类型错误：vertices() 需要 1 个位置参数，但给出了 2 个

问题描述

伙计们，请帮助我编写主类，每当我执行它时都会抛出错误，我不认为有什么问题..在这个程序中，我使用 DFS 和 BFS 从图中打印最短路径，但是 class Graph(object)：请检查主类是否有错误..

def __init__(self,graph_dict=None):
    """ initializes a graph object 
        If no dictionary or None is given,an empty dictionary will be used
    """
    if graph_dict == None:
        graph_dict = {}
    self.__graph_dict = graph_dict

def vertices(self):
    """ returns the vertices of a graph """
    return list(self.__graph_dict.keys())

def edges(self):
    """ returns the edges of a graph """
    return self.__generate_edges()

def add_vertex(self,vertex):
    """ If the vertex "vertex" is not in 
        self.__graph_dict,a key "vertex" with an empty
        list as a value is added to the dictionary. 
        Otherwise nothing has to be done. 
    """
    if vertex not in self.__graph_dict:
        self.__graph_dict[vertex] = []

def add_edge(self,edge):
    """ assumes that edge is of type set,tuple or list; 
        between two vertices can be multiple edges! 
    """
    edge = set(edge)
    (vertex1,vertex2) = tuple(edge)
    if vertex1 in self.__graph_dict:
        self.__graph_dict[vertex1].append(vertex2)
    else:
        self.__graph_dict[vertex1] = [vertex2]

def __generate_edges(self):
    """ A static method generating the edges of the 
        graph "graph". Edges are represented as sets 
        with one (a loop back to the vertex) or two 
        vertices 
    """
    edges = []
    for vertex in self.__graph_dict:
        for neighbour in self.__graph_dict[vertex]:
            if {neighbour,vertex} not in edges:
                edges.append({vertex,neighbour})
    return edges

def __str__(self):
    res = "vertices: "
    for k in self.__graph_dict:
        res += str(k) + " "
    res += "\nedges: "
    for edge in self.__generate_edges():
        res += str(edge) + " "
    return res


def Dfsutil(self,v,visited):

    # Mark the current node as visited
    # and print it
    visited.add(v)
    print(v,end=' ')

    # Recur for all the vertices
    # adjacent to this vertex
    for neighbour in self.graph[v]:
        if neighbour not in visited:
             self.Dfsutil(neighbour,visited)

# The function to do DFS traversal. It uses
# recursive Dfsutil()
def DFS(self,v):

    # Create a set to store visited vertices
     visited = set()
 
    # Call the recursive helper function
    # to print DFS traversal
     self.Dfsutil(v,visited)

打印图形 BFS 的函数

def BFS(self,s):

    # Mark all the vertices as not visited
    visited = [False] * (max(self.graph) + 1)

    # Create a queue for BFS
    queue = []

    # Mark the source node as 
    # visited and enqueue it
    queue.append(s)
    visited[s] = True

    while queue:

        # Dequeue a vertex from 
        # queue and print it
        s = queue.pop(0)
        print (s,end = " ")

        # Get all adjacent vertices of the
        # dequeued vertex s. If a adjacent
        # has not been visited,then mark it
        # visited and enqueue it
        for i in self.graph[s]:
                      if visited[i] == False:
                             queue.append(i)
                             visited[i] = True

if name == "ma​​in":

g = { "a" : ["b","c"],"b" : ["a","d","e"],"c" : ["a","h"],"d" : ["b","c","e","g","f"],"e" : ["b","g"],"f" : ["d","g" : ["e","h" : ["c","f","g"]
    }
graph = Graph(g)

print("Vertices of graph:")
print(graph.vertices())

print("Edges of graph:")
print(graph.edges())

print("Add vertex:")
graph.add_vertex("z")


#call both implemented function here

graph.DFS(graph.vertices(0)) graph.BFS(graph.vertices(0))

解决方法

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