我会反驳说-直截了当的命令并不是最好的选择。
假设您有100个站点，并且有多条非字母和非数字的路线。想想巴黎地铁：
现在尝试使用直接的Python字典来计算FDR和La Fourche之间的时间吗？这涉及两个或多个不同的路线和多个选项。
甲树或某种形式的图形是较好的结构。字典对于1对1映射而言是极好的。树对于相互关联的节点的丰富描述更好。然后，您将使用类似Dijkstra的算法的导航方式。
由于嵌套的dict或list的dict是一个图，因此很容易提出一个递归示例：
def find_all_paths(graph, start, end, path=[]): path = path + [start] if start == end: return [path] if start not in graph: return [] paths = [] for node in graph[start]: if node not in path: newpaths = find_all_paths(graph, node, end, path) for newpath in newpaths: paths.append(newpath) return pathsdef min_path(graph, start, end): paths=find_all_paths(graph,start,end) mt=10**99 mpath=[] print 'tAll paths:',paths for path in paths: t=sum(graph[i][j] for i,j in zip(path,path[1::])) print 'ttevaluating:',path, t if t<mt: mt=t mpath=path e1=' '.join('{}->{}:{}'.format(i,j,graph[i][j]) for i,j in zip(mpath,mpath[1::])) e2=str(sum(graph[i][j] for i,j in zip(mpath,mpath[1::]))) print 'Best path: '+e1+' Total: '+e2+'n'if __name__ == "__main__": graph = {'A': {'B':5, 'C':4}, 'B': {'C':3, 'D':10}, 'C': {'D':12}, 'D': {'C':5, 'E':9}, 'E': {'F':8}, 'F': {'C':7}} min_path(graph,'A','E') min_path(graph,'A','D') min_path(graph,'A','F')印刷品：
All paths: [['A', 'C', 'D', 'E'], ['A', 'B', 'C', 'D', 'E'], ['A', 'B', 'D', 'E']] evaluating: ['A', 'C', 'D', 'E'] 25 evaluating: ['A', 'B', 'C', 'D', 'E'] 29 evaluating: ['A', 'B', 'D', 'E'] 24Best path: A->B:5 B->D:10 D->E:9 Total: 24 All paths: [['A', 'C', 'D'], ['A', 'B', 'C', 'D'], ['A', 'B', 'D']] evaluating: ['A', 'C', 'D'] 16 evaluating: ['A', 'B', 'C', 'D'] 20 evaluating: ['A', 'B', 'D'] 15Best path: A->B:5 B->D:10 Total: 15 All paths: [['A', 'C', 'D', 'E', 'F'], ['A', 'B', 'C', 'D', 'E', 'F'], ['A', 'B', 'D', 'E', 'F']] evaluating: ['A', 'C', 'D', 'E', 'F'] 33 evaluating: ['A', 'B', 'C', 'D', 'E', 'F'] 37 evaluating: ['A', 'B', 'D', 'E', 'F'] 32Best path: A->B:5 B->D:10 D->E:9 E->F:8 Total: 32
