nodes and branching ratio r
nodes=iter(range(n))
parents=[next(nodes)] # stack of max length r
while parents:
source=parents.pop(0)
for i in range(r):
try:
target=next(nodes)
parents.append(target)
yield source,target
except StopIteration:
break
print list(_tree_edges(13,3))
#[(0, 1), (0, 2), (0, 3), (1, 4), (1, 5), (1, 6), (2, 7), (2, 8), (2, 9), (3, 10), (3, 11), (3, 12)]
import networkx as nx
G = nx.Graph(_tree_edges(13,3))
如果你想要整数以外的节点，你可以像这样重新标记或在输入中指定它们
def _tree_edges(nodes,r):
# helper function for trees
# yields edges in rooted tree with given nodes and branching ratio r
nodes=iter(nodes)
parents=[next(nodes)] # stack of max length r
while parents:
source=parents.pop(0)
for i in range(r):
try:
target=next(nodes)
parents.append(target)
yield source,target
except StopIteration:
break
nodes = list('abcdefghijklm')
print list(_tree_edges(nodes,3))
