13

您好,我了解邻接列表和矩阵的概念,但我对如何在 Python 中实现它们感到困惑:

实现以下两个示例的算法可以实现,但从一开始就不知道输入,因为他们在示例中对其进行了硬编码:

对于邻接列表:

    a, b, c, d, e, f, g, h = range(8) 
    N = [ 
     {b:2, c:1, d:3, e:9, f:4},    # a 
     {c:4, e:3},                   # b 
     {d:8},                        # c 
     {e:7},                        # d 
     {f:5},                        # e 
     {c:2, g:2, h:2},              # f 
     {f:1, h:6},                   # g 
     {f:9, g:8}                    # h 
   ] 

对于邻接矩阵:

    a, b, c, d, e, f, g, h = range(8) 
    _ = float('inf') 
    #     a b c d e f g h
    W = [[0,2,1,3,9,4,_,_], # a 
        [_,0,4,_,3,_,_,_], # b 
        [_,_,0,8,_,_,_,_], # c 
        [_,_,_,0,7,_,_,_], # d 
        [_,_,_,_,0,5,_,_], # e 
        [_,_,2,_,_,0,2,2], # f 
        [_,_,_,_,_,1,0,6], # g 
        [_,_,_,_,_,9,8,0]] # h

再次任何帮助将不胜感激,谢谢!

4

3 回答 3

7

假设:

edges = [('a', 'b'), ('a', 'b'), ('a', 'c')]

这是矩阵的一些代码:

from collections import defaultdict

matrix = defaultdict(int)
for edge in edges:
    matrix[edge] += 1

print matrix['a', 'b']
2

对于“列表”:

from collections import defaultdict

adj_list = defaultdict(lambda: defaultdict(lambda: 0))
for start, end in edges:
    adj_list[start][end] += 1

print adj_list['a']
{'c': 1, 'b': 2}
于 2012-11-25T00:52:58.283 回答
2

设置数据结构非常简单。例如,邻接列表示例可以使用如下实现defaultdict

from collections import defaultdict

N = defaultdict(dict)

然后,当您开始获得输入时,只需N[start][end] = weight对每个输入的边执行此操作。如果您有一些没有出站边的节点(您需要将内部字典的键与外部字典结合以确保您拥有它们全部),那么这组节点将更加棘手。但是即使没有完整的节点列表,很多算法也能正常工作。

邻接矩阵稍微复杂一些,因为您需要知道节点的数量才能正确设置其维度。如果您提前知道它,那么它很容易:

number_of_nodes = 8
_ = float("inf")

N = [[_]*number_of_nodes for i in number_of_nodes]

如果不这样做,您可能需要扫描作为输入获得的边以找到编号最高的节点,然后使用上面相同的代码来制作矩阵。例如,如果您的边以(start, end, weight)3 元组列表的形式提供,您可以使用以下命令:

number_of_nodes = max(max(start, end) for start, end, weight in edges)
于 2012-11-25T00:57:16.320 回答
0

我希望下面的示例对您有所帮助,它既有已初始化的图,也有用户自定义的

class Graph:
"""
  Read the Intialized Graph and Create a Adjacency list out of it 
   There could be cases where in the initialized graph <map> link
  issues are not maintained
   for example node 2 to 1 link 
    2->1
   there needs to be a link then since undirected Graph
    1->2
"""

def __init__(self,Graph_init):
    self.edge={}
    for keys,values in Graph_init.items():
         for value in values:
             self.addEdge(keys,value);

"""
Add a vertex to graph map
structure is
int => int list
"""
def addVertex(self,v):
    if v not in self.edge:
        self.edge[v]=[]
"""
Add Edge from both vertex to each other
Make sure the nodes are present   

"""
def addEdge(self,u,v): if u not in self.edge: self.addVertex(u) if v not in self.edge: self.addVertex(v) if u not in self.edge[v ]: self.edge[v].append(u) 如果 v 不在 self.edge[u]: self.edge[u].append(v)

def isEdge(self,u,v):
    if u not in self.edge:
        return False
    if v not in self.edge:
        return False 
    return  u in self.edge[v] 

def display(self):
    for keys,values in self.edge.items():
        print(keys,":=>",values)

"""A initalized Graph (not in form of adjaceny list"""
Graph_init = {1:[2,3,5],
          2:[1,4],
          3:[1,6]};

"""Default constrcutor takes care of making the initialzed map to adjaceny 
list"""                 
g=Graph(Graph_init)
g.addVertex(1)
g.addVertex(2) 
g.addVertex(3)
g.addEdge(1,2)
g.addEdge(3,2)
g.display();
于 2017-06-27T18:43:32.840 回答