我想对元组列表[(0,9),(5,4),(2,9),(7,4),(7,2)]进行排序,得到[(0,9),(2,9),(7,2),(7,4),(5,4)]
更具体地说,我有一组要连接到链(多段线)中的边(线段)。边的顺序是随机的,我想对它们进行排序。每条边都有两个节点,即起点和终点,它们在最终链中的方向可能匹配也可能不匹配。下面的代码可以工作,但速度非常慢,特别是我必须多次调用这个函数。在我的例子中,节点是自定义节点类的实例,而不是整数。平均而言,可能有大约10条边需要连接。我必须用Ironpython运行它。请问有没有什么好办法可以加快这一进程?多谢各位
克里斯
from collections import Counter
class Edge():
def __init__(self,nodeA,nodeB,edges):
self.nodes = (nodeA,nodeB)
self.index = len(edges)
edges.append(self)
def chainEdges(edges):
# make chains of edges along their connections
nodesFlat = [node for edge in edges for node in edge.nodes]
if any(Counter(nodesFlat)[node]>2 for node in nodesFlat): return# the edges connect in a non-linear manner (Y-formation)
# find first edge
elif all(Counter(nodesFlat)[node]==2 for node in nodesFlat):# the edges connect in a loop
chain = [min(edges, key=lambda edge: edge.index)]# first edge in the loop is the one with the lowest index
edges.remove(chain[-1])
nodeLast = chain[-1].nodes[-1]
else:# edges form one polyline
chain = [min([edge for edge in edges if any(Counter(nodesFlat)[node]==1 for node in edge.nodes)], key=lambda edge: edge.index)]# first edge is the one with the lowest index
edges.remove(chain[0])
nodeLast = chain[-1].nodes[0] if Counter(nodesFlat)[chain[-1].nodes[0]]==2 else chain[-1].nodes[1]
# loop through remaining edges
while len(edges)>0:
for edge in edges:
if nodeLast in edge.nodes:
chain.append(edge)
edges.remove(edge)
nodeLast = edge.nodes[0] if nodeLast==edge.nodes[1] else edge.nodes[1]
break
return chain
edges = []
for [nodeA,nodeB] in [(0, 9), (5, 4), (2, 9), (7, 4), (7, 2)]:
Edge(nodeA,nodeB,edges)
print [edge.nodes for edge in chainEdges(edges)]
>>>[(0, 9), (2, 9), (7, 2), (7, 4), (5, 4)]
我发现了问题所在,最慢的部分是当一个节点上有两条以上的边连接时,我是如何检查Y形的。当我运行该方法时,我已经检查了所有边是否以某种方式连接,但我不知道它们是否形成一条线、一个圆或一个Y。以下速度明显更快:
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