Python - 计算Dijkstra中的距离

时间:2011-03-09 22:35:02

标签: python dijkstra

我在确定每个节点距起始节点的距离时遇到了一些麻烦,或者更确切地说是取回任何信息。 我的函数没有输出,附在以下链接中。

#Values to assign to each node
class Node:
     distFromSource = infinity
     previous = invalid_node
     visited = False

#for each node assign default values    
def populateNodeTable(network): 
    nodeTable = []
    index = 0
    f = open('network.txt', 'r')
    for line in f: 
      node = map(int, line.split(',')) 
      nodeTable.append(Node())

      print "The previous node is " ,nodeTable[index].previous 
      print "The distance from source is " ,nodeTable[index].distFromSource 
      index +=1
    nodeTable[startNode].distFromSource = 0 

    return nodeTable

#calculate the distance of each node from the start node
def tentativeDistance(currentNode, nodeTable):
    nearestNeighbour = []
    for currentNode in nearestNeighbour:
      currentDistance == currentNode.distFromSource + [currentNode][nearestNeighbour] #gets current distance from source
      print "The current distance"
      if currentDistance != 0 & currentNode.distFromSource < Node[currentNode].distFromSource:
         nodeTable[currentNode].previous = currentNode
         nodeTable[currentNode].length = currentDistance
         nodeTable[currentNode].visited = True
         nodeTable[currentNode] +=1
         nearestNeighbour.append(currentNode)
         for currentNode in nearestNeighbour:
           print nearestNeighbour

    return nearestNeighbour

至少在我看来,我的逻辑是正确的;但是,在运行代码时,我得到的错误信息并不多。

2 个答案:

答案 0 :(得分:2)

你将nearestNeighbour设置为一个空列表,然后你用for currentNode in nearestNeighbour循环它 - 它什么都不做,因为列表是空的 - 然后你就是从函数返回。

(我假设tentativeDistance是您正在调用的函数,并且什么都没有看到。)

答案 1 :(得分:1)

您应该重新考虑您的算法设计。尝试查找Dijkstra算法的伪代码定义并在Python中实现它。特别是,您应该考虑程序中的控制流程。

您可能希望查看this cookbook recipe以获取Dijkstra的Python实现,看看您是否能够理解它。