Networkx：查找到Graph中多個節點之一的最短路徑

Question

我有一個不同位置的圖：

import networkx as nx

G = nx.Graph()
for edge in Edge.objects.all():
    G.add_edge(edge.from_location, edge.to_location, weight=edge.distance)

位置（節點）具有不同的類型（廁所，建築物入口等），我需要找到從某個給定位置到特定類型的任何位置的最短路徑。 （例如：查找距給定節點最近的入口。）

Networkx庫中是否有某種方法可以解決無循環問題？ 就像是：

nx.shortest_path(
     G,
     source=start_location,
     target=[first_location, second_location],
     weight='weight'
)

如果兩個位置屬於同一類型，則結果將是到達first_location或second_location的最短路徑。

是否有一些方法也返回路徑長度？

Answer 1

我們將分三步進行。

• 步驟1：讓我們創建一個虛擬圖來說明
• 第2步：繪制圖形和顏色節點，以指示邊長和特殊節點類型（廁所，入口等）
• 步驟3：從任何給定節點（源）計算到所有可達節點的最短路徑，然后子集到感興趣的節點類型並選擇具有最小長度的路徑。

可以肯定地優化下面的代碼，但這可能更容易遵循。

步驟1：建立圖表

edge_objects = [(1,2, 0.4), (1, 3, 1.7), (2, 4, 1.2), (3, 4, 0.3), (4 , 5, 1.9), 
(4 ,6, 0.6), (1,7, 0.4), (3,5, 1.7), (2, 6, 1.2), (6, 7, 0.3), 
(6, 8, 1.9), (8,9, 0.6)]

toilets = [5,9] # Mark two nodes (5 & 9) to be toilets
entrances = [2,7] # Mark two nodes (2 & 7) to be Entrances
common_nodes = [1,3,4,6,8] #all the other nodes

node_types = [(9, 'toilet'), (5, 'toilet'),
              (7, 'entrance'), (2, 'entrance')]

#create the networkx Graph with node types and specifying edge distances
G = nx.Graph()

for n,typ in node_types:
    G.add_node(n, type=typ) #add each node to the graph

for from_loc, to_loc, dist in edge_objects:
    G.add_edge(from_loc, to_loc, distance=dist) #add all the edges   

步驟2：繪制圖形

#Draw the graph (optional step)
pos = nx.spring_layout(G)
nx.draw(G, pos, with_labels=True)
edge_labels = nx.get_edge_attributes(G,'distance')
nx.draw_networkx_edge_labels(G, pos, edge_labels = edge_labels)
nx.draw_networkx_nodes(G, pos, nodelist=toilets, node_color='b')
nx.draw_networkx_nodes(G, pos, nodelist=entrances, node_color='g')
nx.draw_networkx_nodes(G, pos, nodelist=common_nodes, node_color='r')
plt.show()

步驟3：創建小函數以找到節點類型的最短路徑

def subset_typeofnode(G, typestr):
    '''return those nodes in graph G that match type = typestr.'''
    return [name for name, d in G.nodes(data=True) 
            if 'type' in d and (d['type'] ==typestr)]

#All computations happen in this function
def find_nearest(typeofnode, fromnode):

    #Calculate the length of paths from fromnode to all other nodes
    lengths=nx.single_source_dijkstra_path_length(G, fromnode, weight='distance')
    paths = nx.single_source_dijkstra_path(G, fromnode)

    #We are only interested in a particular type of node
    subnodes = subset_typeofnode(G, typeofnode)
    subdict = {k: v for k, v in lengths.items() if k in subnodes}

    #return the smallest of all lengths to get to typeofnode
    if subdict: #dict of shortest paths to all entrances/toilets
        nearest =  min(subdict, key=subdict.get) #shortest value among all the keys
        return(nearest, subdict[nearest], paths[nearest])
    else: #not found, no path from source to typeofnode
        return(None, None, None)

測試：

 find_nearest('entrance', fromnode=5)

產生：

 (7, 2.8, [5, 4, 6, 7])

含義：距離5最近的“入口”節點為7，路徑長度為2.8，完整路徑為：[5、4、6、7]。 希望這可以幫助您前進。 請詢問是否不清楚。