python graph-tool访问顶点属性 [英] python graph-tool access vertex properties
问题描述
对于我当前的项目，我想使用图形工具库，因为它们声称是最快的: https ://graph-tool.skewed.de/performance .我有一些算法(最短路径等)可以在非常大的网络上运行，所以速度越快越好！
For my current project I want to use the graph-tool library since they claim being the fastest: https://graph-tool.skewed.de/performance. I have some algorithms (shortest path, etc.) to run on really large networks, so the faster the better!
第一个问题:这是最快"的说法吗? ;)
First question: Is this claim 'being the fastest' true? ;)
在尝试构建适合我需求的图形工具图形时，我发现无法以有效的方式访问顶点属性.也许我错过了什么?
While trying to build a graph-tool graph fitting my needs, I figured out that its not possible to access vertex properties in a efficient way. Maybe I missed something?
我的问题是，现在可以以更有效的方式编写函数"getVertexFromGraph(graph，position)"吗?或更广泛地说:我可以有效地检查顶点(由其position属性赋予)是否在图形中.
My question is now, can the function "getVertexFromGraph(graph, position)" be written in a more efficient way? Or more in general: Can I check efficiently if a vertex (given by its position property) is already in the graph or not.
提前谢谢！
import graph_tool as gt
#from graph_tool.all import *
edgeList = [[(0.5,1),(2.1,4.3)],[(2.1,4.3),(5.4,3.3)],[(5.4,3.3),(1.3,3.5)],[(4.4,3.3),(2.3,3.5)]] #A lot more coordinate values....
# Initialize the graph
routableNetwork = gt.Graph()
# Initialize the vertex property "position" to store the vertex coordinates
vpPosition = routableNetwork.new_vertex_property("vector<double>")
routableNetwork.vertex_properties["position"] = vpPosition
def getVertexFromGraph(graph, position):
"""
This method checks if a vertex, identified by its position, is in the given graph or not.
:param graph: The graph containing all vertices to check
:param position: The vertex/position to check
:return: The ID of the vertex if the vertex is already in the graph, 'None' otherwise
"""
for v in graph.vertices():
if graph.vp.position[v] == position:
return v
return None
def main():
"""
This method creates the graph by looping over all given edges, inserting every:
- non existent vertex in the graph with its coordinates (property 'position')
- edge with its corresponding length (property 'distance')
:return: -
"""
for e in edgeList:
vertex0 = getVertexFromGraph(routableNetwork,e[0])
vertex1 = getVertexFromGraph(routableNetwork,e[1])
if vertex0 == None:
vertex0 = routableNetwork.add_vertex()
routableNetwork.vertex_properties['position'][vertex0] = e[0]
if vertex1 == None:
vertex1 = routableNetwork.add_vertex()
routableNetwork.vertex_properties['position'][vertex1] = e[1]
edge = routableNetwork.add_edge(vertex0,vertex1)
#routableNetwork.edge_properties['distance'][edge] = calculateDistance(e[0][0],e[0][1],e[1][0],e[1][1])
#saveRoutableNetwork(routableNetwork)
#graph_draw(routableNetwork, vertex_text=routableNetwork.vertex_index, vertex_font_size=18, output_size=(200, 200), output="two-nodes.png")
if __name__ == "__main__":
main()
推荐答案
您要查找的功能是find_vertex():
https://graph-tool.skewed .de/static/doc/util.html#graph_tool.util.find_vertex
重要的是要意识到graph-tool通过卸载从性能敏感的循环到Python到C ++来达到其速度.因此，每当您像在代码中那样遍历顶点时，就会失去任何优势.
It is important to realize that graph-tool achieves its speed by off-loading performance-sensitive loops from Python to C++. So whenever you iterate through the vertices, like you did in your code, you lose any advantage.
还要注意，尽管find_vertex()是用C ++实现的，因此比纯Python中的等效速度快许多倍，但它仍然是O(N)操作.对于大型图形，最好创建一个良好的旧python字典，该字典将属性值映射到顶点，而查找成本为O(1).
Note also that, although find_vertex() is implemented in C++, and hence many times faster than the equivalent in pure Python, it is still an O(N) operation. For large graphs, you are better off creating a good old python dictionary that maps property values to vertices, which has an O(1) cost for lookup.
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